Let $\Omega \subset \mathbb{R}^n$ be an open, bounded domain. Let $u : \Omega \to \mathbb{R}$ be measurable with $||u||_\infty < \infty$. For $p \in [1, \infty)$, define
$$\Phi_u : p \mapsto |\Omega|^{-\frac{1}{p}}||u||_{L^p(\Omega)} = \left( \frac{1}{|\Omega|}\int_{\Omega} |u|^p dx \right)^{\frac{1}{p}}, $$
where $|\Omega|$ denotes Lebesgue measure.
I would like to show that the function $\Phi_u$ (fixed $u$) is logarithmically convex in the variable $x = \frac{1}{p} \in (0, 1]$.
This fact is asserted at the beginning of Chapter 7 in Gilbarg and Trudinger's Elliptic Partial Differential Equations of Second Order, and I am attempting to prove it. Here's what I have so far:
Let $x_1, x_2 \in (0,1]$ with $x_1 < x_2$. Take $t \in [0,1]$. Our goal is to show that
$$\log \left( \frac{1}{|\Omega|}\int_{\Omega} |u|^{\frac{1}{tx_1 + (1-t)x_2}} dx \right)^{tx_1 + (1-t)x_2} \le t\log \left( \frac{1}{|\Omega|}\int_{\Omega} |u|^{\frac{1}{x_1}} dx \right)^{x_1} + (1-t)\log \left( \frac{1}{|\Omega|}\int_{\Omega} |u|^{\frac{1}{x_2}} dx \right)^{x_2}. $$
Well, we can start with
$$\log \left( \frac{1}{|\Omega|}\int_{\Omega} |u|^{\frac{1}{tx_1 + (1-t)x_2}} dx \right)^{tx_1 + (1-t)x_2} = (tx_1 + (1-t)x_2) \log \left( \frac{1}{|\Omega|}\int_{\Omega} |u|^{\frac{1}{tx_1 + (1-t)x_2}} dx \right) = t\log \left( \frac{1}{|\Omega|}\int_{\Omega} |u|^{\frac{1}{tx_1 + (1-t)x_2}} dx \right)^{x_1} + (1-t)\log \left( \frac{1}{|\Omega|}\int_{\Omega} |u|^{\frac{1}{tx_1 + (1-t)x_2}} dx \right)^{x_2}. $$
But now comes the messy task of dealing with the integrands, which currently have the the form $|u|^{\frac{1}{tx_1 + (1-t)x_2}}$. But the first integrand needs to become $|u|^{\frac{1}{x_1}}$, and the second needs to be converted to $|u|^{\frac{1}{x_2}}$. And I'm not exactly sure how to proceed from here.
Any help with finishing up this calculation is very appreciated!
This answer uses Lemma 2 (Log-convexity of $L^p$ norms) in the blog post that is linked in the comments. Several proofs of this lemma are given in the blog post.
We can use the lemma in this context as follows. Let's consider the measure space $(\Omega, \mathcal{L}(\Omega) , \frac{m}{|\Omega|})$, where $\mathcal{L}(\Omega)$ denotes the Lebesgue measurable subsets of $\Omega$, and $\frac{m}{|\Omega|}$ denotes Lebesgue measure divided by $|\Omega|$. Then, if $1 \le p \le q \le r \le \infty$ and $\frac{1}{q} = \frac{t}{p} + \frac{1-t}{r}$ for $t \in [0,1]$, the lemma says:
$$||u||_{L^q\left(\frac{dm}{|\Omega|}\right)} \le ||u||^t_{L^p\left(\frac{dm}{|\Omega|}\right)} \cdot ||u||^{(1-t)}_{L^r\left(\frac{dm}{|\Omega|}\right)}.$$
If we take the log of both sides of the inequality, we get the logarithmic convexity that we are after.