# Limit of $L^p$ norm

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
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

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
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
Doesn't your proof assume $\mu(X)<\infty$? –  Eric Auld Apr 22 at 22:24

Let $f:X\to \mathbb{R}$. Assume that $f$ is measurable, and that $\|f\|_p<\infty$ for all large $p$. Suppose for convenience that $f\geq 0$. (If not, just work with $f^*:=|f|$.) We define $$\|f\|_{\infty}:=\sup \{r\in \mathbb{R}: \mu\left( \{x:|f(x)|\geq r\} \right)>0\}.$$

If $\|f\|_{\infty}=0$, we can see that the proposition holds trivially. If $\|f\|_{\infty}\neq 0$, let $M:=\|f\|_{\infty}$.

Fix $\epsilon$ such that $0< \epsilon < M$. Define $D:=\{x:f(x)\geq M-\epsilon\}$. Observe that $\mu(D)>0$ by definition of $\|f\|_{\infty}$. Also, $\mu(D)<\infty$ since $f$ is integrable for all large $p$. Now we can establish $\liminf_{p\to\infty }\|f\|_p\geq M-\epsilon$ by $$\left( \int_{X}f(x)^p dx \right)^{1/p} \geq \left( \int_D (M-\epsilon)^pdx \right)^{1/p} = (M-\epsilon)\mu(D)^{1/p} \xrightarrow{p\to\infty}(M-\epsilon)$$

Now we show $\limsup_{p\to\infty}\|f\|_{p} \leq M+\epsilon$. Let $\tilde{f}(x) := \dfrac{f(x)}{M+\epsilon}$. Observe that $0\leq \tilde{f}(x)\leq M/(M+\epsilon)<1$, and that $$\left( \int_{X} f(x)^p dx \right)^{1/p} = (M+\epsilon)\left( \int_{X} \tilde{f}(x)^p dx \right)^{1/p}.$$

Now it suffices to show that $\int_X \tilde{f}(x)^p dx$ is bounded above by $1$ as $p\to \infty$, since then we have

$$\left( \int_{X} f(x)^p dx \right)^{1/p} = (M+\epsilon)\left( \int_{X} \tilde{f}(x)^p dx \right)^{1/p} \leq M+\epsilon.$$

But observe that $$\int_{X} f(x)^{a+b} dx = \int_{X} f(x)^{a}f(x)^b dx$$ $$\leq \int_{X} f(x)^{a} \left(\frac{M}{M+\epsilon}\right) ^b dx = \left(\frac{M}{M+\epsilon}\right)^b \int_{X} f(x)^{a} dx.$$
Therefore $\int_{X} f(x)^{p} dx$ will eventually be less than one. This shows $\limsup_{p\to\infty}\|f\|_{p} \leq M+\epsilon$ and completes the proof.

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