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I'm working from Rudin's $Real~And~Complex~Analysis$ (3rd Ed.) on my own -- more specifically on Exercise 4 from Chapter 3 which is partially restated below (slightly modified). I only reference part (b) of the exercise since this is the only part I'm struggling with. For the rest of the problem (namely part (a) and parts (c), (d) & (e) of the exercise), refer to this link pointing to a free download of Rudin's book referenced above (see PG. 71 in his book), or see this post, etc. Please note that I'm working out of this book exclusively, and all concepts, definitions, theorems, etc., therein is what I'm working with.

Exercise 3.4 Suppose $(X,\mathfrak{M},\mu)$ is a measure space, where $\mu$ is a positive measure, and suppose $f\!:\!X\!\rightarrow\!\mathbb{C}$ is a measurable function. At any $p_{0}\in(0,+\infty)$ we define $\varphi(p_{0}):=\big(\|f\|_{p_{0}}\big)^{p_{0}}\!\!={\displaystyle{\!\!\int_{X}|f|^{p_{0}}~\!d\mu}}$, as well as we define the set $E:=\big\{\widetilde{p}\in(0,+\infty):\varphi(\!~{\widetilde{p}}\!~)\!<\!+\infty\big\}$. We also assume that $\|f\|_{+\infty}\!>0$.

(b) Prove that $\log(\varphi)$ is convex on the interior of $E$ as well as that $\varphi$ is continuous on $E$.

I'm having difficulty showing that $\varphi$ is continuous at all points in $E$. As far as my dilemma in particular, I graciously need help with an elaboration of the very last, strict, inequality found in this proof for part (b) only. The estimate in that proof gives continuity on all of $E$ (note also that my comment is at the bottom of that webpage also requesting the same elaboration, to which I haven't received a response as of yet, and I figured I'd come here to see what I can find in order to move on in the book overall). In case the link for this proof is broken (or breaks at some point in the future), I will provide the relevant part of the proof posted there below (also slightly modified) in regards to part (b) in particular.

//Proof (b): Begin by letting $r,s\in E\subseteq\mathbb{R}^{+}$ where $r<s$, and then fix $\lambda\in(0,1)$ arbitrarily in order to set $p=p_{\lambda}=(1-\lambda)r+\lambda s\in(r,s)$. By appealing to H$\ddot{\text{o}}$lder's inequality, we can now deduce that $\varphi(p)\leq\big(\varphi(r)\big)^{1-\lambda}\big(\varphi(s)\big)^{\lambda}\!<\!+\infty$. We can use this inequality to show the first part of (b) -- we need $\varphi>0$ on the interior of $E$, which is indeed the case since $\|f\|_{+\infty}\!>0$ (I showed this with a quick proof by contradiction), and so $\log(\varphi)$ is convex on the interior of $E$ yielding $\varphi$ is convex on the interior of $E$ and we can conclude $\varphi$ is continuous on the interior of $E$.

In order to show that $\varphi$ is continuous at all points in $E$, we first let $\varepsilon>0$, and then we can find an $N\in\mathbb{N}$ such that the set $E_{N}:=\big\{\widetilde{x}\in X:|f(\widetilde{x})|>N\big\}\cup\big\{\widetilde{x}\in X:0<|f(\widetilde{x})|<\frac{1}{N}\big\}$ has measure less than $\varepsilon$ -- namely, $0\leq\mu(E_{N})<\varepsilon$, and this is indeed possible since $\|f\|_{+\infty}\!>0$. This being said, choose a $\delta>0$ such that $0\leq\text{min}\big\{|1-N^{\delta}|,\big|1-\frac{1}{N^{\delta}}\big|\big\}<\varepsilon$. Then, for any $x,y\in E$ where $|x-y|<\delta$ we estimate that:

${\displaystyle{|\varphi(x)-\varphi(y)|=\bigg|\int_{X}|f|^{x}-|f|^{y}d\mu\bigg|=\bigg|\int_{E_{N}~\sqcup~X\backslash E_{N}}\!\!\!\!\!\!|f|^{x}-|f|^{y}d\mu\bigg|}}$

${\displaystyle{~~~~~~~~~~~~~~~~~~~~~\leq\bigg|\int_{E_{N}}|f|^{x}-|f|^{y}d\mu\bigg|+\bigg|\int_{X\backslash E_{N}}|f|^{x}-|f|^{y}d\mu\bigg|}}$

${\displaystyle{~~~~~~~~~~~~~~~~~~~~~<2\varepsilon\big(\varphi(x)+\varphi(y)\big)}}$,

which shows that $\varphi$ is continuous on $E$.

