Let $A$ be a $n\times n$-matrix. We let $\|A\|_p$ denote the norm of $A$ when considered as a linear operator on $\ell^p(\{1,2,\ldots,n\})$, that is,
$$
\|A\|_p = \sup_{x\neq 0}\frac{\|Ax\|_p}{\|x\|_p}.
$$
By the Riesz-Thorin theorem, the $p$-norm of $A$, as a function of $p$, is log-convex, meaning that the function $[1,\infty]\ni p\mapsto \log(\|A\|_p)$ is convex.
Question: Is the p-norm of a matrix strictly log-convex (unless it is constant)?
The proof of the Riesz-Thorin theorem uses holomorphic functions, so it seems plausible that if $\log(\|A\|_p)$ is constant on some non-trivial interval, then $\log(\|A\|_p)$ has to be globally constant.
Edit (following Terry Tao's answer):
Can $\|A\|_p$ be constant on some non-trivial interval, yet not be globally constant?
Best Answer
The answer is "no".
First, a trivial counterexample: let $A_n$ be the $n \times n$ matrix with all $1$s on the first row and zeroes elsewhere, then $\frac{1}{p} \mapsto \|A_n\|_p = n^{1/p}$ is log-convex but non-constant and not strictly log-convex.
One can deal with this example by replacing "constant" with "log-linear". However, if we consider the $(n+m) \times (n+m)$ matrix
$$ B := \begin{pmatrix} a A_n & 0\\ 0 & b A_m \end{pmatrix} $$ for some real parameters $a,b > 0$ and distinct natural numbers $n,m > 0$, then $\frac{1}{p} \mapsto \|B\|_p = \max( a n^{1/p}, b m^{1/p})$, which is log-convex but not log-linear and not strictly log-convex.
It seems to me that a similar construction allows any log-convex function to arise as $\frac{1}{p} \mapsto \|A\|_p$, at least if one allows A to be infinite-dimensional.