Composition of convex function and affine function

Let $g: E^{m} \rightarrow E^{1}$ be a convex function, and let $h: E^{n} \rightarrow E^{m} $ be an affine function of the form $h(x)=Ax+b$, where $A$ is an $m \times n$ matrix and $b$ is an $m \times 1 $ vector. Then, show that the composite function $f : E^n \rightarrow E^{1} $ defined as $f(x)=g(h(x))$ is a convex function.

Also, assuming twice differentiability of $g$, derive the expression for the hessian of $f$

Solution 1:

Let $0 < \theta < 1$ and $x_1, x_2 \in E^m$. Note that $h(\theta x_1 + (1-\theta)x_2) = \theta h(x_1) + (1-\theta)h(x_2)$. It follows that \begin{align} f(\theta x_1 + (1-\theta) x_2) &= g(\theta h(x_1) + (1-\theta)h(x_2)) \\ &\leq \theta g(h(x_1)) + (1-\theta) g(h(x_2)) \\ &= \theta f(x_1) + (1-\theta) f(x_2) \end{align} so $f$ is convex.

From the chain rule, $f'(x) = g'(h(x)) h'(x) = g'(h(x))A$ so \begin{align} \nabla f(x) &= f'(x)^T \\ &= A^T g'(h(x))^T \\ &= A^T \nabla g(h(x)). \end{align} The chain rule again now tells us that $\nabla^2 f(x) = A^T \nabla^2 g(h(x)) h'(x) = A^T \nabla^2 g(h(x)) A$.