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.

  • 2
    $\begingroup$ 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
    Commented Nov 22, 2012 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
    Commented Nov 22, 2012 at 20:19
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    $\begingroup$ 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.) $\endgroup$
    – Not Euler
    Commented May 26, 2019 at 12:58
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    $\begingroup$ @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
    Commented Oct 2, 2019 at 14:42
  • $\begingroup$ @fonini Right. I saw the accepted answer and thought that we were assuming it to be finite $\endgroup$
    – Not Euler
    Commented Oct 4, 2019 at 18:51

4 Answers 4


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.

  • 13
    $\begingroup$ Take this time $\limsup_{p\to \infty}$. $\endgroup$ Commented Nov 22, 2012 at 20:28
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    $\begingroup$ Doesn't your proof assume $\mu(X)<\infty$? $\endgroup$
    – Eric Auld
    Commented Apr 22, 2014 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. $\endgroup$ Commented Apr 23, 2014 at 8:17
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    $\begingroup$ @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 $\delta = 0$. $\endgroup$
    – jawheele
    Commented Dec 18, 2017 at 8:25
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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
    Commented Aug 2, 2018 at 17:47

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.


I have found below simple proof.

Let $(\Omega, \mathcal F, \mu)$ be a finite measure space. If $f \in L^\infty (\Omega)$ then $\lim_{p \to \infty} \|f\|_p = \|f\|_\infty$.

We have $|f| \le \|f\|_\infty$ a.s., so $$ \|f\|_p \le (\mu(\Omega))^{1/p} \|f\|_\infty \quad \forall p \in [1, \infty). $$

Then $$ \limsup_{p \to \infty} \|f\|_p \le \limsup_{p \to \infty} \big ((\mu(\Omega))^{1/p} \|f\|_\infty \big ) = \|f\|_\infty. $$

Fix $r < \|f\|_\infty$ and let $C:=\{f>r\} \in \mathcal F$. Then $\mu(C)>0$ and $$ \|f\|_p \ge (\mu(C))^{1/p} r \quad \forall p \in [1, \infty). $$

Then $$ \liminf_{p \to \infty} \|f\|_p \ge \liminf_{p \to \infty} \big ((\mu(C))^{1/p} r \big ) = r. $$

Then $$ \liminf_{p \to \infty} \|f\|_p \ge \|f\|_\infty. $$

This completes the proof.

  • 1
    $\begingroup$ This is the same as @David's answer. $\endgroup$
    – Kroki
    Commented Jun 6, 2023 at 15:36

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.

  • $\begingroup$ choice of M depends on q. I don't think you can take limits in q in the penultimate line. $\endgroup$
    – vishmay
    Commented Apr 10, 2023 at 22:21
  • $\begingroup$ I think this proof is fine. It suffices to generalize the penultimate line for 𝑘 large enough, for which the 𝜖/2 requirement for the $L_\infty$ norm's triangle inequality already holds. However, compared to Davide's more measure-theoretical approach, this proof is just a more tedious constructive version of showing that the essential supremum clearly dominates $L_p$ norm by drowning out the other values $f$ may take almost everywhere as $p\to\infty$. $\endgroup$
    – mtcicero
    Commented Mar 5 at 19:32

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