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. – 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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    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.) – Hrit Roy May 26 '19 at 12:58
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    @HritRoy You are assuming $\mu(X)<\infty$. If you don't limit the size of $\mu$, then there's a simple counterexample, $f:\mathbb R\to\mathbb C,f(x)=1$. – fonini Oct 02 '19 at 14:42
  • @fonini Right. I saw the accepted answer and thought that we were assuming it to be finite – Hrit Roy Oct 04 '19 at 18:51

3 Answers3


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.

Davide Giraudo
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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
  • @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
  • @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? – Bman72 Jan 23 '14 at 16:12
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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. – Davide Giraudo Jan 23 '14 at 21:40
  • @DavideGiraudo Why is $\mu(S_\delta)>0$? – Twink Apr 08 '14 at 03:26
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    We use the definition of the essential supremum. (we probably have to distinguish with the case $f\equiv 0$). – Davide Giraudo Apr 08 '14 at 09:06
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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! – Bachmaninoff Apr 14 '14 at 22:28
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    Doesn't your proof assume $\mu(X)<\infty$? – Eric Auld Apr 22 '14 at 22:24
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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. – Davide Giraudo Apr 23 '14 at 08:17
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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? – poppy3345 Aug 18 '15 at 04:03
  • @DavideGiraudo oops, meant to write $(\int ||f||^p_\infty)^{1/p} = ||f||_\infty \mu(X)^{1/p}$ there – poppy3345 Aug 18 '15 at 04:41
  • @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. – Davide Giraudo Aug 18 '15 at 07:54
  • @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. – poppy3345 Aug 18 '15 at 19:29
  • @DavideGiraudo:How does $\delta $ vanished in $\liminf_{p\to +\infty}\lVert f\rVert_p\geqslant\lVert f\rVert_\infty.$ – P.Styles Apr 30 '17 at 18:44
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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 $\delta = 0$. – jawheele Dec 18 '17 at 08:25
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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 $\mu(S_\delta)<+\infty$. – RozaTh Aug 02 '18 at 17:47
  • @DavideGiraudo I agree with Twink. I don't see why $\mu (S_\delta) > 0$. The norm $||f||_\infty$ is defined to be the infimum of all essential upper bounds of $f$, and I don't see any contradiction in assuming that $||f||_\infty - \delta$ is an essential upper bound. If it were an essential upper, then $||f||_\infty - \delta < ||f||_\infty$, but where's the contradiction in that? – user193319 Apr 28 '19 at 15:41
  • This is an old question, but I want to highlight that you need the assumption that $\mu(\Omega)$ is finite. Here is an example where the inequality doesnt hold when this is violated: Assume $\mu$ is a Lebesgue-Measure on $\mathscr{B}(\mathbb{R})$ and $f(x) = 1$ for $n \le x \le n + 1/n$ and $0$ elsewhere, then $\|f\|_p=\infty$ for $p < \infty$ but $\|f\|_{\infty} = 1$ – Antonio Horta Ribeiro Nov 11 '19 at 12:56
  • Also, the question asked here is similar to exercise 9 on Section 2-4 of Robert B. Ash "Probability and Measure Theory" 2nd edition. The solution is given on the solution manual (which appear on page 470 of the book, including the counter example I mentioned above). I would recommend to anyone interested in this question to take a look over there... – Antonio Horta Ribeiro Nov 11 '19 at 13:01
  • @AntonioHortaRibeiro the hypothesis explicitly states that $f$ belongs to $L^\infty$ and some $L^q$, and implicitly implies $q < \infty$. – Umberto P. Jan 05 '20 at 19:45
  • Thank you for pointing that out @UmbertoP. I missed that when reading the question statement. – Antonio Horta Ribeiro Jan 06 '20 at 12:47
  • @DavideGiraudo Studying this myself and while I think there are a few details to fill in, this solution and its comments were very elucidating. – Antoine Love Jun 01 '20 at 15:49
  • Very Beautiful proof – Rabee Tourky Aug 11 '20 at 07:47
  • @DavideGiraudo - Quick question: Why are these two expressions equal in $(\int_{S_{\delta}} (||f|| - \delta)^p_\infty d\mu)^{1/p} = (||f||_\infty - \delta) \mu(S_{\delta})^{1/p}$? – Taylor Rendon Apr 02 '21 at 21:03
  • @TaylorRendon You integrate a constant function over $S_\delta$. – Davide Giraudo Apr 03 '21 at 09:59

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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Eric Auld
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Possibly another solution. Please let me know if there is an error.

