# convexity of certain function, alternative proof

I want to show that $f:\mathbf{R} \to \mathbf{R}$ given by $f(x) = |x|^p$ is convex for $p \geq 1$. But I don't want to use the derivative of $f$. How can I do this?

Really, it boils down to showing $f(\lambda x + (1-\lambda)y) \leq \lambda f(x) + (1-\lambda) f(y)$. Then I would have $|\lambda x + (1-\lambda)y|^p \leq (|\lambda x| + |(1-\lambda) y|)^p$. I thought about taking a taylor expansion here, but that doesn't seem to be a good idea.

• Even if you wanted to use derivatives you can not because $|x|^p$ is not differentiable at zero. – Sergio Parreiras Nov 19 '13 at 3:58

Every supremum of affine functions is convex hence the following remarks yield the result. First, for $p=1$, $$|x|=\sup\{x,-x\}.$$ Second, for each $p\gt1$, let $\alpha_t^{(p)}$ denote the affine function defined by $$\alpha_t^{(p)}(x)=a_ptx-b_p|t|^q,$$ where $q$ is the exponent conjugate of $p$, defined by $\frac1p+\frac1q=1$. Then, for some suitable values of $a_p$ and $b_p$ that I will let you discover, $$|x|^p=\sup\{\alpha_t^{(p)}(x)\,;\,t\in\mathbb R\}.$$ This can be generalized to the function $x\mapsto\|x\|^p$ in every $\mathbb R^d$, using, for $p=1$, $$\|x\|=\sup\{\langle t,x\rangle\,;\,t\in\mathbb R^d,\|t\|=1\},$$ and, for $p\gt1$, $$\|x\|^p=\sup\{\alpha_t^{(p)}(x)\,;\,t\in\mathbb R^d\},\qquad \alpha_t^{(p)}(x)=a_p\langle t,x\rangle-b_p\|t\|^q.$$