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Let $f(x)=\|x\|^\alpha,\alpha\geq 1,x\in\mathbb{R}^n$, show that $f$ is convex.

I tried to prove the equivalent condition $$f(x+y)\geq f(x)+f'(x)\cdot y$$ and I get (if $x\neq 0$) $$\|x+y\|^\alpha\geq^?\alpha\|x\|^{\alpha-2}x\cdot y$$

And by definition is something similar, I think there must be an inequality that could help, but I don't remember, things.

José
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  • $t \mapsto t^\alpha$ is non-decreasing for $t \in [0, +\infty]$ (check the derivative) and $x \mapsto |x|_2$ is convex on $\mathbb{R}^n$. The composition of a non-decreasing function and a convex function is again convex. – dohmatob Sep 18 '15 at 03:34

1 Answers1

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The real valued function $f(t)=t^{\alpha}$ is convex for $t\geq 0$ because the second derivative is non-negative everywhere. The function $t^{\alpha}$ is increasing, so:

$$||\lambda x+(1-\lambda) y||^{\alpha}\leq (\lambda ||x||+(1-\lambda)||y||)^{\alpha}\leq \lambda||x||^{\alpha}+(1-\lambda)||y||^\alpha$$

The first inequality follows from the triangle inequality and monotonicity of $t^{\alpha}$. The second from the convexity of $t^{\alpha}$ for $\alpha \geq 1$.

uniquesolution
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