I would like you to evaluate the following proof that $f:\mathbb{R}^{n}\rightarrow \mathbb{R}$ is affine if and only if its convex and concave:
From the convexity of f we have for $x_1, x_2 \in \mathbb{R}^{n}$ and $\lambda \in [0,1] f(\lambda x_1+ (1−\lambda)x_2)\leq\lambda f(x_1)+(1 - \lambda)f(x_2)$ and for concavity we have $f(\lambda x_1+ (1−\lambda)x_2)\geq\lambda f(x_1)+(1 - \lambda)f(x_2)$
Thus it follows that f convex and concave is $f(\lambda x_1+(1−\lambda)x_2)=\lambda f(x_1)+(1 - \lambda)f(x_2)$
And F is affine if $f(x)=ax+b$
Going: $$ f(\lambda x_1+ (1−\lambda)x_2)=\lambda f(x_1)+(1 - \lambda)f(x_2)=\lambda (ax_1+b)+(1-\lambda)(ax_2+b)=\lambda ax_1+\lambda b+ax_2-\lambda ax_2+b-\lambda b=a(\lambda x_1+x_2-\lambda x_2)+b=a(\lambda x_1+(1-\lambda)x_2)+b $$
Back: $$ f(\lambda x_1 + (1-\lambda)x_2)=a(\lambda x_1+(1-\lambda)x_2)+b=a(\lambda x_1+x_2-\lambda x_2)+b=\lambda ax_1+ax_2-\lambda ax_2+b=\lambda ax_1+ax_2-\lambda ax_2+b+\lambda b-\lambda b=\lambda(ax_1+b)+(1-\lambda)(ax_2+b)=\lambda f(x_1)+(1-\lambda)f(x_2) $$