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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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    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.2012-11-22
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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.2012-11-22

2 Answers 2

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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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    How does that last step give us the reverse inequality?2012-11-22
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    Take this time $\limsup_{p\to \infty}$.2012-11-22
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    @DavideGiraudo could you elaborate on just how the limit superior of the right side of your last inequality ends up being $||f||_\infty$?2013-10-18
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    When $p\to \infty$, $\frac{p-q}p\to 1$ and $q/p\to 0$.2013-10-18
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    @DavideGiraudo Nice Proof. I could prove by myself that $limsup_{p\to \infty} |f|_p \le |f|_{\infty}$. I understand the proof that shows $\liminf_{p\to +\infty}\lVert f\rVert_p\geqslant\lVert f\rVert_\infty.$ but I don't understand the Idea behind it, do you know it? Could you explain it?2014-01-23
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    Assume that we work on a probability space. Then $\lVert f\rVert_p\leqslant \lVert f\rVert_\infty$, and we are interested about a control of $\lVert f\rVert_\infty$ in terms of the $L^p$ norm. We thus fix an arbitrary small real number and look at the set where $|f|$ is greater than the supremum norm minus the small number.2014-01-23
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    @DavideGiraudo Why is $\mu(S_\delta)>0$?2014-04-08
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    We use the definition of the essential supremum. (we probably have to distinguish with the case $f\equiv 0$).2014-04-08
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    @DavideGiraudo Hi Davide, in your proof, can you help me to understand how we can conclude that $\text{lim inf}_{p \to \infty} || f ||_p \geq || f ||_\infty$. I'm not sure how to obtain that part. Thanks!2014-04-14
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    Doesn't your proof assume $\mu(X)<\infty$?2014-04-22
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    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.2014-04-23
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    I know this is an old question, but I'm confused on your last step, hoping you can explain - if $|f| \le ||f||_\infty$ a.e., is there a problem with writing $||f||_p = (\int |f(x)|^p)^{1/p} \le (\int ||f||_\infty^p)^{1/p} = ||f||_\infty$, and taking the limit there?2015-08-18
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    @DavideGiraudo oops, meant to write $(\int ||f||^p_\infty)^{1/p} = ||f||_\infty \mu(X)^{1/p}$ there2015-08-18
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    @gesa Your inequality $\lVert \rVert_p\leqslant \mu(X)^{1/p}\lVert f\rVert_\infty$ is correct. Maybe then it is preferable to take the $\limsup_{p\to \infty}$ instead of the limit since we do not know yet that the limit exists.2015-08-18
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    @DavideGiraudo right, I should have said $\limsup$. Thanks for confirming this! Just wanted to make sure I hadn't made a huge logical error or something.2015-08-18
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    @DavideGiraudo:How does $\delta $ vanished in $\liminf_{p\to +\infty}\lVert f\rVert_p\geqslant\lVert f\rVert_\infty.$2017-04-30
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    @PKStyles: $\delta$ does not "vanish" under the application of $\liminf$. The immediate statement is that $\liminf ||f||_p \ge ||f||_\infty - \delta $, and since this is true for arbitrarily small $\delta$, it is true for $\delta = 0$.2017-12-18
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    @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 $\mu(S_\delta)<+\infty$.2018-08-02
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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.