Question

Let S be a convex subset of \({\mathbb{R}^n}\) and suppose that \(f:{\mathbb{R}^n} \to {\mathbb{R}^m}\) is a linear transformation. Prove that the set \(f\left( S \right) = \left\{ {f\left( {\bf{x}} \right):{\bf{x}} \in S} \right\}\) is a convex subset of \({\mathbb{R}^m}\).

Short answer

\(f\left( S \right)\) is a convex subset of \({\mathbb{R}^m}\).

Step-by-step solution

Find the value of the function for two different constants

Let, \({\bf{q}} \in f\left( S \right)\), then for \({\bf{r}},{\bf{s}} \in S\)

\(f\left( {\bf{r}} \right) = {\bf{p}}\) and \(f\left( {\bf{s}} \right) = {\bf{q}}\)

 Step 2: Check that \(f\left( S \right)\) is convex

Check whether the line segment \(y = \left( {1 - t} \right){\bf{p}} + t{\bf{q}}\) is in \(f\left( S \right)\):

\(\begin{aligned}{}y &= \left( {1 - t} \right){\bf{p}} + t{\bf{q}}\\ &= \left( {1 - t} \right)f\left( {\bf{r}} \right) + tf\left( {\bf{s}} \right)\\ &= f\left( {1 - t} \right){\bf{r}} + tf\left( {\bf{s}} \right)\\ &= f\left\{ {\left( {1 - t} \right){\bf{r}} + t{\bf{s}}} \right\}\end{aligned}\)

As S is a convex set, \(\left( {1 - t} \right){\bf{r}} + t{\bf{s}}\)is also in S, therefore

\(f\left\{ {\left( {1 - t} \right){\bf{r}} + t{\bf{s}}} \right\} \in f\left( S \right)\)

So, \(f\left( S \right)\) is aconvex subset of \({\mathbb{R}^m}\).