LP Spaces – Rudin’s RCA Q3.4 Explained

lp-spaces

I'm trying to solve the following question from Rudin's Real & Complex Analysis. (Chapter 3, question 4) :

Suppose $f$ is a complex measurable function on $X$, $\mu$ is a positive measure on $X$, and
$$\varphi(p) ~=~ \int_X |f|^p \; d\mu ~=~ \|f\|_p^p,~~~~~~~~~~(0 < p < \infty).$$
Let $E := \big\{ p :~ \varphi(p) < \infty\big\}$. Assume $\|f\|_\infty > 0$.

(a) If $r < p < s$, $r \in E$, and $s \in E$, prove that $p \in E$.

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

(c) By (a), $E$ is connected. Is $E$ necessarily open ? Closed ? Can $E$ consist of a single point ? Can $E$ be any connected subset of
$(0,\infty)$ ?

(d) If $r < p < s$, prove that $\|f\|_p \leq \max\big( \|f\|_r, \|f\|_s\big)$. Show that this implies the inclusion
$$ \mathcal{L}_r(\mu) \cap \mathcal{L}_s(\mu) ~\subseteq~\mathcal{L}_p(\mu).$$

(e) Assume that $\|f\|_r < \infty$ for some $r < \infty$ and prove that $$ \|f\|_p \xrightarrow[p \rightarrow \infty]{}\|f\|_\infty.$$

I got a solution to (a), (d) and (e). While typing my question, MSE suggested me to look at $a\mapsto \log\left(\lVert f\lVert_{1/a}\right)$ is a convex map
which seems to be related to (b). However I'm clueless about (c). Where should I start from ?

Edit

I believe the idea from $a\mapsto \log\left(\lVert f\lVert_{1/a}\right)$ is a convex map can be applied to get a proof that
$p \mapsto \log\|f\|_{\frac{1}{p}}$ is continuous on the interior of $E$. But this is not quite what we are asked to demonstrate… I'm puzzled.

Second Edit

Based on zhw's answer, it appears that $E$ is not necessarily open, nor necessarily closed and that it can be a singleton. The question whether or not $E$ can be any connected subset of $(0,\infty)$ remains. But I think I can come up with a proof that it can.

Best Answer

For (c ) consider the following Lebesgue measure situations:

  1. $X=(0,1)$ with $f(x) = 1/x.$

  2. $X=(0,1/2)$ with $f(x) = 1/[x(\ln x)^2].$

  3. $X=(0,\infty),f(x) = [1/(x(\ln x)^2](\chi_{(0,1/2)}(x) + \chi_{(2,\infty)}(x)).$