How to prove $\|\bf x\|^{\it p}_{\it p}$ can be a convex function on $ \mathbb R^n $

Question

How to show that $ \|\mathbf{x}\|^p_p $ can be a convex function on $ \mathbb R^n $ for $ p>1 $,
where $ \displaystyle \|\mathbf{x}\|^p_p = \sum_{i=0}^n |\mathcal{x_i}|^p $ is the pth power of the $ l_{p} $-norm of the vector $ \mathbf x \in \mathbb R^n $.

Answer 1

In fact, you have strict convexity for $1 < p < \infty$. See proof here, e.g.

Answer 2

Let $f$ be that function. Let $\mathbf x=(x_1,\ldots,x_n$ be a point of $\mathbb R^n$. For all $i\neq j$ in $\{1,\dots,n\}$, we have $$ \frac{\partial^2 f}{\partial x_i\partial x_j}(\mathbf x) = 0 $$ (since the derivative w.r.t. to $x_j$ exists because $p>1$, is $px_j|x_j|^{p-2}$).

For $i=j$ however we have $$ \frac{\partial^2 f}{{\partial x_i}^2}(\mathbf x) = p|x_i|^{p-2} + p(p-2)x_i^2|x_i|^{p-4} = p|x_i|^{p-2}(1+p-2) = p(p-1)|x_i|^{p-2} \geq 0. $$

Thus $f$'s Hessian matrix $\nabla^2f$ at any point $\mathbf x$ is diagonal with nonnegative eigenvalues thus positive semidefinite, which means $f$ is convex.