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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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    $\begingroup$ I guess you should deal with the fact the differentiability of a Brownian motion implies Hölder continuity of and exponent $1$. $\endgroup$ – Ilya Jan 30 '12 at 12:39
  • $\begingroup$ Thanks. What's the full statement of your fact? $\endgroup$ – Ben Derrett Jan 30 '12 at 13:09
  • $\begingroup$ If $\exists B'_t = b$ then $$ B(t+h)-B(t) = bh+\varepsilon(h)h$$ where $\varepsilon(h)\to 0$ with $h\to 0$. It means that we can take $h$s.t. $|\varepsilon(h')|\leq 1$ for all $|h'|\leq h$ and so $$|B(t+h')-B(t)|\leq (b+1)|h'|$$ $\endgroup$ – Ilya Jan 30 '12 at 13:22
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    $\begingroup$ I managed to catch your comment ) seems that the problem now is that $t$ is fixed and hence there is no Holder continuity on the whole interval. I am not sure that differentiability would imply it (see for example $x^2\cdot1_\mathbb Q(x)$). So the argument with Holder continuity only tells that if $B_t$ cannot have bounded derivative on the open interval. It does not tell anything about differentiability in only open point though. $\endgroup$ – Ilya Jan 30 '12 at 13:31
  • $\begingroup$ @Ilya The trouble is that this doesn't imply Hölder continuity, as far as I can tell, on intervals. We need some kind of uniformity condition to avoid examples like $x^2\sin(1/x)$, which is differentiable, but not Hölder continuous with exponent 1 (Lipschitz) on any interval containing the origin. $\endgroup$ – Ben Derrett Jan 30 '12 at 13:31
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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 $a<x_1<y_1<b$ 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 $x_1<x_2<y_2<y_1$ and $|f(y_2)-f(x_2)|>2(y_2-x_2)$. Continuing inductively, we obtain two sequences $(x_n)$ and $(y_n)$ with $$a<x_1<x_2<\ldots<y_2<y_1<b$$ 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 $x^*<y^*$, 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 $x^*$.

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