Q:

If  $f , g \in L^2 \cap L^\infty$, then $ \| fg \|_2 \leqslant c ( \|f \|_\infty + \| g \|_\infty ) $?

Prove that there exists $c >0$ such that for  $f , g \in L^2 ( \mathbb R) \cap L^\infty ( \mathbb R)$ , $$ \| fg \|_{L^2(\mathbb R)} \leqslant c ( \|f \|_\infty + \| g \|_\infty ). $$ 
Is this true? If so, would you give me a proof for this?

A:

Suppose it's true. Take $f$, $g$ such that $\| fg \|_{L^2(\mathbb R)}>0\;$. Then for any $a>0$ where whould hold the inequality
$$
a^2\| fg \|_{L^2(\mathbb R)} \leqslant c a ( \|f \|_\infty + \| g \|_\infty ).
$$
A contradiction.

A:

More generally, we can't find a function $F\colon \Bbb R^2\to \Bbb R$ such that for all $f,g\in L^2(\Bbb R)\cap L^{\infty}(\Bbb R)$, $$\lVert fg\rVert_{L^2(\Bbb R)}\leq F(\lVert f\rVert_{\infty},\lVert g\rVert_{\infty}).$$
Indeed, consider for a fixed $n$, $f_n=g_n=\chi_{(0,n)}\in L^2\cap L^{\infty}$. Then 
$$\lVert f_ng_n\rVert_{L^2}=\sqrt n,\quad \lVert f_n\rVert_{\infty}=\lVert g_n\rVert_{\infty}=1,$$
and we would have for each integer $n$,
$$\sqrt n\leq F(1,1),$$
which is impossible. 

