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I am working on the following task:

If $f: I \to \mathbb{R}$, $I=\mathbb{R}$ or $I=(-R,R), R>0$, is a convex function, then $f(-x)$ is convex too.

I have already shown that this is true for even functions. But I don't know what to do if $f$ is an odd function or has no symmetry. Thanks for your help!

Gary
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hannah2002
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    Let $g(x)=f(-x)$. Then you can prove that $g(-x)$ is convex. This is true because a composition $g\circ h$, where $h$ is linear is always convex. It's easy to see that $g\left(h\left(\frac{x_1+x_2}2\right)\right)=g\left(\frac{h(x_1)+h(x_2)}2\right)$. You can just show that $(g\circ h)\left(\frac{x_1+x_2}2\right)\le\frac{(g\circ h)(x_1)+(g\circ h)(x_2)}2$. – PinkyWay May 06 '21 at 14:39
  • How odd... this is practically an exact duplicate of https://math.stackexchange.com/questions/4129492/give-all-odd-convex-functions asked on May 6th as well. Homework? – postmortes May 12 '21 at 05:38
  • Does this answer your question? Give all odd convex functions – postmortes May 12 '21 at 05:39

1 Answers1

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Let $a, b\in I$ and $t\in [0,1]$. You have $$f(-(ta+(1-t)b)) =f(t \times (-a)+(1-t)\times (-b)) \leq t f(-a) + (1-t)f(-b)$$

using the convexity of $f$. So the function $x \mapsto f(-x)$ is also convex.

TheSilverDoe
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