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Given that $f$ and $g$ are essentially bounded, show that their sum $f+g$ is also essentially bounded and furthermore show that the triangle inequality holds $||f+g||_{\infty}\leq ||f||_{\infty}+||g||_{\infty}$, where the norm in question is the $L_{\infty}$ norm. [The hint provided is that it is enough to show the inequality holds almost everywhere, which makes sense because equality in the $L_{\infty}$ norm can differ at most by a set of measure zero.]

I know that the definition of an essentially bounded function $f:E\to\overline{\mathbb{R}}$ is essentially bounded if $\exists A\in [0,\infty)$, where $|f|\leq A$ almost everywhere.

I know how to show the triangle inequlaity using Holder's inequality for $p-$norms, where $p<\infty$, but I'm not sure how to show it when $p=\infty$

José Carlos Santos
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Marshant
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  • The results for finite $p$ might actually be a distraction rather than help here. I would recommend first proving that, on the space of continuous functions $f$ (say, on $[0, 1]$), the supremum of $|f|$ (in fact, it turns out to be a max) is a norm on that space.

    To address $|| \cdot ||_{\infty}$ (i.e., to handle the behavior of a function on sets of small measure), you have to go from supremum to "essential supremum".

    http://math.stackexchange.com/questions/462006/essential-supremum-vs-uniform-norm

     – avs Aug 08 '16 at 17:46

2 Answers2

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So $|f(x)|\le \|f\|_\infty$ for a.e. $x$, and $|g(x)|\le\|g\|_\infty$ for a.e. $x$. Let $A$ be the set where the first inequality holds and $B$ the set where the second inequality holds. Then the complement of $A\cap B$ has measure $0$. Try to estimate $|f(x)+g(x)|$ for $x\in A\cap B$.

John Dawkins
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It's not as complicated as Hoelder's inequality. Start by showing $\sup_\alpha{x_\alpha+y_\alpha}\leq \sup_\alpha x_\alpha + \sup_\alpha y_\alpha$ for any set of $x_\alpha$ and $y_\alpha$. Then incorporate the "almost everywhere" part.

mathematician
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