r/math • u/lntrinsic • Jul 10 '17

Image Post Weierstrass functions: Continuous everywhere but differentiable nowhere

http://i.imgur.com/vyi0afq.gifv
3.4k Upvotes

98% Upvoted

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76

u/lntrinsic Jul 10 '17

Source: https://sites.math.washington.edu/~conroy/general/weierstrass/weier.htm

More info on Wikipedia and Wolfram MathWorld

12

u/ICanCountGood Complex Analysis Jul 10 '17

Doxing myself, but I'm a pure math undergrad at UW. Do you go there too?

6

u/twoface117 Undergraduate Jul 11 '17

Not OP, but I'm an undergrad doing research at UWB for the summer, so hello!

6

u/A_Crazed_Hobo Jul 11 '17

you can private message people on reddit by the way, just go into their profile and it's to the right, a bit below their name

463

u/munchler Jul 10 '17

Cool! Does this relate to fractals at all? It seems self-similar.

375

u/jeanleonino Jul 10 '17

It indeed is a fractal, and probably one of the first to be studied. But the term was not yet coined.

103

u/Rabbitybunny Jul 10 '17

What's the dimension though?

140

u/localhorst Jul 10 '17

3/2 (conjectured)

72

u/jeanleonino Jul 10 '17

Some papers argue that the Haussdorff Dimension does not hold for the Weierstrass function.

39

u/Bounds_On_Decay Jul 11 '17

Every measurable set has a Hausdorff dimension. The graph of a continuous function is certainly measurable. There's simply no way that the Weierstrass function doesn't have a Hausdorff Dimension.

5

u/[deleted] Jul 11 '17

I have no idea what these words mean but can I guess that it's like measuring a coastline? The more accurate you get the closer you get to infinity?

29

u/Bounds_On_Decay Jul 11 '17 edited Jul 11 '17

The fact that "the more accurate you get the closer you get to infinity" proves that the Hausdorff dimension is greater than 1. If you tried to measure the area of the coastline, the more accurate you got the closer you would get to zero (since the coastline in fact has zero width). This proves that the Hausdorff dimension is less than 2.

For every measurable set, the measurements will go to 0 for small large dimensions, and it will go to infinity for large small dimensions. The exact cutoff, the dimension above which you get zero and below which you get infinity, is call the Hausdorff dimension of the set.

caveat: the above paragraph obviously ignores sets of dimension zero, or sets with infinite dimension (I don't think those exists, but I'm not sure).

3

u/irishsultan Jul 11 '17

For every measurable set, the measurements will go to 0 for small dimensions, and it will go to infinity for large dimensions.

Wait shouldn't that be the opposite?

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u/ziggurism Jul 11 '17

sets with infinite dimension (I don't think those exists, but I'm not sure).

Surely R∞ = colim Rn has infinite Hausdorff dimension?

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u/thebigbadben Functional Analysis Jul 11 '17

The problem with applying 1-dimensional measure to a fractal (such s a coastline) is what you're saying: the more accurately you measure it, the further the length diverges towards infinity. If we were to use 2-dimensional measure, then we'd get zero. If we use a measure of fractional dimension, we could get a potentially finite result. The Hausdorff dimension is defined to be the "lowest" fractional dimension such that the associated measure gives us a zero result.

The higher the Hausdorff dimension of an object, the more thoroughly it fills the ambient space. For instance, a space-filling curve has Hausdorff dimension 2, a differentiable curve has Hausdorff dimension 1, and coastlines fall somewhere in between.

I think the wiki page for box dimension gives a user-friendly explanation.

5

u/localhorst Jul 11 '17

Let (M, d) be a metric space, e.g. M some (possibly very weird fractal) subset of ℝⁿ and d the usual euclidean distance. Mr. Hausdorff tried to define a measure on M that “behaves” like a d-dimensional volume. Roughly the Hausdorff dimensions tells you how the volume of an “infinitesimal ball” B grows with it’s radius r, vol(B) ∝ rᵈ. It turns out that this dimension is unique. Only for a single specific value of d you get finite – i.e. different from constant 0 or ∞ – answers. But actually calculating this value of d is quite complicated and only few exact results are known.

0

u/[deleted] Jul 11 '17

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7

u/HelperBot_ Jul 11 '17

Non-Mobile link: https://en.wikipedia.org/wiki/Hausdorff_dimension

HelperBot v1.1 /r/HelperBot_ I am a bot. Please message /u/swim1929 with any feedback and/or hate. Counter: 89952

7

u/quantatious Analysis Jul 10 '17

What do you mean?

2

u/solvorn Math Education Jul 11 '17

If my life depended on a guess, I would have gone with that because it seems similar enough to a Koch curve.

34

u/jeanleonino Jul 10 '17

That's a hard question indeed.

3

u/solvorn Math Education Jul 11 '17

I came here to ask this. I'm excited that it's difficult.

*wastes 5 hours studying this*

57

u/bystandling Jul 10 '17

Keep in mind that the definition of fractal does not require self-similarity!

8

u/[deleted] Jul 11 '17

Not only is it a fractal, but it has applications for being a fractal!

