# 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 '12 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. – Parakee Nov 22 '12 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.) – Hrit Roy May 26 '19 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, $f:\mathbb R\to\mathbb C,f(x)=1$. – fonini Oct 2 '19 at 14:42
• @fonini Right. I saw the accepted answer and thought that we were assuming it to be finite – Hrit Roy Oct 4 '19 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.

• How does that last step give us the reverse inequality? – Parakee Nov 22 '12 at 20:27
• Take this time $\limsup_{p\to \infty}$. – Davide Giraudo Nov 22 '12 at 20:28
• Doesn't your proof assume $\mu(X)<\infty$? – Eric Auld Apr 22 '14 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. – Davide Giraudo Apr 23 '14 at 8:17
• @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 2 '18 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.