Let $f \in C^0(\mathbb{R},\mathbb{R})$, is there's $\alpha>0$, and $M>0$ such that for all $y,y'\in L^2(0,1)$ such that, $\|y-y'\|_2\leq \alpha$ we have: $$\sup_{\|z\|_2\leq 1}\|[f(y)-f(y')]z\|_2\leq M \|y-y'\||_2$$
My idea: $$\sup_{\|z\|_2\leq 1}\|[f(y)-f(y')]z\|_2\leq \|f(y)-f(y')\|_{\infty}$$ (Is there a better way than using the $L^{\infty}(0,1) ?$ Now we need to find $M>0$ such that $$\|f(y)-f(y')\|_{\infty} \leq M \|y-y'\|_2$$