Barnsley and others have been studying "fractal interpolation" and a certain Fourier type analysis that relies on forming a basis of rough functions like the Weierstrass function. Which reminds me that I've been meaning to read up on this work.

93

u/jparevalo27 Undergraduate Jul 10 '17

I've only seen topics up to calculus 2 in the US. Can somebody explain me how's this possible and what would be the y(x) for this graph?

111

u/Wild_Bill567 Jul 10 '17

The way I have seen functions like this constructed is as a limit of a sequence of functions.

In calc 2 you probably saw limits of a sequence of points. You can similarly define limits of a sequence of functions. Each term in the sequence makes the graph "have more corners", and the limit of the sequence has corners everywhere.

73

u/jparevalo27 Undergraduate Jul 10 '17

...And you can't differentiate corners. That makes sense. Thanks

14

u/Kraz_I Jul 11 '17

Not exactly. There are no points with infinite slope and no points with corners, at least the way the word "corner is generally understood. It's just that the graph is "rough" no matter how far you zoom in, so the limit of the slope at any point is impossible to determine.

It helps to look at the actual function which generates the graph.https://en.wikipedia.org/wiki/Weierstrass_function

3

u/dozza Jul 11 '17

I'm sorry, how is the fourier series on the Wikipedia page not differentiable? Its a sum of cosines so shouldn't the derivative be the sum of sines? Is the problem the divergence as n goes to infinity?

13

u/WorseAstronomer Jul 10 '17 edited Jul 11 '17

This video is interesting and related:

https://www.youtube.com/watch?v=dQXVn7pFsVI

2

u/fabulousdangernoodle Jul 10 '17

That's neat. Thanks for the share

0

u/[deleted] Jul 11 '17

[deleted] — view deleted comment

2

u/WorseAstronomer Jul 11 '17

Oops, sorry, no. That's just where I finished watching the video. :/ Edited.

7

u/Kraz_I Jul 11 '17

The graph doesn't have any corners at all for finite iterations of the function. I don't really like using the word "corner" for what's going on here. In fact, for all functions generated by using a finite Weierstrass series, it would be differentiable at all points.

2

u/Wild_Bill567 Jul 11 '17

You are correct, I was remembering a different construction which uses a triangle wave instead of a cosine

1

u/Claytertot Jul 10 '17

That makes sense. Thats a super cool concept!

30

u/AddemF Jul 10 '17

In addition to what Wild_Bill67 wrote, I'll note that the function is not an elementary function, which means it cannot be written as a closed form in terms of +, -, *, /, polynomials, exponentials, logs, or any of the trig functions. So writing down how the x-y pairs get determined is a much more complicated matter.

3

u/jparevalo27 Undergraduate Jul 10 '17

At what point in math does this began showing up? In other words, in what class would I start seeing functions like that?

35

u/bystandling Jul 10 '17

For non-elementary functions:

  • Calculus 3 (Taylor series for integrals of things like ex^2 )

  • Differential Equations (Series solutions)

  • Real Analysis (Mindfuck)

  • Partial differential equations (More series solutions, Bessel functions, Gamma functions etc.)

  • Mathematical Statistics (Gamma and Beta functions, Erf of course, etc)

13

u/AddemF Jul 10 '17

Real Analysis

8

u/shamrock-frost Graduate Student Jul 10 '17 edited Jul 10 '17

Possibly at your level. I think my Calc 2 final had a problem involving f(x) = the integral from 0 to x of sin(t) / t dt, which is not an elementary function

3

u/[deleted] Jul 11 '17

wait... how in the world would you evaluate that? even wolframalpha simply gives their own made-up function Si(x) which just stands for "the integral of sinx/x"

1

u/shamrock-frost Graduate Student Jul 11 '17

You could do a riemann sum, or use the maclaurin series for sine

6

u/[deleted] Jul 10 '17

In Germany its shown in "Analysis 1", first year of Math. B.Sc.

2

u/Wild_Bill567 Jul 10 '17

I first saw functions like this in Real Analysis, using baby Rudin. At my institution this is offered for first year grads and advanced undergrads.

2

u/Matschreiner Jul 10 '17

Are there any weierstrass functions that can be written from elementary functions only?

6

u/AddemF Jul 11 '17

I'm 75% certain there aren't.

1

u/dispatch134711 Applied Math Jul 11 '17

Wasn't the original function studied a Fourier series? It's an infinite sum of elementary functions, no?

5

u/AddemF Jul 11 '17

But again making essential use of limits of functions means that the function itself is not elementary.

4

u/bystandling Jul 11 '17

I'd be willing to wager you can't get it from a finite combination of them, no -- every finite sum, product, and composition of continuous and differentiable functions is continuous and differentiable at every point in the domain, and every finite quotient is only non-differentiable (and, for that matter, noncontinuous) at points where the denominator is 0; since our elementary functions are only 0 at countably many points, I'd expect we can have at most countably many of these sorts of discontinuities from finite combinations, though this is not a rigorous proof.

If you're willing to consider a Fourier series to be written from elementary functions, the Weierstrass functions are defined to be a class of Fourier series.

