Rudin's RCA Q3.4 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.
 A: For (c ) consider the following Lebesgue measure situations:

*

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


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


*$X=(0,\infty),f(x) = [1/(x(\ln x)^2](\chi_{(0,1/2)}(x) + \chi_{(2,\infty)}(x)).$
A: This is an answer for the continuity of $\varphi.$
WLOG $f\ge 0.$ Let $A=\{f\le 1\},$ $B= \{f> 1\}.$ Suppose $[p,q]\subset E.$ Let $p_n\to p$ within $[p,q].$ Clearly $f^{p_n}\to f^p$ pointwise everywhere.
Now
$$\tag 1\int_X f^{p_n} = \int_X f^{p_n}\chi_{A} + \int_X f^{p_n}\chi_{B}.$$
Observe $f^{p_n}\chi_{A} \le f^{p}\chi_{A}$ and $f^{p_n}\chi_{B} \le f^{q}\chi_{B}.$ Since both $f^{p}\chi_{A}, f^{q}\chi_{B} \in L^1(X),$ the DCT implies the right side of $(1)$ converges to
$$\int_X f^{p}\chi_{A} + \int_X f^{p}\chi_{B} = \int_X f^{p}.$$
This implies $\varphi$ is continuous from the right at $p.$
Similarly, $\varphi$ is continuous from the left at $q.$ It follows that $\varphi$ is continuous on $E.$
A: I am going to add something for $b)$ because I have been working on this problem too and want anyone else that searches for this problem to have easy access to. The first fact that we need is for any function $f\in L^\alpha$ and for any $\beta > 0$ that
$$||f||_\alpha^{\beta}= \Big(\int_X |f|^{\alpha}d\mu\Big)^{\beta/\alpha}=\Big(\int_X (|f|^{\beta})^{\alpha/\beta}d\mu\Big)^{\beta/\alpha} = || |f|^\beta||_{\alpha/\beta}$$
Thus $|f|^\beta\in L^{\alpha/\beta}$ if $f\in L^\alpha$. With this fact in mind, let $p,q\in int(E)$ and $t\in (0,1)$. Thus $f\in L^p$ and $f\in L^q$. Using the previous fact, we have that
\begin{align*}
||f||_p^{(1-t)p} &= || (|f|^{(1-t)p})||_{p/[(1-t)p]}= || (|f|^{(1-t)p})||_{1/(1-t)}\\
||f||_q^{tq} &= || (|f|^{tq})||_{q/tq}= || (|f|^{tq})||_{1/t}\\
\end{align*}
So we have that $|f|^{(1-t)p}\in L^{1/(1-t)}$ and $|f|^{tq}\in L^{1/t}$. Then as $$\frac{1}{1/(1-t)}+\frac{1}{1/t}=1-t+t = 1 $$
We can use Holder's inequality to see that $|f|^{(1-t)p+tq}\in L^1$ and that $$||(|f|^{(1-t)p+tq})||_1 \leq || (|f|^{(1-t)p})||_{1/(1-t)} || (|f|^{tq})||_{1/t}=||f||_p^{(1-t)p}||f||_q^{tq} = \varphi(p)^{1-t}\varphi(q)^t$$
At the same time we have that
$$\varphi((1-t)p+tq)=||f||_{(1-t)p+qt}^{(1-t)p+qt} = || (|f|)^{(1-t)p+qt}||_1$$
So if we take the $\log$ on both sides and use properties of the $\log$ then
$$
\begin{align*}\log(\varphi((1-t)p+tq)) &\leq \log(\varphi(p)^{1-t}\varphi(q)^t)\\
&=\log(\varphi(p)^{1-t}) + \log(\varphi(q)^t) \\
&= [1-t]\log(\varphi(p)) + t \log(\varphi(q))
\end{align*}$$
Since $p,q \in int(E)$ and $t\in (0,1)$ were arbitrary, we may conclude that $\log \varphi$ is a convex function on the interior of $E$. Now, $\log\varphi$ being convex implies that $\varphi$ is convex and hence is also continuous on $int(E)$.
