179
$\begingroup$

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.

$\endgroup$
5
  • 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
    Nov 22, 2012 at 20:10
  • 18
    $\begingroup$ 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
    Nov 22, 2012 at 20:19
  • 3
    $\begingroup$ 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.) $\endgroup$
    – Hrit Roy
    May 26, 2019 at 12:58
  • 1
    $\begingroup$ @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
    Oct 2, 2019 at 14:42
  • $\begingroup$ @fonini Right. I saw the accepted answer and thought that we were assuming it to be finite $\endgroup$
    – Hrit Roy
    Oct 4, 2019 at 18:51

3 Answers 3

179
$\begingroup$

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.

$\endgroup$
27
  • 11
    $\begingroup$ Take this time $\limsup_{p\to \infty}$. $\endgroup$ Nov 22, 2012 at 20:28
  • 9
    $\begingroup$ Doesn't your proof assume $\mu(X)<\infty$? $\endgroup$
    – Eric Auld
    Apr 22, 2014 at 22:24
  • 9
    $\begingroup$ 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. $\endgroup$ Apr 23, 2014 at 8:17
  • 2
    $\begingroup$ @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
    Dec 18, 2017 at 8:25
  • 9
    $\begingroup$ @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
    Aug 2, 2018 at 17:47
36
$\begingroup$

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.

$\endgroup$
0
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.