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Could someone help me prove that given a measure space $(X, \mathcal{M}, \sigma)$ and a measurable function $f:X\to\mathbb{R}$ in $L^{\infty}$ and some $L^{q}$, $\lim_{p\to\infty}||f||_p=||f||_\infty$? I dont know where to start.

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Why are you taking the limit as p goes to infinity? (i.e. what is the motivation?) I've often seen people use the limit as p goes to 1, since certain optimizations aren't unique in taxicab space. –  Ryan Nov 22 '12 at 20:10
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Its an exercise in a book I'm reading. I don't have any real motivation, except maybe to justify the definition of the $L^{\infty}$ norm. –  Parakee Nov 22 '12 at 20:19
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up vote 18 down vote accepted

Fix $\delta>0$ and let $S_\delta:=\{x,|f(x)|\geqslant \lVert f\rVert_\infty-\delta\}$ for $\delta<\lVert f\rVert_\infty$. We have $$\lVert f\rVert_p\geqslant \left(\int_{S_\delta}(\lVert f\rVert_\infty-\delta)^pd\mu\right)^{1/p}=(\lVert f\rVert_\infty-\delta)\mu(S_\delta)^{1/p},$$ since $\mu(S_\delta)$ is finite and positive. This gives $$\liminf_{p\to +\infty}\lVert f\rVert_p\geqslant\lVert f\rVert_\infty.$$ As $|f(x)|\leqslant\lVert f\rVert_\infty$ for almost every $x$, we have for $p>q$, $$ \lVert f\rVert_p\leqslant\left(\int_X|f(x)|^{p-q}|f(x)|^qd\mu\right)^{1/p}\leqslant \lVert f\rVert_\infty^{\frac{p-q}p}\lVert f\rVert_q^{q/p},$$ giving the reverse inequality.

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How does that last step give us the reverse inequality? –  Parakee Nov 22 '12 at 20:27
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Take this time $\limsup_{p\to \infty}$. –  Davide Giraudo Nov 22 '12 at 20:28
    
Very nice! Thank you! –  Bombyx mori Dec 24 '12 at 19:21
    
@DavideGiraudo could you elaborate on just how the limit superior of the right side of your last inequality ends up being $||f||_\infty$? –  alonso s Oct 18 '13 at 11:29
    
When $p\to \infty$, $\frac{p-q}p\to 1$ and $q/p\to 0$. –  Davide Giraudo Oct 18 '13 at 11:31
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