I can't seem to simplify everything properly in order to show the very last, strict inequality above ${\displaystyle{\bigg|\int_{E_{N}}|f|^{x}-|f|^{y}d\mu\bigg|+\bigg|\int_{X\backslash E_{N}}|f|^{x}-|f|^{y}d\mu\bigg|<2\varepsilon\big(\varphi(x)+\varphi(y)\big)}}$ holds. I was hoping, not only if this is correct, but to determine how to establish this, since this actually shows $\varphi$ is uniformly continuous on $E$ implying continuity on $E$ (right [?] -- it looks correct as $\varphi(x)+\varphi(y)<+\infty$ whenever $x,y\in E$). I figured we need to take the first integral ${\displaystyle{\bigg|\int_{E_{N}}|f|^{x}-|f|^{y}d\mu\bigg|}}$ and use our assumption that $\mu(E_{N})<\varepsilon$ in order to get ${\displaystyle{\bigg|\int_{E_{N}}|f|^{x}-|f|^{y}d\mu\bigg|<\varepsilon\big(\varphi(x)+\varphi(y)\big)}}$; additionally, I figured our choice of $\delta>0$ will enable us to show ${\displaystyle{\bigg|\int_{X\backslash E_{N}}|f|^{x}-|f|^{y}d\mu\bigg|<\varepsilon\big(\varphi(x)+\varphi(y)\big)}}$ as well -- I think I can do this alone, but, if I'm correct, then my difficult arises with the first integral...I'm stumped. Any help is greatly appreciated!

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This approach is somewhat more simple. Let $H=\{x\in X:|f(x)|\leq1\}$. Then, $|f(x)|^p\chi_H(x)$ is non increasing and $|f(x)|^p\chi_{X\backslash H}(x)$ is non decreasing in $p$, at every $x\in X$. Then, if $p_0,p_1\in E,\ p_0<p_1$, by the dominated convergence theorem: $$\lim_{p\to p_0,p\in E}\int_{X\backslash H}|f|^pd\mu = \int_{X\backslash H}|f|^{p_0}d\mu\quad (1)$$ Note that the limit $\lim_{p\to p_0^-}$ holds even if there is no $p_1>p_0$ in $E$. Similarly, if $p_0,p_1\in E,\ p_0>p_1$, then $$\lim_{p\to p_0,p\in E}\int_{H}|f|^pd\mu = \int_{H}|f|^{p_0}d\mu\quad (2)$$ and the limit $\lim_{p\to p_0^+}$ always holds. I left the details to you.

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  • $\begingroup$ Ty for you post, and leaving out details; I was too busy trying to come up with the mechanics to drive out the inequality (then work further once doing so) -- in other words you feel that the proof linked in my question could be incorrect for this part overall? I was starting to think this myself after posting. I started to realize that, even after showing the strict inequality holds (if possible) from the linked-proof in my question, getting rid of the quantity $\varphi(x)+\varphi(y)$ presents difficulty also. I thought of considering the $\sup(E)$, for example, but this need not be finite. $\endgroup$ – Procore Dec 26 '17 at 5:41
  • $\begingroup$ I thought about it and I realized that $2\varepsilon\big(\varphi(x)+\varphi(y)\big)$ is enough because of the convexity of $\log \varphi$ $\endgroup$ – Veridian Dynamics Dec 27 '17 at 1:19
  • $\begingroup$ I'll keep working. My intent was to come back a few days after posting (I'm new to the chapter). I think your comment hints upon the elaboration I'm after, but I'm curious how to bring convexity of $\log(\varphi)$ into everything -- if you can elaborate that would help. So we've shown $\log(\varphi)$ is convex on the interior of $E$, but I'm confused on how the strict inequality was obtained initially on top of using your comment. I'll work through your initial solution first; then I can always go back secondarily based on anything posted afterwards, but at least I have a way out overall. $\endgroup$ – Procore Dec 27 '17 at 4:28
  • $\begingroup$ Veridian, I had a brief set back, but I've returned to this problem. As an upfront quick question, shouldn't we distinguish your set $E$, from the given set $E$ defined in the problem-statement to be precise? In other words (I figured to ask first), but shouldn't we let, say, $E^{*}=\big\{x\in X:|f(x)|\leq 1\big\}$ since $E:=\big\{\widetilde{p}\in(0,+\infty):\varphi(\widetilde{p})<+\infty\big\}$ in the problem-statement? I figured to bring this up as I'm working through your solution above now, and I didn't want confusion to arise. $\endgroup$ – Procore Jan 12 '18 at 1:05
  • $\begingroup$ uuuppps I didn't notice that I used the same letter for both sets. I edited the post! $\endgroup$ – Veridian Dynamics Jan 12 '18 at 23:45

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