# Hölder type of inequality?

Is the following inequality true? I can't find a counterexample so I start to believe it is true but I do not manage to prove it :) Any ideas?

Let $f$ be a compactly supported bounded twice continuously differentiable function with bounded derivatives, then $$\int_0^t \int_{\mathbb{R}} f''(z) \frac{1}{\sqrt{2\pi s}} e^{- \frac{z^2}{2s}}dz ds \leq C [f]_{\alpha}\leq C [f]_{\alpha} + C\|f\|_{\infty} = C \|f\|_{\alpha}$$ for some finite constant $C>0$ where $$[f]_\alpha := \sup_{x\neq y} \frac{|f(x)-f(y)|}{|x-y|^{\alpha}}$$and here $$\|f\|_{\alpha} := [f]_{\alpha} + \|f\|_{\infty}$$ is the Hölder norm.

Observation: one can upperbound the integral by $C\|f'\|_{\infty}$ by a simple use of integration by parts. Using integration by parts twice is too much, the time integral explodes, so the question remains if there is a middle step in order to obtain the Hölder norm in the right hand side.

Thanks!

• I am confused, do you mean $f''(z)$? – Lost1 Oct 23 '15 at 20:23
• Yes sorry. Changed. – Martingalo Oct 23 '15 at 20:25
• And what should the $\int_0^t$ at the start really be? – David C. Ullrich Oct 23 '15 at 20:38

Let $W$ be the standard Wiener process. Then, by the Ito formula $$E[f(W_t) -f(0)] = E\left[\int_0^t f'(W_s)dW_s\right] + \frac12 E\left[\int_0^t f''(W_s)ds \right]\\ = \frac12\int_0^t \int_{\mathbb{R}} f''(z) \frac{1}{\sqrt{2\pi s}} e^{- \frac{z^2}{2s}}dz\, ds,$$ whence $$\left|\int_0^t \int_{\mathbb{R}} f''(z) \frac{1}{\sqrt{2\pi s}} e^{- \frac{z^2}{2s}}dz\, ds\right| \le 2 E\big[|f(W_t)-f(0)|\big]\le 4||f||_\infty.$$
• This is what I was in the process of typing out, except I think the top line should be $F(W_t)-F(0)$? – Lost1 Oct 23 '15 at 20:41
• @Lost1, sorry, it should be $f(W_t) - f(0)$ (it is already). – zhoraster Oct 23 '15 at 20:42
• Oh... Does this actually imply that one can upperbound $\int_0^t \int_{\mathbb{R}} f^{(n)}(z) \frac{1}{\sqrt{2\pi s}}\exp (-z^2/(2s) ) dzds$ for any $n$-th derivative using the argument above iteratively? – Martingalo Oct 23 '15 at 20:53
• @Martingalo, but partially, yes. You can integrate $E[f^{(n-2)}(W_t)] = \int_{\mathbb{R}} f^{(n-2)}(\sqrt{t}x) \varphi(x) dx$ by parts many times. You will end up with an estimate like $|f^{(n-2)}(0)| + C(t) ||f||_\infty$. Unfortunately, $C(t)$ explodes as $t\to 0+$. – zhoraster Oct 23 '15 at 20:59