Is it a convex function? Let $f(.)$ be a function. If $f(X)$ is a convex function of $X$, where $X$ is a matrix. Is $f(AXB)$ also a convex function of $X$? ($A$ and $B$ are fixed matrices).
 A: The answer by MGA is correct, but I'll try to explain without referring to outside materials. 


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*The transformation  $X\mapsto AXB$ is linear, which you can see by distributing the multiplication here: $$A(\alpha X+\beta Y)B=\alpha AXB+\beta AYB$$

*Convexity of a function is preserved under linear transformation of the domain. (In other words, if $f$ is convex then $f\circ T$ is convex for any linear map $T$.) You can check this directly using the definitions of convexity and of a linear map. Here is an argument without computations: convexity of $f$ is equivalent to the statement that for every $X_0$ there is a an affine function $g$ such that $g(X_0)=f(X_0)$ and $g(X)\le f(X)$ for all $X$. A linear transformation preserves the above property, because it preserves the class of affine functions. 

A: See here: http://www.stanford.edu/class/ee364a/lectures/functions.pdf
Slide 14 makes your statement and slide 18 has your proof (you may have to read through the rest to understand it).
To adapt it to your case, keep in mind that, if $f$ is a scalar-valued function, then you can represent $X$ as a vector instead of a matrix, so $f(AXB) = f(C\mathbf{x})$, where $C$ is a new matrix calculated from $A$ and $B$ depending on how you choose to map $X$ onto $\mathbf{x}$. 
The rest should be clear.
