Relationship Between Affine and Convex Functions

Question

I'm trying to prove that if $\Omega \subset \Bbb R^n$ is an open set, a function $f: \Omega \to \Bbb R$ is affine if and only if $f$ and $-f$ are both convex. I've managed to prove the forward direction with the following argument:

Suppose that $f$ is affine and let $\mathbf a \in \Bbb R^n$ and $b \in \Bbb R$ such that $f(\mathbf x) = \langle \mathbf a, \mathbf x \rangle + b$. If $\mathbf x_1, \mathbf x_2 \in \Bbb R^n$ and $\lambda \in [0,1]$, $$\begin{align*} f(\lambda \mathbf x_1 + (1 - \lambda) \mathbf x_2) &= \langle \mathbf a, \lambda \mathbf x_1 + (1- \lambda) \mathbf x_2\rangle + b \\ &= \lambda \langle \mathbf a, \mathbf x_1 \rangle + (1 - \lambda) \langle \mathbf a, \mathbf x_2 \rangle + (\lambda + (1-\lambda)) b \\ &= \lambda \left(\langle \mathbf a, \mathbf x_1 \rangle + b\right) + (1 - \lambda) \left(\langle \mathbf a, \mathbf x_2 \rangle + b\right) \\ &= \lambda f(\mathbf x_1) + (1 - \lambda) f(\mathbf x_2) \end{align*}$$ implies that $f$ is convex. The argument that $-f$ is convex is similar.

I need help proving the converse.

Answer 1

If $f$ and$-f$ are both convex we get $f(\lambda x_1+(1-\lambda x_2))=\lambda f(x_1)+(1-\lambda) f(x_2)$ Let$g(x)=f(x)-f(0)$. Check that $g(\lambda x_1+(1-\lambda x_2))=\lambda g(x_1)+(1-\lambda) g(x_2)$ and $g(0)=0$. Put $x_2=0$ to get $g(\lambda x_1)=\lambda g(x_1)$ and put $\lambda =\frac 1 2$ to get $g(\frac {x_1+x_2} 2)=\frac {g(x_1)+g(x_2)} 2$. Conclude that $g$ is a linear map from $\mathbb R^{n} $ to $\mathbb R$. This implies that $g(x)= \langle a, x \rangle$ for some $a$. Can you finish?