Proof that the image of an Itō integral is convex if the driving Wiener process is in a metric ball

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Let $(\Omega, \mathcal{F}, P)$ be a probability space and $A := \int_0^1 f(t)\,d W_t$ be the Itō integral of an $L_2([0,1])$ deterministic function $f$ with respect to the Wiener process $W$. Additionally, define the set $B\subset\mathbb{R}$ by $$ \textstyle B:= \{A(\omega) \,|\, \omega\in C \}, $$ where $$ \textstyle C:= \{\omega\in\Omega \,|\, \sup_{t\in[0,1]}|W_t(\omega)|< 1\}. $$ How can I prove that the set $B$ is convex and symmetric?

Context

This statement (that $B$ is convex) is used by Shepp and Zeitouni (1992, p. 654). They mention it as something obvious in less than a line and proceed to use it to obtain their result. I'm trying to generalize their result slightly and am not sure where this statement comes from, and why it holds. While I could simply say that Shepp and Zeitouni say it is true, I'm stubborn and would like a formal proof.

Obs.: the article by Shepp and Zeitouni is open access and can be freely downloaded.

Research

This seems to follow trivially if the Itō integral of deterministic $L_2$ functions is linear with respect to the integrator $W$. I know that if $\{f_i\}_{i=1}^\infty$ is a sequence of piecewise constant (simple) functions such that $f_i\to f$ in $L_2([0,1])$ then $A_i:=\int_0^1 f_i(t)\,dW_t$ are linear with respect to the integrator $W$. Although $A_i\to A$ in $L_2(\Omega,\mathcal{F}, P)$, I'm not sure if this implies anything in terms of the linearity of $A$ with respect to $W$.

Also, since $A$ is formally defined as the $L_2(\Omega, \mathcal{F}, P)$-limit of $A_i$, it is defined as an element of $L_2$, which is a space of equivalence classes of functions, identified by equality $P$-almost everywhere. That is, if some $A'\colon\Omega\to\mathbb{R}$ is such that $A'=A$ almost surely, then it is the same element of $L_2(\Omega, \mathcal F, P)$. This seems to leave some ambiguity in the definition of $B$, allowing it to be non-convex and non-symmetric. Consequently, it seems more appropriate to say that there exists some $B'$ which is convex and symmetric but differs from $B$ by a measure zero set.

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Hi this is not a full answer but I think that with some little work you should get to the answer using those ideas.

First remark, if $f=1$ then $A=W_1$ so if $x\in B$ then $-x\in B$ as the path $-\omega \in C$. Convexity follows by the same line of reasoning.

Now try to extend this to simple functions of the form $f(t)=\sum_{i=1}^n \lambda_i 1[t\in O_i ] $ where $O_i$ form a borelian partition of $[0,1]$, and $\lambda_i$ are real. Then extend the result by a density argument to $f\in L^2([0,1])$.

Best regards