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Suppose $f$ is a complex measurable function on $X$, $\mu$ is a positive measure on $X$, and $$\varphi(p) ~=~ \int_X |f|^p \; d\mu \quad (0 < p < \infty)$$ Let $E :=\{ p : \varphi(p) < \infty\}$. Assume $\|f\|_\infty > 0$. Prove that $\log\varphi$ is convex in the interior of $E$ and that $\varphi$ is continuous on $E$.

I have proved that $\log\varphi$ is convex on $\operatorname{int}E$, using Holder's inequality. We know that log-convexity implies convexity, so $\varphi$ is also convex on $\operatorname{int}E$, and hence continuous on $\operatorname{int}E$. How do we establish the continuity of $\varphi$ on $E$, though? I am missing exactly the points in $E\setminus\operatorname{int}E$.

I know this question has been previously answered here, but I'm unable to understand the proof. Moreover, I am trying to take a different approach to prove continuity. Let $\{p_n\}_{n\in\mathbb N}$ be a sequence in $E$, such that $p_n\to p$ (may or may not be in $E$ - we don't know if $E$ is closed, right?). I wish to show $\varphi(p_n)\to \varphi(p)$, i.e. $$\lim_{n\to\infty} \int_X |f|^{p_n}\ d\mu = \int_X |f|^p \ d\mu$$ This looks like a possible application of Lebesgue's Dominated Convergence Theorem, but I could not find a dominating function yet.

I would appreciate any help, thanks a lot!

I have already seen this answer - it did not help much. The proof was too convoluted, and I needed more details. @OliverDiaz and @robjohn were able to provide convincing detailed arguments, hence this post.

Xander Henderson
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stoic-santiago
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    In $A={|f|>1}$, if $p,q\in E$ and $p<q$ you have $0\leq |f|^p\mathbb{1}_A\leq |f|^q\mathbb{1}_A$, and both $|f|^p\mathbb{1}_A$ and $|f|^q\mathbb{1}_A$ are integrable. Here you can apply monotone convergence as $q\nearrow r$. In $B=X\setminus A={|f|\leq 1}$, if $p,q\in E$ and $p<q$, then $|f|^q\mathbb{1}_B\leq |f|^p\mathbb{1}_B$ and both $0\leq |f|^q\mathbb{1}_B$ and $ |f|^p\mathbb{1}_B$ are integrable. Again, here you can apply monotone convergence (decreasing version) as $q\nearrow r$, or dominated convergence since $|f|^p\mathbb{1}_B$ dominates all $|f|^q\mathbb{1}_B$ with $E\ni q>p$. – Mittens Jun 03 '21 at 12:19
  • From this, you also see why we take powers $p$ in $E$. – Mittens Jun 03 '21 at 12:21
  • Yes, that clears it up @OliverDiaz. In fact, I found an alternative way to apply DCT without splitting the integral into two parts. For example, if $E = [a,\infty)$, and we are concerned about continuity at $r = a$. Then, fix some $b>a$ such that $b\in E$. Consider a sequence ${p_k}$ such that $p_k\to p$. Then, there exists $N\in\mathbb N$ such that for all $k > N$, $p_k < b$. Clearly, $|f|^{p_k} \le |f|^a + |f|^b$. So, ${|f|^{p_k}}_{k=N}^\infty$ satisfy the hypothesis of DCT, and the result follows. For all other cases (i.e. $E = [a,b], (a,b]$, etc.) similar reasoning applies. Looks good? – stoic-santiago Jun 03 '21 at 15:44
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    Yes, it seems to work. But don't get too caught up in getting the best dominating functions. The key idea here is that you can get to the endpoint power ($|f|^r$) monotonically by breaking the function in pieces, one where $f$ is small (${|f|\leq1}$) and another where $f$ is large (${|f|>1}$). This, though simple in appearance, will get you very far when you study convolution operators. Best.. – Mittens Jun 03 '21 at 16:26
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    Does this answer your question? Rudin's RCA; Exercise 3.4 Part (b) Only – amWhy Jun 16 '21 at 16:33
  • @amWhy I had seen that answer before - it did not help much. The proof was too convoluted, and I needed more details. Oliver and robjohn were able to provide convincing detailed arguments, hence this post. Please consider retracting the downvotes, I noticed this post has been attracting unnecessary downvotes. – stoic-santiago Jun 17 '21 at 11:09

2 Answers2

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Since $E$ is a convex subset of $[0,\infty)$, $E$ is an interval (possibly open, half open, or closed). If $E$ has no interior then either $E$ is empty or $E$ is a singleton $\{p\}$ (for example, for $f(x)=\frac{1}{x(1+|\log x|)^2}\mathbb{1}_{(0,\infty)}(x)$, and $(X,\mu)=(\mathbb{R},\lambda)$, then $E=\{1\}$). In this case there is nothing else to prove.

