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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.

  • 1
    I guess you should deal with the fact the differentiability of a Brownian motion implies Hölder continuity of and exponent $1$.2012-01-30
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    Thanks. What's the full statement of your fact?2012-01-30
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    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 $$|B(t+h')-B(t)|\leq (b+1)|h'|$$2012-01-30
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    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.2012-01-30
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    @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.2012-01-30
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    agree - so my point is that lack of Holder continuity for $\alpha>1/2$ on any open interval *alone* is not sufficient to prove the lack of differentiability in a single point (like in our examples)2012-01-30
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    Does this book prove that almost surely the Brownian motion is nowhere Holder continuous for $\alpha > \frac{1}{2}$ or does it prove that the Brownian motion is not Holder continuous at time $t$ almost surely? Depending on the proof used in the book (assuming the former), the relationship they are looking for may be that the same proof can be modified to show both results. This is the case for the proof of nowhere differentiability in Morters and Peres for example.2012-02-13
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    @Chris Thanks. This is from a rather old exam - I don't know what was proved in the corresponding class. What's your definition of "not Holder continuous at time $t$"?2012-02-13
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    It's just a question of whether you fixed $t$ and then asked if the Brownian motion was Holder continuous there or if you looked at the set of points where it was Holder continuous.2012-02-13
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    @Chris Here, we're using the definition of Holder continuity on a set (as Wikipedia does). There are several possible interpretations of the statement "Holder continuous there", hence my question.2012-02-13
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    Yes sorry that was a silly comment and I said it wrong. The issue is whether you are looking at a fixed region and asking if the BM is Holder there or if you are looking at the collection of sets for which the BM can be Holder. The difference in statements is "almost surely the BM is not Holder on (a,b)" versus "almost surely for all (a,b) the BM is not Holder on (a,b)". If I recall, the proof of the latter claim is virtually identical to the proof of non-differentiability. Again I would recommend looking at Morters and Peres.2012-02-13
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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

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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 $x_12(y_2-x_2)$. 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$$ for all $n$. Then $x_n \to x^*$ and $y_n \to y^*$ with $x^* \le y^*$. If $x^*$f$ is unbounded on $[x_1,y_1]$, so there exists a point of non-differentiability in $[x_1,y_1] \subset (a,b)$. Otherwise if we assume that $f$ is differentiable at $x^*=y^*$ with $f'(x^*)=L$, we get$$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 $x^*$.