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I'm stuck at proving this statement:

Let $f$ be a convex real-valued function in $\mathbb R^d$.

Then, for every $n\in\mathbb N$, the function $x\mapsto\frac{f(nx)}n$ is convex as well.

(In my case $f$ is a cumulant generating function $f_n(x)=\log(\mathbb E[\exp(\langle x,Z_n\rangle)])$ of a random variable $Z_n$, but I suspect this is generally true, even outside of the $\mathbb R^d$ case).

This seems easy, but searching for the title yielded nothing of importance.

Did
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Formyer
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    https://en.m.wikipedia.org/wiki/Convex_function. Look in the section on convex function calculus, which should have what you need. – Zach Boyd Sep 24 '17 at 15:49
  • @ZachBoyd Oh, thanks! So I totally missed the link to a requirement that I had: $f_n$ is assumed to be finite in a neighbourhood of the origin, which ties in nicely with the rule: – Formyer Sep 24 '17 at 15:57
  • If $f(x)$ is convex, then its perspective $g(x,t) = t f(x/t)$ is convex. – Formyer Sep 24 '17 at 15:58
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    Have you tried verifying the conditions for a convex function with $nf(nx)$ given that $f(x)$ satisfies those conditions? – robjohn Sep 24 '17 at 16:18
  • I did, but made a stupid mistake. It really is nothing more than using the convexity of $f$. Thanks for making me re-check, @robjohn! – Formyer Sep 24 '17 at 16:50
  • @Formyer, When you're done. Post your solution to make the question answered. – Royi Sep 24 '17 at 16:51
  • Humm; the $1/n$ factor can be omitted. It adds considerable 'noise" to the statement of the problem. – dohmatob Sep 25 '17 at 03:21
  • Well, so does the factor of $n$ inside the argument. If your argument for why it is 'noise' is that scaling isn't going to change the convexity of a function, I might as well omit both $n$'s. They're both just scaling of a kind, which doesn't affect convexity. – Formyer Sep 25 '17 at 09:39

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Okay, so this is nothing more than checking the definition:

We have to prove:

$$\forall x,y\in\mathbb R^d\forall t\in[0,1]:\frac{f(n(tx+(1-t)y))}n\leq t\frac{f(nx)}n+(1-t)\frac{f(ny)}n.$$

But we already know that $f$ is convex and defined on the whole $\mathbb R^d$, so:

$$\frac{f(n(tx+(1-t)y))}n=\frac{f(t(nx)+(1-t)(ny))}n\leq\frac{tf(nx)+(1-t)f(ny)}n=\\=t\frac{f(nx)}n+(1-t)\frac{f(ny)}n.$$

I was stuck thinking I'd have to get from $f(nx)$ to $f(x)$ somehow, so I didn't find this simple solution.

Formyer
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