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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.

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    $\begingroup$ Even if you wanted to use derivatives you can not because $|x|^p$ is not differentiable at zero. $\endgroup$ – Sergio Parreiras Nov 19 '13 at 3:58
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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 $p\gt1$, $$ \|x\|^p=\sup\{\alpha_t^{(p)}(x)\,;\,t\in\mathbb R^d\},\qquad \alpha_t^{(p)}(x)=a_p\langle t,x\rangle-b_p\|t\|^q. $$

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  • $\begingroup$ Your profile says this is answer number 4000. Nicely done! :) $\endgroup$ – Neal Nov 19 '13 at 15:39
  • $\begingroup$ @Neal True (I hadn't even noticed...). $\endgroup$ – Did Nov 19 '13 at 15:48
  • $\begingroup$ I had clicked through to your profile from another answer on a whim and noticed you had 4001 answers. I'm not stalking you, I swear! $\endgroup$ – Neal Nov 20 '13 at 12:02
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    $\begingroup$ @Neal No problem... :-) $\endgroup$ – Did Nov 20 '13 at 12:22

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