Transformation that preserves convexity of domain result in convex function

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?

• The affine transform result holds for any convex $f$, differentiable or not. This can be seen from the definition of convexity. Mar 1 '20 at 18:23
• Thanks. That clarifies the second part. Mar 1 '20 at 18:48

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.