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!
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 $$ \left|\int_0^t \int_{\mathbb{R}} f''(z) \frac{1}{\sqrt{2\pi s}} e^{- \frac{z^2}{2s}}dz\, ds\right| \le 2 E\big[|f(W_t)-f(0)|\big]\le 4||f||_\infty. $$
So even the supremum norm alone is enough to bound the integral.