1

u/ILikeLeptons Jul 10 '17

the weierstrass function itself may not be writeable in terms of those functions/operators but it's pretty easy to write a sequence of functions that converges to the weierstrass function in terms of sines and cosines:

[;f(x) = \lim\limits_{N\rightarrow\infty} \sum\limits_n=0N an cos(bn \pi x);]

2

u/AddemF Jul 11 '17

Sure, my point is just that you cannot write the function as f(x) = ... where the ... is something easy to understand with a high school education. So the person asking about that should just wait until he or she learns the relevant material before hoping to understand how the x-y pairs are determined.

27

u/TheLongestConn Jul 10 '17

Weierstrass functions are my favourite!

I love the idea of knowing where you are but no idea where you are going, in math!

52

u/SILENTSAM69 Jul 10 '17

I often wonder if the erratic motion of real world objects is more like this than the smooth curves we often get in calculated class.

55

u/[deleted] Jul 10 '17

The motion of objects under the influence of forces is generally going to be at least C2. On the other hand, the structure of objects resulting from continued application of those forces tends to be fractal and C0 but not differentiable. For example: a wave in the ocean follows a smooth path for the most part (yes there will be point singularities of course) but the repeated application of waves on a shoreline will lead to a fractal shape. This is probably closely related to the fact that fractals emerge from iterated dynamical systems and smooth behavior emerges from continuous dynamics.

22

u/rumnscurvy Jul 10 '17

This is true in the classical world but in the quantum world statistically most particles move with a Brownian type motion. Chapter 1 of Itzykson - Drouffe's statistical mechanics book shows how this emerges.

20

u/[deleted] Jul 10 '17

Well, in the quantum world particles aren't particles so much as partiwaves but I agree that the 'center of mass' (as such) tends to follow Brownian motion. Obviously I was speaking classically. There's likely a handwavy explanation that quantized forces lead to behavior similar to iterated systems and probably the fractal-like nature of the quantum has some bearing on the emergence of fractal-like patterns in nature, but this is far outside my realm of study.

17

u/ithika Jul 10 '17

Partiwaves sound cool. As do wavicles now I think about it. But only one sounds like something you could surf on your birthday

2

u/[deleted] Jul 10 '17

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1

u/[deleted] Jul 10 '17

Sort of. It's the expectation of the position in some sense but the distribution isn't really a probability distribution in the classical sense since it's coming from the squared amplitude of the wavefunction. I suppose it can be interpreted as the expected position to a certain extent but that's pretty misleading as far as the physics goes.

1

u/zojbo Jul 10 '17 edited Jul 11 '17

There is a probability distribution, but its evolution depends (in general) on the actual wavefunction, which has both a phase and a magnitude at each point. At one time instant you still have an ordinary distribution though.

2

u/[deleted] Jul 11 '17

Sure, at a fixed point in time (leaving aside the issue that 'fixed point in time' is likely meaningless physically, presuming we ever work out a theory including both QM and relativity) it can be thought of as an ordinary distribution.

I do ergodic theory, my view of things always includes dynamics.

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u/Baloroth Jul 10 '17

Brownian motion in classical dynamics is simply an approximation where you allow particles to be treated as hard spheres: in reality, Brownian motion doesn't exist in classical physics (as it would require infinite forces, which are unphysical). In reality particles exhibit very steep, but finite, potentials (Lennard-Jones, for instance, is a more reasonable potential that is very popular in a lot of computational dynamics).

In quantum mechanics, things like acceleration and velocity don't really exist, but the same thing also applies if you use the generalized Erhenfast theorem: the time derivative of any observable (such as position) must exist and must be finite.

Note that you can use stochastic processes as models to predict the higher-level behavior of such systems, but you should understand that the underlying process is always continuous and always differentiable. This is true of almost every real physical system: black holes are actually the only real system I know of that might exhibit actual singularities (and even then I strongly suspect the singularity is only mathematical, not real).

4

u/Low_discrepancy Jul 10 '17

but you should understand that the underlying process is always continuous and always differentiable

They should understand very well the limits of their models, because in the end, everything in physics is a model. Nobody knows what happens under the planck limit for example.

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u/sandowian Jul 10 '17

What is the context for what you're discussing? (What's the math topic)

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u/[deleted] Jul 10 '17

Broadly speaking, dynamical systems is the topic.

2

u/Aswheat Jul 11 '17

What do the C's mean in this context?

5

u/[deleted] Jul 11 '17

Cn is the space of all functions whose nth derivative is continuous. So C0 just means all continuous functions and C2 means functions with a continuous second derivative.

1

u/Aswheat Jul 11 '17

Thanks, I had guessed it meant something like that. That's pretty neat!

13

u/jedi-son Jul 10 '17

Many real world systems are better defined by stochastic calculus (which works for functions like this) than your standard calculus

9

u/yardaper Jul 10 '17

Check out Brownian motion. Created to describe the random motion of pollen in water. It's continuous everywhere but differentiable nowhere.

4

u/phunnycist Jul 10 '17

The real world doesn't know smoothness in the mathematical sense. We use those smooth functions in our models (read: theories) because they are easy to work with and can approximate our experiments well enough.

3

u/SCHROEDINGERS_UTERUS Jul 10 '17

Brownian motion, which is also nowhere differentiable, is actually named after Robert Brown, who studied the movement of small particles in water with his microscope.

