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?
If $f , g \in L^2 \cap L^\infty$, then $ \| fg \|_2 \leqslant c ( \|f \|_\infty + \| g \|_\infty ) $?
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real-analysis
functional-analysis
inequality
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0It should be true for $f=g\neq 0$ hence $\lVert f^2\rVert_{L^2}\leq c'\lVert f\rVert_{\infty}$. But it's not homogeneous (as andrew did, replace $f$ by $af$, where $a$ is a real number). – 2012-07-08
2 Answers
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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.
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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.