Let $A$ is a $n\times n$ symmetric positive definite matrix and $1<p<\infty$. Is the map $T:\mathbb R^n\to\mathbb R$ given by $T(x)=|Ax\cdot x|^\frac{p}{2}$ convex?
Where $|\cdot|$ is the usual inner product.
As I see for $p=2$, it is convex but I can't do for the general case.
For $p \geq 2$, $T(x)$ is convex.
Notice that $T(x) = (x^\top A x)^{p/2}$ is the composition of $f(x) = x^\top A x$ and $g(t) = \max\{0, t^{p/2}\}$. For $p \geq 2$, $g$ is a nondecreasing function on $\mathbb{R}_+$ and also convex as the pointwise maximum of two convex functions. Additionally, $f$ is convex as a positive-definite quadratic form, and $\forall x \in \mathbb{R}^n, f(x) \geq 0$. Then, we obtain:
$$ f(\lambda x + (1 - \lambda) y) \leq \lambda f(x) + (1 - \lambda) f(y) \Rightarrow \\ \underbrace{g(f(\lambda x + (1 - \lambda) y))}_{T(\lambda x + (1 - \lambda)y)} \overset{g \uparrow}{\leq} g(\lambda f(x) + (1 - \lambda) f(y)) \overset{g \text{ convex}}{\leq} \underbrace{\lambda g(f(x))}_{\lambda T(x)} + \underbrace{(1 - \lambda) g(f(y))}_{(1 - \lambda) T(y)} $$ which is the defining inequality for convexity of a function.