Let $f:\mathbb{R} \to \mathbb{R}$ be a continuous bounded function, $(X_k)_k$ a sequence of i.i.d random variables such that $$\forall x \in \mathbb{R},f(x)=\int_{\mathbb{R}}f(x+y)dP_{X_1}(y).$$ Let $x \in \mathbb{R},Y_k=f\left(x+\sum_{q=1}^kX_q\right),\mathcal{F}_k=\sigma(X_1,...,X_k).$
Verify that: $$\tag{1}E\left[(Y_1-f(x))^2\right] \leq E\left[(Y_2-Y_1)^2\right] $$ and deduce that for every $y \in \text{supp} P_{X_1}:=\{u,\forall r>0,P_{X_1}([y-r,y+r])>0\},f(x+y)=f(x).$
One way to prove the first question is to note that $$\int_{\mathbb{R}}(f(x+y)-f(x))^2dP_{X_1}(y)=\int_{\mathbb{R}}\left(\int_{\mathbb{R}}(f(x+y+u)-f(x+u))dP_{X_2}(u)\right)^2dP_{X_1}(y) \leq \int_{\mathbb{R}}\left(\int_{\mathbb{R}}(f(x+y+u)-f(x+u))^2dP_{X_2}(u)\right)dP_{X_1}(y) \leq E[(Y_2-Y_1)^2]$$ For part 2 we can deduce that for $P_{X_1}$-almost every $y \in \mathbb{R},f(x+y)=f(x),$ how to conclude using continuity that $f(x+y)=f(x)$ for every $y \in \text{supp}P_{X_1}$ ?
How to deduce part 2 ?
You have already found that $f(x+y)=f(x)$ for $P_{X_1}$-a.e. $y\in\mathbb{R}$. Now, let $y\in\mathrm{supp}P_{X_1}$. Then $P_{X_1}(y-1/n,y+1/n)>0$ for every $n\in\mathbb{N}$. Therefore, there must exist $y_n\in(y-1/n,y+1/n)$ such that $f(x+y_n)=f(x)$. Obviously $y_n\to y$ and since $f$ is continuous we have $$ f(x+y)=\lim_{n\to\infty}f(x+y_n)=f(x). $$