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We know that if$\ f$ is a twice-differentiable convex function. Then,$\ g$ defined as follows is also a convex function. $$\ g(y) = f(Ay+b) $$ I was wondering if all transformations that preserve the convexity give us the same property. Formally, $$\ g(y)=f(T(y)) $$ Also, does the first result of Affine Transforms hold for non-differentiable convex functions? That is, $\ f$ is not differentiable?

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    $\begingroup$ The affine transform result holds for any convex $f$, differentiable or not. This can be seen from the definition of convexity. $\endgroup$ Mar 1 '20 at 18:23
  • $\begingroup$ Thanks. That clarifies the second part. $\endgroup$ Mar 1 '20 at 18:48
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In case $T:\mathbb R^n\to\mathbb R^n$ and $f:\mathbb R^n\to\mathbb R$, let $\{f_1,f_2,\ldots,f_n\}$ be the dual basis of the standard basis $\{e_1,e_2,\ldots,e_n\}$ of $\mathbb R^n$, so that $f_i(e_j)=\delta_{ij}$. Since each $f_i$ is linear, both $f_i$ and $-f_i$ are convex. Therefore, if $T$ preserves convexity for all convex functions, then both $f_i\circ T$ and $-(f_i\circ T)$ are convex. Hence $f_i\circ T$ is affine, i.e. there exists a constant vector $a_i\in\mathbb R^n$ and a constant scalar $b_i$ such that $(f_i\circ T)(x)=a_i^\top x+b_i$ for all $x\in\mathbb R^n$. It follows that $T(x)=\left(f_1(Tx),f_2(Tx),\ldots,f_n(Tx)\right)^\top=Ax+b$ where $A$ is a matrix with $a_1^\top,a_2^\top,\ldots,a_n^\top$ as rows.

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