# Limit of $L^p$ norm

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.

• 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, 2012 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. Nov 22, 2012 at 20:19
• 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.) May 26, 2019 at 12:58
• @HritRoy You are assuming $\mu(X)<\infty$. If you don't limit the size of $\mu$, then there's a simple counterexample, $f:\mathbb R\to\mathbb C,f(x)=1$. Oct 2, 2019 at 14:42
• @fonini Right. I saw the accepted answer and thought that we were assuming it to be finite Oct 4, 2019 at 18:51

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.

• Take this time $\limsup_{p\to \infty}$. Nov 22, 2012 at 20:28
• Doesn't your proof assume $\mu(X)<\infty$? Apr 22, 2014 at 22:24
• 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. Apr 23, 2014 at 8:17
• @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 $\delta = 0$. Dec 18, 2017 at 8:25
• @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 $\mu(S_\delta)<+\infty$. 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.

• This is the same as @David's answer. Jun 6 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.

• choice of M depends on q. I don't think you can take limits in q in the penultimate line. Apr 10 at 22:21