# 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 $\mathbb{R}$ (or $\mathbb{C}$), and suppose $f\colon X\to Y$ is a linear map.

1. 1.

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

2. 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

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

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

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

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

Title if $A$ is convex and $f$ linear then $f(A)$ and $f^{-1}(A)$ are convex IfAIsConvexAndFLinearThenFAAndF1AAreConvex 2013-03-22 14:36:18 2013-03-22 14:36:18 matte (1858) matte (1858) 8 matte (1858) Theorem msc 52A99 InverseImage DirectImage