How can I show that the maximum is 0?

You are almost there, you just need to notice that the right-hand side is $p_U(0)$ :)

$$\begin{split} p_U(u) &= \sum_{v\in\mathcal{R}_V} p(v)\cdot p(u-v)\\ &= \sum_{v\in\mathcal{R}_V} p(v)\cdot p(v-u) &\,\,&\text{(as $p$ is even)}\\ &\leq \sqrt{\sum_{v\in\mathcal{R}_V} p(v)^2}\sqrt{\sum_{v\in\mathcal{R}_V} p(v-u)^2}& \,&\text{(by Cauchy-Schwarz)}\\ &\leq \sqrt{\sum_{v\in\mathcal{R}_V} p(v)^2}\sqrt{\sum_{v\in\mathcal{R}_V} p(v)^2}& \,&\text{(change of index $v\rightarrow v+u$)}\\ &\leq \sum_{v\in\mathcal{R}_V} p(v)^2& \,&\text{(rearranging)}\\ &\leq\sum_{v\in\mathcal{R}_V} p(v)p(-v)& \,&\text{(as $p$ is even)}\\ &\leq p_U(0) \end{split} $$

Note: Some people get confused that I keep using $\leq$ between two lines that are equal (like the last one and the one before). The reason I do so is because $\leq$ is relative to the left-hand side, ie. $p_U(u)$.