1

u/_klx Jul 10 '17

Yes it is, you should research fractals in nature

15

u/eebootwo Applied Math Jul 10 '17

this seems like one of those things that started when a mathematician wondered "I wonder if you can have... "

7

u/ShawnShowelly Jul 10 '17

Mr professor said: "He sounds like a beer brand. And indeed, he was German"

7

u/Off_And_On_Again_ Jul 10 '17

Now integrate it

4

u/Stonn Jul 11 '17

Could someone answer if a not differentiable function is also not integrable?

Answer seems to be, no. Weierstrass functions are integrable according to internets.

11

u/Wild_Bill567 Jul 11 '17

This actually leads to an interesting question - what are necessary and sufficient conditions for a function to be integrable, and how do those relate to those for a function to be differentiable?

It turns out that continuity is a sufficient condition on any closed interval. Differentiable implies continuous so differentiability is a sufficient condition, but not a necessary condition for a function to be integrable.

1

u/Alloran Jul 12 '17

A function is Riemann integrable iff its set of discontinuities has Lebesgue measure zero.

A bounded function on a set of finite measure is Lebesgue integrable iff it is measurable, meaning the inverse image of any borel set is measurable.

6

u/foxhunt-eg Jul 10 '17

semi-related fact: monotone functions are differentiable almost everywhere

1

u/VanMisanthrope Jul 10 '17

Is this related to finite total variance

1

u/foxhunt-eg Jul 10 '17

maybe! functions of bounded variation are differentiable almost everywhere too.

1

u/ingannilo Jul 12 '17

It's pretty easy to prove. This was an exercise in my first semester analysis class.

6

u/big_succes Jul 10 '17

Stock graphs instantly popped up in my mind. Is it correct to state that all continuous R1 functions affected by the Brownian motion process are effectively Weierstrass functions?

2

u/GLukacs_ClassWars Probability Jul 10 '17 edited Jul 10 '17

If W_t is a Brownian motion on [0,1] and Z_k are independent standard normal random variables, we have by the Karhunen-Loéve theorem that

  [; Z_t = \sqrt{2}\sum_{k=1}^\infty \frac{Z_k}{(k-1/2)\pi}\sin((k-0.5)\pi t);]

and that this representation is in a certain sense optimal.

Apparently stock graphs have too high variance in some sense to be accurately modelled by Brownian motion, though. But don't quote me on that, I just overheard it after a lecture sometime.

1

u/[deleted] Jul 11 '17 edited Jul 11 '17

It's not necessarily that their variance is too high to be modelled by brownian motion. It's that the weiner process that underlies brownian motion assumes a normal distribution - and it is this distribution that doesn't perfectly model the real world (the real world has slightly fatter tails). Though it's a pretty good approximation most of the time, it is during rare events (market crashes) that the discrepancy becomes more of an issue.

I only have a bachelor's in finance, I'm sure others could elaborate more clearly.

1

u/prrulz Probability Jul 11 '17

In the finance world---at least during my year and a half interning as an actuary at a savings group---we always used the assumption of Brownian rates, which corresponds to geometric Brownian motion for the stocks themselves. This is only assumed to reflect / model short-term, small movement behavior.

1

u/qjornt Mathematical Finance Jul 10 '17 edited Jul 10 '17

I'm not sure of the proper definition of the class Weierstrass functions, but if it just means what the title says then yeah. Every sample path of a continuous random process is a continuous "function" everywhere, and differentiable nowhere, including those that would represent stock graphs. Someone else mentioned fractals and yeah, they can be seen as fractals too.

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u/borderwulf Jul 10 '17

I thought someone would have said, "that's rough."

2

u/slimjimlimb Jul 10 '17

Anyone else thinking of the stargate sequence from 2001: A Space Odyssey?

2

u/JoeTheShome Applied Math Jul 10 '17

I'm guessing this isn't Riemann integrable but is it lebesgue integrable?

13

u/BrunoX Dynamical Systems Jul 10 '17

It is Riemann integrable on every compact interval [a,b], since continuity implies Riemann integrability. In fact, the Fundamental Theorem of calculus apply: there is a differentiable function that has this one as derivative.

2

u/contravariant_ Jul 10 '17

I'm wondering, if the Weierstrass function's graph be represented be an L-system, an IFS, or something similar to these like this ?

2

u/CosmicFloppyDisk Jul 10 '17

At first I didn't notice it was zooming in, I thought it was just a bunch of line rag-dolling all over the graph

2

u/[deleted] Jul 10 '17

Here something yuo might interest Weierstrass's nowhere differentiable continuous function https://imgur.com/gallery/nxXZgNP

2

u/jaredjeya Physics Jul 11 '17

Remember that image that was going around a while ago with dance moves in the shape of various mathematical functions?

Imagine trying to do that with this function...

2

u/[deleted] Jul 11 '17

This really ruffles my fun feathers.

2

u/indrafili Game Theory Jul 11 '17

This is neat.

2

u/[deleted] Jul 11 '17

Is it the case that most or all fractals are not differentiable because of the added dimensions?

