5
$\begingroup$

This post is partly inspired by Fourier Coefficients and Hölder Continuity.

Typical proofs of the nowhere differentiability of Brownian paths is by contradiction using binary expansion from real analysis. For example, Prop 11. of S.Lally's notes.

But usually one can prove a function $f$ is nowhere differentiable by examining their Fourier coefficients $\hat{f}$. For example Simple Proofs of Nowhere-differentiability Weierstrass's Function and Cases of Slow Growth

Can we prove the nowhere differentiability (w.p.1) of Brownian paths by examining its Karhunen–Loève coefficients?

Moreover, what is the relationship between Karhunen–Loève coefficients $\hat{X_t}$ and the smoothness of the path of an $L^2$ stochastic process $X_t$?

If there is such a relationship, is this relationship similar to the relationship between Fourier coefficients and original function?

In response to Christian's comment,

The Fourier coefficients don't look like the right thing to look at to me if you want to establish that Brownian paths are nowhere differentiable, since the FT responds to lack of smoothness somewhere by lack of decay.

A natural thing to ask is that what object characterizes/detects global non-smoothness in in situation?

Improve this question
$\endgroup$
7
  • $\begingroup$ For a start, I think the Kolmogorov 0-1 law together with the fact that all the coefficients are smooth should get you that nowhere differentiability has probability 0 or 1 (because tweaking any finite number of your iid random variables will not change the differentiability). $\endgroup$
    – Nate Eldredge
    Commented Apr 18, 2017 at 5:39
  • $\begingroup$ @NateEldredge Yes, we can observe that convergence of KL series is a tail event but do you mean that differentiability can be determined by the convergence rate of KL series itself and hence it's a tail event too? $\endgroup$
    – Henry.L
    Commented Apr 18, 2017 at 10:55
  • 2
    $\begingroup$ The Fourier coefficients don't look like the right thing to look at to me if you want to establish that Brownian paths are nowhere differentiable, since the FT responds to lack of smoothness somewhere by lack of decay. $\endgroup$
    – Christian Remling
    Commented Apr 22, 2017 at 18:22
  • 1
    $\begingroup$ @ChristianRemling So what is the correct thing to look at here? And can't we discover nowhere differentiability by showing lack of decay with high probability? $\endgroup$
    – Henry.L
    Commented Apr 25, 2017 at 17:29
  • 1
    $\begingroup$ No, not that easily because lack of decay of the FT just implies that the function is non-smooth somewhere. For me, the natural intuitive explanation of the fact is that typically $B_{t+h}\simeq h^{1/2}$. $\endgroup$
    – Christian Remling
    Commented Apr 25, 2017 at 19:27

0

Reset to default

You must log in to answer this question.