Triangle inequality for Ito integral?

Question

For Lebesgue integrals one has the triangle inequality saying that for continuous functions let's say

$$\left\vert\int_0^t f(s) \ ds\right\vert \le \Vert f \Vert_{\infty} \int_0^t \ ds$$

Now if we consider an Ito integral, then

$$\left\vert\int_0^t f(s) \ dW(s)\right\vert \le \Vert f \Vert_{\infty} \vert \int_0^t \ dW(s)\vert$$

does not hold pointwise, but I was wondering whether this one holds probabilitically, i.e.

does there exist a constant $c(t)>0$ such that for all deterministic continuous $f$

$$\mathbb P\left(\vert W(t) \vert \Vert f \Vert_{\infty}\ge a\right)\ge c(t)\mathbb P\left(\left\vert\int_0^t f(s) \ dW(s)\right\vert \ge a\right)?$$

Answer 1

If the function $f$ is indeed deterministic, with $M:=\|f\|_\infty$ and $\sigma^2:=\int_0^t f(s)^2\,ds$, then $X:=\int_0^t f(s)\,dW(s)\sim N(0,\sigma^2)$, whereas $Y:=MW(t)\sim N(0,M^2t)$, and $k^2:=\sigma^2/(M^2t)\le1$. So, $X$ equals $kY$ in distribution, and so, $P(|Y|\ge a)\ge P(|X|\ge a)$ for all real $a$; that is, your inequality holds with $c(t)=1$.