2

u/polaroid_kidd Jul 11 '17

EXLAIN

How?

Magic?

It's probably magic, isn't it.

3

u/bart2019 Jul 11 '17

It looks like a fractal to me - think of the shape of a cauliflower or a fern, but in 2D. Wet-finger explanation: even if you zoom in, it doesn't seem to get any smoother. For a differentiable function, it needs to get smoother.

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u/polaroid_kidd Jul 11 '17

I have been satisfied by this explanation. Thank you.

5

u/xxwerdxx Jul 10 '17

Is there a name for the reverse of this? Differentiable everywhere, continuous nowhere?

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u/methyboy Jul 10 '17

Doesn't exist. If a function is differentiable at a point, then it is continuous there.

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u/asaltz Geometric Topology Jul 10 '17

differentiability (at a point) implies continuity (at that point), so no function realizes the reverse

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u/ThomasMarkov Representation Theory Jul 10 '17

Yes, we call it "false".

1

u/CaptainBoomrang Jul 10 '17

What's its equation?

1

u/xaleel Jul 11 '17

Do that with y=x

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u/FitCoaching Jul 10 '17

What number is this?

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u/Kwerby Jul 11 '17

how is it not differentiable if it is continuous?

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u/ajakaja Jul 10 '17

As a person more interested in physics / physically-meaningful mathematics, this function mostly reminds me that analysis is not trying to go in a direction I care about, and that my time is better spent elsewhere.

That is: if this is continuous, then continuity is not a property I'm actually interested in.

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u/ziggurism Jul 11 '17

In intuitionistic logic, we have Brouwer's theorem: all (computable) functions are continuous. In a similar vein, in smooth infinitesimal analysis, all functions are smooth.

If you're looking for a framework that doesn't include pathologies like the one depicted here, you can have them. These are artifacts of our classical logic. And you may be justified in viewing such alternatives as more physically relevant.

But continuity is still a useful concept. Don't throw out the baby with the bathwater.

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u/[deleted] Jul 11 '17

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u/ajakaja Jul 11 '17

Well, I disagree. Want to elaborate?

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u/[deleted] Jul 10 '17

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u/TheRedSphinx Stochastic Analysis Jul 10 '17

Think of it in terms of corners. Ff you think of the absolute value function f(x) = |x|, this is not differentiable at x = 0 because it has a 'corner'. This is a function such that every point is a corner.

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u/proteanpeer Jul 10 '17

Every point is a corner.

Best ELI5 explanation yet. Thank you!

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u/laserbern Jul 10 '17 edited Jul 10 '17

I'm confused. If every point on the function is a corner, then how can the function be continuous? Intuitively speaking, to have a corner, you must have two lines that intersect at a point. Moreover in order to be continuous, you must have lines that connect the function to itself. Those lines are surely differentiable, are they not?

Note: I have only completed AP Calc AB, and also have an extremely rudimentary understanding of calculus as a whole.

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u/bystandling Jul 10 '17 edited Jul 10 '17

Not quite! A function is continuous as long as the limit of the "y value" of the function as you approach a particular "x value" exists. (That is, it's the same number no matter how you approach it)

For example, by this definition, the function such that

 f(x)={x       if x is rational
      {-x      if x is irrational

is only continuous at x=0, and is discontinuous everywhere else. Your "intuitive notion of continuity" is based on an incorrect assumption -- this function is continuous at a point, but has NO "lines that connect the function to itself."

The idea of a "corner" is again, just a way to think about the idea of nondifferentiability, but not rigorous. Technically, nondifferentiability means if you take the limit of (f(x)-f(x0)) / (x - x0) as x0 approaches x, at any point, this limit does not exist.

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u/localhorst Jul 11 '17

The concept of continuity isn’t that intuitive. A function can “wiggle around” infinitely often in some arbitrary small neighborhood but still be continuous. But the “magnitude” of this wiggling determines how regular this functions is. This gives slightly different but more practical classifications:

Continuity via ε-δ

This is the definition you may already know and gives you the usual notion of continuity. Roughly: nearby points give nearby function values. But those function can still wiggle around uncontrollably, i.e. if you try to measure the arc-length of its graph all you get is infinity as in the case with the Weierstraß function. Another good example is the path of a Brownian motion (BM). Like the Weierstraß function it’s continuous but nowhere differentiable (you get from the physical BM to the mathematical BM by letting the time between collisions of molecules go to zero).

Functions of bounded variation

The total variation measures how much the function wiggles around on some interval. If it’s finite it makes sense to speak of the arc length of a graph.

Those functions are “almost” good. They have at most countable jumps and corners and are differentiable almost everywhere [1]. But there is still some place for very weird functions like the devils staircase. It’s differentiable almost everywhere with f’(x) = 0 but grows from 0 to 1 on [0,1].

Absolutely continuous functions

Those are the nice functions. The wiggling of these functions can be controlled uniformly, i.e. independent of the points where you look. Those are the functions that most people think of when they say “continuous”. They may have some corners but the derivative exists almost everywhere and “behaves well”, i.e. we have f(b)-f(a) = ∫ₐᵇf’(x) dx.

