Let p ∈ [1,+∞[. Show that f: X → R: x → ∥x∥^p is convex.

Question

I am stuck on this question. Here is what I have so far:

Take any $x_1,x_2 \in \mathbb{R}^n$ and $λ \in (0,1)$

$$f(λx_1 + (1- λ)x_2) \leq λf(x_1) + (1- λ)f(x_2)$$

then I would plug the function $||x||^p$ into the above, and I'm not sure where to go from there.

Answer 1

As $x \mapsto \Vert x \Vert$ is a norm, you have for $\lambda \in [0,1]$ and $x,y \in X$: $$\Vert (1-\lambda)x +\lambda y\Vert \le (1-\lambda) \Vert x \Vert + \lambda \Vert y \Vert.$$

Now as $p >0$ $x \mapsto x^p$ is increasing, you also get $$\Vert (1-\lambda)x +\lambda y\Vert^p \le [(1-\lambda) \Vert x \Vert + \lambda \Vert y \Vert]^p.$$

Finally, as $p \ge 1$, $x \mapsto x^p$ is convex, so: $$\Vert (1-\lambda)x +\lambda y\Vert^p \le [(1-\lambda) \Vert x \Vert + \lambda \Vert y \Vert]^p \le (1-\lambda) \Vert x \Vert^p +\lambda \Vert y \Vert^p,$$ which enables to conclude.