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Let $L^{\infty}$ denote the set of all essentially bounded functions and suppose that $f \in L^{\infty}$ and $g \in L^{\infty}$. Then the product $fg$ is in $L^{\infty}$. Now the question is: prove or disprove: $\|fg\|_\infty \leq \|f\|_\infty \|g\|_\infty$.

Any hints?

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Ignore the set of measure zero on which f is unbounded, then ignore the set of measure zero on which g is unbounded. – Qiaochu Yuan Nov 10 '10 at 2:15

What Qiaochu said, and use the definition of the $L^\infty$-norm along with a similar result regarding absolute values.

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Thanks. elementary question: if f,g are bounded functions then $||fg||_{\sup} = ||f||_{\sup} ||g||_{\sup}$ ? Here sup denotes the supremum norm (not the essential norm) If this is true I got it. – student Nov 10 '10 at 2:38
@Nicolas: No, they need not be equal. Take $f$ to be the characteristic function of $[0,1]$ and $g$ to be the characteristic function of $[2,3]$ in the real line. Si $||fg||_{\mathrm{sup}} = ||f||_{\mathrm{sup}}||g||_{\mathrm{sup}}$? On the other hand, there is something you can say to relate $||fg||_{\sup}$ with $||f||_{\sup}$ and $||g||_{\sup}$. – Arturo Magidin Nov 10 '10 at 2:40
Aha, thanks to all! – student Nov 10 '10 at 2:54

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