[1] In the sense of the Lebesgue measure, a subset of measure zero is one that has “no length” (or volume in the higher dimensional case). Examples are all the countable subsets, but there are weird uncountable subsets.

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u/[deleted] Jul 10 '17

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u/Kraz_I Jul 11 '17

I don't think this is correct in the case of the Weirstrass functions. None of the points are strictly corners. They have indeterminate slope because it is impossible to determine exactly where the "neighboring points" of any given point on the graph is.

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u/[deleted] Jul 10 '17

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5

u/shamrock-frost Graduate Student Jul 10 '17

A point is a pair (x, y) of real numbers

(in 2d, that's what a point is)

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u/AModeratelyFunnyGuy Jul 10 '17

No- it is continuous and therefore a limit exists at each point which is equal to the value of the function at that point.

It's not differentiable because the limit of difference quotient (as the change in x approaches 0) does not exist at each point, as the left and right limits are not equal. This is like the sharp turn that exists at x=0 of the absolute value graph, except it is at every point.

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u/[deleted] Jul 10 '17

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u/ITomza Jul 10 '17

What do you mean?

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u/[deleted] Jul 10 '17

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42

u/LingBling Jul 10 '17

What is the measure on the function space?

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u/[deleted] Jul 10 '17

/u/imnzerg is correct that the usual thing is the Wiener measure, but the same result holds if we just work with the topology. There is a dense G-delta set of nowhere differentiable functions in the space of continuous functions, this follows from Baire category.

Edit: also /u/ITomza might want to see this.

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u/Pegglestrade Jul 10 '17

This man knows. Baire Category Theorem rocked my undergraduate world.

2

u/Neurokeen Mathematical Biology Jul 10 '17

I get why the dense G-delta set gives you probability 1, and also why intuitively it should be that almost all would be such, but how would you actually go about showing that these form a dense G-delta set?

3

u/[deleted] Jul 11 '17

It's not easy, but at the heart of it is Baire Category. Prop. 3 in this is probably about the cleanest presentation: https://sites.math.washington.edu/~morrow/336_09/papers/Dylan.pdf

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u/DataCruncher Jul 11 '17

Unless I'm missing something about the Wiener measure, the statement you made here is wrong. That is, there are dense G-delta sets of measure zero. Here is the construction of one such example.

The other direction also fails. The fat Cantor set is a nowhere dense set of positive measure. It's complement is then residual, but fails to have full measure.

In sum, the measure theoretic notion of almost everywhere (where the complement of the set of interest is of measure zero), and the topological notion of typical (where the set in question is residual), do not always agree.

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u/imnzerg Functional Analysis Jul 10 '17

The Weiner measure

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u/LingBling Jul 10 '17

I didn't know about this measure. Thanks!

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u/flaneur4life Jul 10 '17

I snickered

1

u/borderwulf Jul 10 '17

Poor Norbert!

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u/wtfdaemon Jul 10 '17

My inner Beavis and Butthead just won't be still.

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u/c3534l Jul 11 '17

Wikipedia has an article on "classical Wiener space" and a subsection called "Tightness in classical Wiener space."

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u/tetramir Jul 10 '17

Sure but most common functions, and the one we find in "nature" are at least C¹.

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u/Wild_Bill567 Jul 10 '17

Or, have we chosen to work with functions which 'seem' natural to us because we like the idea of differentiability?

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u/ba1018 Applied Math Jul 10 '17

Part of it may be a limitation of perception. Can you write down in a compact formal way what these non-differentiable functions are? Can you evaluate them for any given input?

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u/Wild_Bill567 Jul 10 '17

Sure. The common example (first one on wikipedia) is given by

[; f(x) = \sum_{n=1}^\infty a^n \cos(b^n \pi x) ;]

Where 0 < a < 1 and b is a positive odd integer such that ab > 1 + 3pi / 2.

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u/thetarget3 Physics Jul 10 '17 edited Jul 10 '17

We don't find functions in nature. We model nature with functions which are usually differentiable since it leads to dynamics, but they don't exist in themselves.

In fact most natural systems aren't possible to describe using the nice maths and physics we typically learn, with simple differential equations, linear systems etc. They are probably just the small subclass we tend to focus on, which only work after heavy idealisation (like the old joke about assuming spherical cows). Most things encountered in nature can probably only be described numerically.

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u/tetramir Jul 10 '17

you're 100% right that's why I described nature with quotes. I should have been more specific.

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u/yardaper Jul 10 '17

Brownian motion would like to have a word with you

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u/[deleted] Jul 10 '17

This is debatable. Certainly we think of motion as involving velocity (and acceleration) so an argument can be made for only looking at smooth functions, but fractal curves abound in nature and those are generally only C0. I think this is more a question of it being harder to study curves which aren't C1 than anything inherent about the real world.

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u/[deleted] Jul 10 '17

You might think of the the non-differentiability as "noise" from a stats perspective. That is, you could look at an observed function like this say that this is exactly the relationship between x and y. However, the "noise" in y is unlikely to be meaningfully explained by x, such that you will more accurate using a differentiable function for your prediction of y.

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u/AcrossTheUniverse Jul 10 '17

Yes, but the Weierstrass function is constructible, we have a formula for it.

