I have to prove to things about strong convexity and I want to know if my demonstrations are corrects:
1: All of it in $R^n$.I have $f(x)$ strongly convex in $R^n$ and $g(x)=x-w$ with $w$ different from $0$. Then $f \circ g$ is strongly convex.
2:I have $f(x)$ strongly convex in $R^n$ and $g(x)=Ax-w$ where $g(x)$ belong to $R^m$, the matrix $A$ is $n\times m$ and different from $0$, and $w$ belongs to $R^n$.Then $f\circ g$ is strong convexity.
Prove:
1- I thought that when I do $f \circ g$ what I´m doing is a translation, and geometric properties of $f$ will not change with it. So $f \circ g$ is strongly convex like $f$.
2- I use the idea that $f \circ g$ can be a rotation if I take $w=0$ and $A$ like the matrix rotation. So if I take $A$ like the matrix rotation that rotates $f$ through $180$ degrees I can said it would not be a convex function and then will not be strongly convex.
If some explanation is not so clear just tell me and I will try to solve it.
Thanks in advance for your help.
Your argument for (2) is not correct. If $g$ is a rotation and $f$ (strongly) convex then $f \circ g$ is (strongly) convex as well.
Generally, $g(x) = Ax + w$ with $A \in \Bbb R^{n\times m}$ and $w \in \Bbb R^n$ is an affine transformation from $ \Bbb R^m$ to $ \Bbb R^n$. In particular, $$ g(\lambda x + (1-\lambda) y) = \lambda g(x)+ (1-\lambda) g(y) $$ for $x, y \in \Bbb R^m$ and $0 < \lambda < 1$. Therefore, if $f : \Bbb R^n \to \Bbb R$ is convex then $$ f(g(\lambda x + (1-\lambda) y)) = f( \lambda g(x)+ (1-\lambda) g(y)) \\ \le \lambda f(g(x))+ (1-\lambda) f(g(y)) $$ which shows that $f \circ g$ is convex as well.
If $f$ is strongly convex and $A$ has full column rank then $f \circ g$ is strongly convex as well: $g$ is injective, so that $x \ne y$ implies $g(x) \ne g(y)$, and therefore $$ f(g(\lambda x + (1-\lambda) y)) = f( \lambda g(x)+ (1-\lambda) g(y)) \\ < \lambda f(g(x))+ (1-\lambda) f(g(y)) $$ for $x \ne y \in \Bbb R^m$ and $0 < \lambda < 1$.