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Consider a strongly convex function $g$, that is, for all $x,y$ in the domain and $t\in[0,1]$ we have $$ g(t x + (1-t)y) \le tg(x)+(1-t)g(y) - \frac{1}{2}mt(1-t)||x-y||_2^2 $$ for some $m>0$. Also, let $A$ be a full rank linear tranformation. Define $f = g\circ A$.

Is $f$ strongly convex?

jonem
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  • Is $m$ fixed, depending only on $g$, or depending on $x,y$? And is the domain finite-dimensional? – Hagen von Eitzen Aug 20 '14 at 06:19
  • Yes, there exists a fixed $m>0$, such that the inequality holds for all $x,y$ in the domain. For example, $g(x)=x^2$ is strongly convex, where $g(x)=e^x$ is not (when the domain is $\mathbb{R}$) because we can always find an $x$ and $y$ that will violate the inequality for a fixed $m>0$. The function $g(x)=x^4$ is another example that is not strongly convex. We can assume the domain is finite-dimensional. – jonem Aug 20 '14 at 06:24

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Yes. Note that there are constants $b, c > 0$ such that $b \|x\|_2 \le \|Ax\|_2 \le c \|x\|_2$ for all $x$. So $f \circ A$ is strongly convex if $f$ is strongly convex (but with a different $m$); if $A$ is onto this is if and only if.

Robert Israel
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