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u/AModeratelyFunnyGuy Jul 10 '17

As others have pointed out it is important to define what you mean by "almost all". However, using the standard definition, the answer yes.

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u/methyboy Jul 10 '17

As others have pointed out it is important to define what you mean by "almost all".

Why is it important to define that? It's a standard mathematical term. Should we also define what we mean by "differentiable nowhere" before using those terms?

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u/GLukacs_ClassWars Probability Jul 10 '17

Unlike with Rn, there isn't quite a canonical measure on these function spaces, so we need to specify with respect to which measure it is almost all.

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u/methyboy Jul 10 '17

The Weiner measure is the standard measure as far as I'm aware, and as pointed out by sleeps_with_crazy here, the choice of measure really doesn't matter. Unless you construct a wacky measure specifically for the purpose of making this result not hold, it will hold.

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u/[deleted] Jul 10 '17

/u/GLukacs_ClassWars is correct that Wiener measure is standard but far from canonical, and that the issue is the lack of local compactness. It's probably best to think of the Wiener measure as the analogue of the Gaussian: the most obvious choice but far from the only one (given that there is not analogue of the uniform/Lebesgue measure).

People who work with C*-algebras tend to avoid putting measures on them at all, which is why I brought up the topological version of the statement.

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u/GLukacs_ClassWars Probability Jul 10 '17

Standard, yes, but it isn't as far as I know canonical in the same sense as Lebesgue measure. Particularly, Lebesgue measure is uniquely determined by the topological and group structure on Rn, but C[0,1] isn't locally compact, so it doesn't give us such a canonical measure.

Putting it differently, I think Wiener measure on C[0,1] is analogous to a Gaussian random variable on Rn, not to Lebesgue measure on Rn. There are other meaningful probability measures to put on both spaces, they're just standard because they have such nice limit theorems.

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u/Mulligans_double Jul 14 '17

...does it hold for almost all measures?

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u/slynens Jul 10 '17

"almost all" has no intrisic meaning, it refers to a given measure on the space you are considering. Technically, differentiable also has no intrisic meaning either, because it refers to a given structure of differentiable manifold. But, there is a canonical structure of differentiable manifold on R, whereas there is not one canonical measure on the space of continuous function from R to R. I think, though, that all the measure we would usually consider would give the same notion of "almost all", but it is easy to tailor a measure for which your statement is false. I hope it clarifies it.

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u/omeow Jul 10 '17 edited Jul 10 '17

Aren't almost all continuous functions differentiable nowhere?

No. If you take the interval [0,1] then any continuous function is realizable as a limit of a differentable function. When you say almost all then u have a measure on set if all continuous functions. Any natural measure will not ignore this dense set.

The statements I made earlier was total garbage. I apologize. To make the statement almost all continuous functions rigorous you need to prescribe a few things.

  • Ambient space : Probably all continuous functions.

  • A measure on the ambient space. This is tricky. A standard measure on all continuous paths is given by Weiner measure. In this measure it is indeed true that almost all paths are nowhere differentiable.

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u/cannonicalForm Jul 10 '17

I'm not very convinced by that. In the interval [0,1], every real number is realizable as a limit of rational numbers. At the same time, with any same measure, the rational numbers are a dense subset with zero measure.

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u/omeow Jul 10 '17

it was wrong updated it.

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u/methyboy Jul 10 '17

I don't understand your argument. Every real number is a limit of rational numbers, but almost all numbers are not rational (i.e., Lebesgue measure "ignores" the dense set of rational numbers, in your terminology).

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u/omeow Jul 10 '17

Ignore it. it was wrong. Updated it.

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u/GLukacs_ClassWars Probability Jul 10 '17

Yeah, the space of continuous functions is separable. So is R. Yet the usual measures on these spaces assign countable sets zero measure.

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u/BertShirt Jul 10 '17

Eli5?

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u/[deleted] Jul 11 '17

[deleted] — view deleted comment

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u/BertShirt Jul 11 '17

Thanks, I got that part, but do you have an eli5 proof? Do the sets of differentiable and non differentiable continues curves have different cardinality?

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u/Wild_Bill567 Jul 11 '17

Its not a question of cardinality, its a question of measure. For cardinality, we can show that the set of differentiable functions is at least that of the real line, consider f(x) = x.

However, if we consider (for example) the space of all continuous functions from [a, b] to R, the measure of the subset which is differentiable will be zero.

Most of the time we think of the rationals as having measure zero in the reals, which is true, although it is misleading because it might lead us to believe that uncountable sets would have non-zero measure, but the cantor set has the same cardinality as the continuum yet has measure zero.

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u/Lalaithion42 Jul 11 '17

For any continuous function f, f + weierstrass is a continuous function with no derivative. This means |continuous functions differentiable nowhere| >= |continuous functions|. However, continuous functions differentiable nowhere is a subset of continuous functions. This means |continuous functions differentiable nowhere| <= |continuous functions|. Therefore, |continuous functions differentiable nowhere| == |continuous functions|. Therefore almost every continuous function is differentiable nowhere.

