if $A$ is convex and $f$ linear then $f(A)$ and $f^{-1}(A)$ are convex

Proposition 1.

Suppose $X$, $Y$ are vector spaces over $\mathrm{R}$ (or $\mathrm{C}$),and suppose $f\mathrm{:}X\mathrm{\to }Y$ is a linear map.

1. If $A\subseteq X$ is convex, then $f(A)$ is convex.

2. If $B\subseteq Y$ is convex, then $f^{-1}(B)$ is convex,where $f^{-1}$ is the inverse image.

Proof.

For the first claim,suppose $y,y^{\prime }\in f(A)$, say,$y=f(x)$ and $y^{\prime }=f(x^{\prime })$ for $x,x^{\prime }\in A$, and suppose $\lambda \in (0,1)$.Then

$\lambda y+(1-\lambda )y^{\prime }$$=$$\lambda f(x)+(1-\lambda )f(x^{\prime })$

$=$$f(\lambda x+(1-\lambda )x^{\prime }),$

so $\lambda y+(1-\lambda )y^{\prime }\in f(A)$ as $A$ is convex.

For the second claim, let us first recall that $x\in f^{-1}(B)$if and only if $f(x)\in B$. Then, if $x,x^{\prime }\in f^{-1}(B)$,and $\lambda \in (0,1)$, we have

$f(\lambda x+(1-\lambda )x^{\prime })$$=$$\lambda f(x)+(1-\lambda )f(x^{\prime }).$

As $B$ is convex, the right hand side belongs to $B$,and $\lambda x+(1-\lambda )x^{\prime }\in f^{-1}(B)$.∎