# Triangle inequality for Ito integral?

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 $$f$$

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

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 $$c(t)=1$$.