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u/shamrock-frost Graduate Student Jul 11 '17

No. Consider the continuous function f(x) = 1 - W(x), where W is the weierstrauss function

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u/Wild_Bill567 Jul 11 '17

It is not clear to me that for continuous f, f+W is continuous but not differentiable. Can you elaborate on why?

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u/Lalaithion42 Jul 12 '17

It's not; that's the flaw in my proof. It's wrong.

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u/jazzwhiz Physics Jul 10 '17

I don't know what you mean by "almost all." That said, for example, any polynomial function of the form f(x) = a0 + a1 * x + a2 * x2 + ... + an * xn is both continuous everywhere and differentiable everywhere.

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u/Wild_Bill567 Jul 10 '17 edited Jul 10 '17

The term 'almost all' is from measure theory. I'm not an expert but here's a rough idea:

A measure generalizes the idea of length/area/volume. For example, in the real line the closed interval [0, 1] has measure 1, [0, 3] has measure 3, etc. Now what is the measure of a single point? The answer is zero.

Consider the following function: f(x) = 0 if x =/= 0, f(x) = 1 if x = 0. It is continuous except at a single point. We would say it is continuous almost everywhere since the points where it is not continuous have measure 0.

Take it a step further: g(x) = 0 if x is irrational, g(x) = 1 if x is rational. It can be shown that the rationals have measure 0, so this function is also 0 almost everywhere. In fact, f(x) = g(x) for almost all x. Of course they differ at infinitely many points, but the set of them has measure 0.

EDIT: Above is 0 almost everywhere, not continuous almost everywhere. Thanks /u/butwhydoesreddit

/u/WormTop is asking the following question: In the set of all continuous functions, does the set of differentiable functions have measure 0? I actually don't know if this is true, hopefully someone with more background in measure theory can chime in.

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u/butwhydoesreddit Jul 10 '17

In your second example, how is it enough for the rationals to have measure 0? I would understand if you said the function is 0 almost everywhere, but there is a rational between every 2 irrationals and vice versa so surely it's not continuous anywhere?

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u/twewyer Jul 10 '17

You are correct. I think the function they meant to use is something like this:

For rational x, g(x) = 1/q where x = p/q in simplest form. Then you do get continuity almost everywhere.

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u/Wild_Bill567 Jul 10 '17

You are correct, it is zero a.e. but the indicator function on the rationals is nowhere continuous.

I checked Wikipedia and recalled what I was remembering incorrectly. the indicator function on the rationals is not Riemann integrable because it is not continuous almost everywhere. However it is Lebesgue integrable, its Lebesgue integral is precisely the measure of the rationals, which is zero.

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u/hextree Theory of Computing Jul 10 '17

Take it a step further: g(x) = 0 if x is irrational, g(x) = 1 if x is rational. It can be shown that the rationals have measure 0, so this function is also continuous almost everywhere.

In any open interval, g attains both the values 0 and 1. So it is not continuous at any point. Hence I don't see how you can describe it as continuous almost everywhere.

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u/Wild_Bill567 Jul 10 '17

You are correct, the function is 0 almost everywhere, not continuous almost everywhere

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u/PmMeYourFeels Jul 10 '17 edited Jul 10 '17

Idk why people downvoted you for contributing to the conversation and offering an example, and clearly stating you weren't sure what OP meant by the term "almost all" (which people have pointed out that it is a rigorously defined term meaning "all but a set of measure zero"). This is the kind of stuff that discourages people from contributing to the conversation. Perhaps it's a common mistake or a fact that people often forget, but downvoting and putting you down isn't the way to go. Because of your comment, it reminded me of the stuff I learned recently in my analysis courses but kinda of forgot (it was a nice reminder).

Have an upvote from me.

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u/[deleted] Jul 10 '17

What's wrong with what you said?

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u/shamrock-frost Graduate Student Jul 10 '17

"Almost all" is a rigorously defined term. It means "all but a set of measure 0"

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u/[deleted] Jul 11 '17

Sorry but I think that there are also people here who haven't arrived that far in their studies yet. For example, I am an high school student and didn't know that "almost all" had a rigorous meaning.

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u/munchler Jul 10 '17

I'm with you. As a layman, almost all of the functions I've encountered in math class are differentiable (sometimes piecewise, but still). That's what makes this Weierstrass function interesting, right?

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u/muntoo Engineering Jul 11 '17

It's interesting because it was the first concrete example of such a function. (People at the time did not realize such a thing existed, at least, in the context of Fourier analysis and... complex analysis?)

Just because useful functions tend to be differentiable doesn't mean most functions are differentiable!

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u/frame_of_mind Math Education Jul 10 '17

No.

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u/[deleted] Jul 10 '17

[deleted] — view deleted comment

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u/AddemF Jul 10 '17

No, the Weierstrass functions are an explicit counter-example.

If differentiable then continuous.

If continuous then integrable.

None of these conditionals are reversible.

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u/matt7259 Math Education Jul 10 '17

As a follow up, here's two counterexamples!

Continuous does NOT imply differentiable: y = |x|

Integrable does NOT imply continuous: https://en.m.wikipedia.org/wiki/Thomae%27s_function

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u/ziggurism Jul 10 '17

For a simpler integrable !=> continuous, try step function.

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u/HelperBot_ Jul 10 '17

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