In the following, we assume that $f \in L^p \cap L^{\infty}$ and thus $f\in L^q$ for all $q \in [p,\infty]$. The assumption of finite measure is not used.

First we show the result for simple functions. Suppose $f$ is simple such that $f \in L^q$ for $q \in [p,\infty]$. Let the standard representation of $f$ be, \begin{align} f = \sum_{1}^{n}a_j\chi_{E_j}. \end{align} Since $f \in L^q$, $\mu(E_j) < \infty$ for all $j \in \{1,\ldots,n\}$. Without loss of generality, assume that $a_j \neq 0$ and $\mu(E_j)>0$ for all $j$. Also, assume that $|a_j| \leq |a_n|$ for all $j$ and denote by $\eta = \sum_{1}^{n}\mu(E_j)$ (note that $0 < \eta < \infty$). Then, \begin{align} \left\|f\right\|_q &= \left(\sum_{1}^{n}|a_j|^q\mu(E_j)\right)^{1/q}\\ &= |a_n|\eta^{1/q}\left(\frac{\mu(E_n)}{\eta} + \sum_{1}^{n-1}\frac{|a_j|^q\mu(E_j)}{|a_n|^q\eta}\right)^{1/q}\\ \lim\inf_{q\rightarrow \infty} \left\|f\right\|_q &\leq \lim\inf_{q\rightarrow \infty}|a_n|(n\eta)^{1/q} = |a_n| = \left\|f\right\|_{\infty} \end{align} Also, since $|a_j|^q\mu(E_j) \leq \left\|f\right\|^q_q \implies |a_j|\mu(E_j)^{1/q} \leq \left\|f\right\|^q$ for all $j$, therefore, $\left\|f\right\|_{\infty} = |a_n| = \lim\sup_{q\rightarrow \infty}|a_n|\mu(E_n)^{1/q} \leq \lim\sup_{q\rightarrow \infty}\left\|f\right\|_q$. This concludes the result for the simple functions.

Now, let $f \in L^p \cap L^{\infty}$ be an arbitrary measurable functions. Then there exists sequence of simple functions $\{\phi_k\}$ such that $|\phi_1\| \leq \ldots \leq |f|$, $\phi_i \rightarrow f$. Let $q > \max\{p, 1\}$. Given $\epsilon > 0$, there exist $M, N \in \mathbb{N}$ such that for all $m \geq M$ and for all $n \geq N$ we have, \begin{align} \left\|f-\phi_m\right\|_q \leq \epsilon/2 \text{ and } \left\|f-\phi_n\right\|_{\infty} \leq \epsilon/2. \end{align} Since $\left\|\cdot \right\|_q$ and $\left\|\cdot \right\|_{\infty}$ are norms, by triangle inequality, we also have, \begin{align} |\left\|f\right\|_q-\left\|\phi_m\right\|_q| \leq \left\|f-\phi_m\right\|_q \text{ and } |\left\|f\right\|_\infty-\left\|\phi_n\right\|_\infty| \leq \left\|f-\phi_n\right\|_\infty. \label{folland7_1} \end{align} Now, take $K = \max\{M,N\}$, then for all $k \geq K$, we have, \begin{align} \left\|\phi_k\right\|_{q} - \epsilon/2 \leq \left\|f\right\|_{q} \leq \left\|\phi_k\right\|_{q} + \epsilon/2. \end{align} Take limit $q \rightarrow \infty$ (and use the above result for simple functions) to obtain, \begin{align} \left\|\phi_k\right\|_{\infty} - \epsilon/2 \leq \lim_{q\rightarrow \infty}\left\|f\right\|_{q} \leq \left\|\phi_k\right\|_{\infty} + \epsilon/2 \end{align} and then, \begin{align} \left\|f\right\|_{\infty} - \epsilon \leq \lim_{q\rightarrow \infty}\left\|f\right\|_{q} \leq \left\|f\right\|_{\infty} + \epsilon. \end{align} Since $\epsilon > 0$ is arbitrary, we conclude the result.

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