If $E$ has interior, then $E$ could be of the form $(a, b)$, $(a,b]$, $[a,b)$, $[a,b]$, $(a,\infty)$, or $[a,\infty)$, where $0\leq a<b<\infty$. Hence, if $r$ is an endpoint of the interval $E$, then either $r$ is finite or $r=\infty$.

Case $r$ is finite: consider the set $A=\{|f| > 1\}$. Then $|f|^p\xrightarrow{E\ni p\rightarrow r}|f|^r$ monotonically on $A$ and on $X\setminus A$. Applying monotone (or dominated convergence) gives \begin{align} \int_A|f|^p&\xrightarrow{E\ni p\rightarrow r}\int_A|f|^r\\ \int_{X\setminus A}|f|^p&\xrightarrow{E\ni p\rightarrow r}\int_{X\setminus A}|f|^r \end{align} From this, we obtain $\lim_{E\ni p\rightarrow r}\varphi(p)=\|f\|^r_r=\phi(r)$. Notice that if $r\notin E$, then $\varphi(r)=\infty$.

Case $r=\infty$: Although this case is not necessary to study continuity of $\varphi$ on $E$, a few observations can be made.

  • (1) $\lim_{p\rightarrow\infty}\|f\|_p=\lim_{p\rightarrow\infty}(\varphi(p))^{1/p} =\|f\|_\infty$; a proof can be found here.
  • (2) One can also see that $\lim_{p\rightarrow\infty}\varphi(p)$ exists in $[0,\infty]$. Consider $A$ as before and let $C=\{|f|<1\}$, and $p_0\in E$. Then $|f|^p\mathbb{1}_C\leq |f|^{p_0}\mathbb{1}_C$ for all $p>p_0$. Another application of monotone convergence and dominated convergence gives \begin{align} \lim_{p\rightarrow\infty}\int_X |f|^p\,d\mu&=\lim_{p\rightarrow\infty}\Big(\int_C|f|^p\,d\mu + \mu(|f|=1) +\int_A|f|^p\,d\mu\Big)\\ &=\mu(|f|=1)+\infty\cdot\mu(\{|f|>1\})\end{align} If $\mu(|f|>1)>0$, then $\|f\|_\infty>1$ and $\lim\limits_{E\ni p\rightarrow\infty}\varphi(p)=\infty$. If $\mu(|f|>1)=0$, then $\lim\limits_{E\ni p\rightarrow\infty}\varphi(p)=\mu(|f|=1)$.

Notice that unless $\|f\|_\infty<1$ or $\|f\|_\infty>1$, $\varphi(\infty)$ is not well defined.

Mittens
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  • Just one clarification to make, is it okay if I use the same ideas as yours but consider a sequence ${p_n}\subset E$ instead? I can still use MCT and DCT for the finite cases on $A$ and $X\setminus A$ respectively. Also the proof for $r = \infty$ should be the same. I ask this because my original attempt was to use the definition of sequential continuity to show continuity of $\varphi$. Thank you! – stoic-santiago Jun 02 '21 at 16:22
  • In the $r = \infty$ case, you comment that "this is not necessary to study the continuity of $\varphi$ on $E$" - why is it not necessary? It's one of the two possible cases. – stoic-santiago Jun 02 '21 at 18:16
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    Because $\varphi$ is continuous on $E={0<p<\infty: \varphi(p)<\infty}$. $E$ is a subset of $[0,\infty)$ (not of $[0,\infty]$). Also, notice that $\varphi$ is not defined at $p=\infty$. – Mittens Jun 02 '21 at 18:20
  • Apologies for missing this earlier, but how do the following translate to continuity? \begin{align} \int_A|f|^p&\xrightarrow{E\ni p\rightarrow r}\int_A|f|^r\ \int_{X\setminus A}|f|^p&\xrightarrow{E\ni p\rightarrow r}\int_{X\setminus A}|f|^r \end{align} i.e. how are they the same as saying: for every sequence $p_n\to p$, we have $\varphi(p_n)\to\varphi(p)$? Just to clarify, we are proving continuity at the endpoints only, right? Continuity over rest of the interval follows from the argument I have presented in the post? – stoic-santiago Jun 02 '21 at 19:31
  • Also, $r\in E$ is obvious since you mention it is an endpoint of $E$. "Notice that it could well happen that $\int_X |f|^r = \infty$..." - No, this cannot happen since $r\in E$. – stoic-santiago Jun 02 '21 at 19:59
  • Could you please clarify why it suffices to only consider elements $p\in E$? In general, sequential continuity requires that for any sequence $(p_n)\to p$, we have $\varphi(p_n)\to\varphi(p)$. Shouldn't we be working with $\lim_{p\to r}$ in place of $\lim_{E\ni p\to r}$? Also, I'm unable to see or locate the chat that was created by you a few hours ago, and that comment appears to be deleted as well. Sorry for the late response, but it was very late at night where I'm at. – stoic-santiago Jun 03 '21 at 05:31
  • What's wrong with \begin{align} \lim_{p\rightarrow r}\int_A|f|^p&\longrightarrow \int_A|f|^r\ \lim_{p\rightarrow r}\int_{X\setminus A}|f|^p&\longrightarrow\int_{X\setminus A}|f|^r \end{align} – stoic-santiago Jun 03 '21 at 05:41
  • It is clear, but I have trouble seeing what dominating function to choose in different cases (assuming all your conclusions follow from DCT). – stoic-santiago Jun 03 '21 at 11:07
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Since $\varphi$ is log-convex, $E$ is an interval.

Least Upper Bound is Included

Suppose that $b$ is the least upper bound of $E$. If $b\not\in E$, there is nothing to prove, so suppose that $b\in E$. Let $b_n\in E$ and $b_n\to b$ monotonically (increasing).

Define $$ Y=\{x\in X:|f(x)|\le1\} $$ Since $b_n\le b$, $$ \int_{X\setminus Y}|f(x)|^{b_n}\,\mathrm{d}\mu\le\int_{X\setminus Y}|f(x)|^b\,\mathrm{d}\mu\lt\infty $$ and since $b_1\le b_n\le b$, $$ \int_Y|f(x)|^b\,\mathrm{d}\mu\le\int_Y|f(x)|^{b_n}\,\mathrm{d}\mu\le\int_Y|f(x)|^{b_1}\,\mathrm{d}\mu\lt\infty $$ Thus, $|f(x)|^b[x\in X\setminus Y]+|f(x)|^{b_1}[x\in Y]$ is a dominating function proving that $$ \lim_{b_n\to b}\int_X|f(x)|^{b_n}\,\mathrm{d}\mu=\int_X|f(x)|^b\,\mathrm{d}\mu $$

Greatest Lower Bound is Included

Suppose that $a$ is the greatest lower bound of $E$. If $a\not\in E$, there is nothing to prove, so suppose that $a\in E$. Let $a_n\in E$ and $a_n\to a$ monotonically (decreasing).

Define $$ Y=\{x\in X:|f(x)|\le1\} $$ Since $a_n\ge a$, $$ \int_Y|f(x)|^{a_n}\,\mathrm{d}\mu\le\int_Y|f(x)|^a\,\mathrm{d}\mu\lt\infty $$ and since $a_1\ge a_n\ge a$, $$ \int_{X\setminus Y}|f(x)|^a\,\mathrm{d}\mu\le\int_{X\setminus Y}|f(x)|^{a_n}\,\mathrm{d}\mu\le\int_{X\setminus Y}|f(x)|^{a_1}\,\mathrm{d}\mu\lt\infty $$ Thus, $|f(x)|^{a_1}[x\in X\setminus Y]+|f(x)|^a[x\in Y]$ is a dominating function proving that $$ \lim_{a_n\to a}\int_X|f(x)|^{a_n}\,\mathrm{d}\mu=\int_X|f(x)|^a\,\mathrm{d}\mu $$

robjohn
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