Let $W$ be a classical Wiener process on $[0,1]$ and let $$ \mathcal{I}\colon a\mapsto \int_0^1a(t) dW(t) $$ be the stochastic integral with respect to $W$. Ito isometry states that $\mathcal{I}$ is an isometry between $L^2([0,1])$ and $L^2(\Omega, \mathcal{P})$. I have a question concerning the pointwise estimate.
Question. It is true that for almost all $w\in \Omega$ there exists $C(\omega)$ such that $$ |I(a)(w)|\le C(w)\|a\|_{L^2} $$ holds for all $a\in L^2([0,1])$?
The answer seems to be no.
Assume that $f$ is a smooth function then Ito's formula reads as $$ d(f(t)W(t)) = W(t) f'(t) + f(t) dW(t). $$ Integration on $[0,1]$ we get the formula for integration by parts: $$ \int_0^1 f(t)dW(t) = f(1)W(1) - \int_0^1 W(s) f'(s)\,ds. $$ Now we want to choose $f$ to be a trigonometric polynomial. The Fourier coefficients of $W(s)$ behave as $\xi_n/n$, where $\xi_n$ are i.i.d, standart Gaussian variables (Wiener representation). Hence the integral on the right rewrites as $$ \int_0^1 W(s) f'(s)\,ds = \sum_{n\in \mathbb{Z}} \xi_nf_{n+1}, $$ where $f_{n}$ are the Fourier coefficients of $f$. Thus the required estimate rewrites to (need to get rid of $f(1)W(1)$ term somehow) $$ |\sum_{n\in \mathbb{Z}} \xi_n(w)f_{n+1}|\le C(w) (\sum |f_n|^2)^{1/2}. $$ Now we can choose $f$ such that $f_{n + 1} = \xi_n(w)$ for all $n\le N$ and $f_{n + 1} = 0$ for $n > N$. This will give $$ \sum_{n= 0 }^N |\xi_n(w)|^2\le C(w) (\sum_{n= 0 }^N |\xi_n(w)|^2)^{1/2}, \\ (\sum_{n= 0 }^N |\xi_n(w)|^2)^{1/2} \le C(w). $$ However the left-hand side is a.s. not bounded in $N$.
