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

I don't know where to start.

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    $\begingroup$ 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. $\endgroup$ – Ryan Nov 22 '12 at 20:10
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    $\begingroup$ 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. $\endgroup$ – Parakee Nov 22 '12 at 20:19
  • $\begingroup$ But why do we need the condition $f\in L_q?$ If $f\in L_{\infty}$ then $|f(x)|\leq ||f||_{\infty}$ for almost all $x$; so we can say $|f(x)|\leq ||f||_{\infty}$ for all $x\in N^c$ with $\mu(N)=0$. Then $\int |f|^pd\mu = \int_{N^c} |f|^p d\mu \leq ||f||_{\infty}^p\mu(X) < \infty$. So that $f\in L_p$ for all $1\leq p<\infty$. (Correct me if I am wrong.) $\endgroup$ – Hrit Roy May 26 at 12:58
  • $\begingroup$ @HritRoy You are assuming $\mu(X)<\infty$. If you don't limit the size of $\mu$, then there's a simple counterexample, $f:\mathbb R\to\mathbb C,f(x)=1$. $\endgroup$ – fonini Oct 2 at 14:42
  • $\begingroup$ @fonini Right. I saw the accepted answer and thought that we were assuming it to be finite $\endgroup$ – Hrit Roy Oct 4 at 18:51
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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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    $\begingroup$ How does that last step give us the reverse inequality? $\endgroup$ – Parakee Nov 22 '12 at 20:27
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    $\begingroup$ Take this time $\limsup_{p\to \infty}$. $\endgroup$ – Davide Giraudo Nov 22 '12 at 20:28
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    $\begingroup$ Doesn't your proof assume $\mu(X)<\infty$? $\endgroup$ – Eric Auld Apr 22 '14 at 22:24
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    $\begingroup$ Yes. However, I think we can get rid of this assumption: we only have to consider the case where the measure space is $\sigma$-finite, hence $X=\bigcup_n A_n$ where $A_n\uparrow X$ and each $A_n$ is of finite measure. Then we pick $n$ such that $\mu(A_n\cap S_{\delta})$ is a positive real number. $\endgroup$ – Davide Giraudo Apr 23 '14 at 8:17
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    $\begingroup$ @DavideGiraudo: Correct me if I am wrong pls, but I don't think we need the assumption of finite or even $\sigma-$finite measure. $f$ is in some $L^q$ so the given inequality holds for $p=q$ and $$+\infty>\lVert f\rVert_q\geqslant (\lVert f\rVert_\infty-\delta)\mu(S_\delta)^{1/q},$$ so $\mu(S_\delta)<+\infty$. $\endgroup$ – RozaTh Aug 2 '18 at 17:47
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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 $$ \|f\|_{\infty}:=\sup \{r\in \mathbb{R}: \mu\left( \{x:|f(x)|\geq r\} \right)>0\}. $$

I claim without proof that $\|f\|_p < \infty$ for large $p$ implies that $\|f\|_\infty < \infty$.

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 $\displaystyle\limsup_{p\to\infty}\|f\|_{p} \leq M+\epsilon$ and completes the proof.

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