TITLE: Show that $f(x)=||x||^p, p\ge 1$ is convex function on $\mathbb{R}^n$
QUESTION [8 upvotes]: Show that $f(x)=||x||^p, p\ge 1$  is convex function on $\mathbb{R}^n$. I have tried to use Holder's inequality, but I still cannot solve this problem. Could you help me with this problem? Thank you so much.

REPLY [11 votes]: This actually holds for any norm $||\cdot||$ on $\mathbb{R}^n$. Let $h:[0,\infty)\to[0,\infty)$ and $g:\mathbb{R}^n\to[0,\infty)$ be defined as
$$h(x):=x^p,\quad\quad g(x):=||x||.$$
where $p\geq 1$. From the triangle inequality we have that $g$ is convex. Also, it is easy to see that $h$ is non-decreasing and convex (since anywhere on $[0,\infty)$ the first and second derivatives of $h$ are non-negative).
So, for any $x,y\in\mathbb{R}^n$ and $\lambda\in [0,1]$,
$$f(\lambda x+(1-\lambda)y)=h(g(\lambda x+(1-\lambda)y))\leq h(\lambda g(x)+(1-\lambda)g(y))$$
$$\leq \lambda h(g(x))+(1-\lambda)h(g(y))=\lambda f(x)+(1-\lambda)f(y).$$
where the first inequality follows from convexity of $g$ and "increasingness" of $h$ and the second from convexity of $h$.
Edit: I just realised that the above was basically already stated in @GiuseppeNegro's comment. You can also generalise the above by substituting $\mathbb{R}^n$ with any convex normed vector space (over the reals).