$\|fg\|_{L^2(\Omega)}\leq \|f\|_{L^2(\Omega)}\|g\|_{L^\infty(\Omega)}$

Question

is true that $\|fg\|_{L^2(\Omega)}\leq \|f\|_{L^2(\Omega)}\|g\|_{L^{\color{blue}\infty}(\Omega)}$ ?

I can't see a proof for this :/

( of course, $\|fg\|_{L^2(\Omega)}\leq \|f\|_{L^2(\Omega)}\|g\|_{L^{\color{blue}2}(\Omega)}$ )

I'm using the usual notation for $L^p(\Omega)$ spaces (like $L^p$ space in wikipedia)

Answer 1

The inequality you are having trouble with is actually trivial. The "of course" one is actually false. $$\|fg\|_2^2 = \int_\Omega |fg|^2 \, dx \le \|g\|_\infty^2 \int_\Omega |f|^2 \, dx = \|f\|_2^2 \|g\|_\infty^2.$$

On the other hand, if $\Omega = (0,1)$ and $f(x) = g(x) = x^{-1/3}$, you have $\|fg\|_2 = \infty$ but $\|f\|_2 = \|g\|_2 < \infty$.

Answer 2

Yes, in fact if $p \ge 1$, $f \in L^p(\Omega)$ and $g \in L^\infty(\Omega)$, then $\|fg\|_p \le \|f\|_p \|g\|_\infty$. For $|f(x)g(x)|\le |f(x)|\|g\|_{\infty}$ for almost every $x\in \Omega$, so

$$\|fg\|^p_p = \int_\Omega |fg|^p \le \int_\Omega |f|^p \|g\|_\infty^p = \|f\|_p^p\, \|g\|_\infty^p,$$

which implies $\|fg\|_p \le \|f\|_p\|g\|_\infty$.

Answer 3

Yes, because $g(x) \leq ||g||_{\infty}$ for a.e $x$, so $$||fg||_2 = \bigg( \int |f(x)|^2|g(x)|^2 d\mu(x) \bigg)^{\frac{1}{2}} \leq \bigg( \int |f(x)|^2||g||_{\infty}^2 d\mu(x) \bigg)^{\frac{1}{2}}=\bigg( ||g||_{\infty}^2 \int |f(x)|^2 d\mu(x) \bigg)^{\frac{1}{2}}=||g||_{\infty} \bigg( \int |f(x)|^2 d\mu(x) \bigg)^{\frac{1}{2}}=||g||_{\infty} ||f||_2$$