[Math] Show that the inverse image of convex set is convex

convex-analysislinear algebralinear-transformations

From S.L Linear Algebra:

Let $L: V \rightarrow W$ be a linear map. Let $S'$ be a convex set in
$W$. Let $S$ be the set of all elements $P$ in $V$ such that $L(P)$ is
in $S'$. Show that $S$ is convex.

Remark. … The set $S$ in the exercise above is called the inverse image of $S'$ under $L$ …

I'm not completely aware how is inverse image applicable in this case, but I have been able to find few points that might lead to the proof.

It can be seen that $S'$ is set of elements $L(P)$ and $L(Q)$ where $P, Q \in S$, such that the line segment $L((1-t)P+tQ))$ where $t \in [0, 1]$, belongs to $S'$ (since $S'$ is a convex set).

Due to the axioms of the linear transformations:

  1. $L(X+Y)=L(X)+L(Y)$, for $X, Y \in V$.
  2. $cL(X)=L(cX)$, for $X \in V$ and $c \in \mathbb{R}$

It can be seen that:

$$L((1-t)P+tQ))=L((1-t)P)+L(tQ)=(1-t)L(P)+tL(q)$$

I believe the result above might correlate with the proof, since $(1-t)$ and $t$ remains unchanged under linear transformation $L$.

How can I show that for any two points $P,Q$, $(1-t)P+tQ \in S$? (or more generally, how can I show that $S$ is a convex set?)

Best Answer

If $P,Q\in S$ and $t\in[0,1]$, then $f\bigl((1-t)P+tQ\bigr)=(1-t)f(P)+tf(Q)\in S'$. