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Let $h(x)$ be defined for $x \ge 0$. Suppose that $h(x)$ is non-negative and convex with $h(0) = 0$. I need to show that for $x_2 \ge x_1$ $h(x_2)\ge h(x_1)$.

I need to do this from the definition of convexity: for $\lambda \in[0,1]$, and $x_1,x_2 \in[0, \infty]$ then $h(\lambda x_1 + (1 - \lambda)x_2) \le \lambda h(x_1) + (1 - \lambda)h(x_2)$.

I drew a picture of this, and it seems obvious that the statement holds. My argument is that because $h(0) = 0$, and $h(x)$ is non-negative, then $h'(0) \ge 0$. Since $h(x)$ is convex, $h'(x)$ is increasing, and therefore $h'(x) \ge 0$ for all $x$, which means $h(x)$ is an increasing function.

But, even though the claim is clear, I need to be rigorous about this, and not mention that it is obvious from looking at derivatives.

meta_warrior
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user7348
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1 Answers1

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HINT: I will prove it using contradiction, i.e. assume that $h(x_2)<h(x_1)$. We can express $x_1$ as $x_1=\lambda x_2+(1-\lambda)0$, where $0\le\lambda\le1$. By the definition of convexity, $$h(x_1)=h(\lambda x_2+(1-\lambda)0)\le \lambda h(x_2).$$

So, this is a contradiction.

meta_warrior
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