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I came across the following exercise:

Let $(B_t)_{t\geq 0}$ be a Brownian motion. Show that, almost surely, there is no interval $(r,s)$ on which $t\to B_t$ is Hölder continuous of exponent $\alpha$ for any $\alpha>\frac{1}{2}$. Explain the relation of this result to the differentiability properties of $B$.

I'm happy about the first part, but am wondering how this relates to the differentiability of $B$. This property alone isn't strong enough to ensure that the paths are nowhere differentiable (which they are). Does it perhaps imply that $B$ is almost surely not differentiable on any open interval?

So it would be informative to answer the question:

If $f:\mathbb{R}\to\mathbb{R}$ is not Lipschitz on any open interval, then does every open interval contain a point at which $f$ is non-differentiable?

Thank you.

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    These two statements are probably equivalent for Holder continuity (by taking rational endpoints), but they are very different for differentiability, which can be stated pointwise (i.e. "almost surely the BM is not differentiable at$t$vs. almost surely the BM is nowhere differentiable") and the latter statement is the one with the proof that works for both results.2012-02-13

1 Answers 1

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The answer to your second question is yes. Assuming that $f:\mathbb{R}\to\mathbb{R}$ is not Lipschitz on any open interval, let $(a,b)$ be an arbitrary non-empty open interval. By assumption there exist $x_1,y_1$ with $a such that $|f(y_1)-f(x_1)| > y_1-x_1$. Since $f$ is not Lipschitz on $(x_1,y_1)$, there exist $x_2,y_2$ with $x_1 and $|f(y_2)-f(x_2)|>2(y_2-x_2)$. Continuing inductively, we obtain two sequences $(x_n)$ and $(y_n)$ with $a such that $\frac{|f(y_n)-f(x_n)|}{y_n - x_n}>n$ for all $n$. Then $x_n \to x^*$ and $y_n \to y^*$ with $x^* \le y^*$. If $x^*, then the function $f$ is unbounded on $[x_1,y_1]$, so there exists a point of non-differentiability in $[x_1,y_1] \subset (a,b)$. Otherwise if we assume that $f$ is differentiable at $x^*=y^*$ with $f'(x^*)=L$, we get$f(x_n) = f(x^*) + (x_n-x^*)a_n\quad \text{ and } \quad f(y_n) = f(x^*) + (y_n-x^*)b_n $with $L = \lim\limits_{n\to\infty} a_n = \lim\limits_{n\to\infty} b_n.$ Subtracting these we get $|f(y_n) - f(x_n)| \le (y_n-x_n)(|a_n|+|b_n|), $ so $ \frac{|f(y_n)-f(x_n)|}{y_n-x_n} \le |a_n|+|b_n| \to 2|L|$ as $n\to\infty$, contradicting the previous estimate that this sequence of quotients is unbounded. This shows that $f$ is not differentiable at $x^*$.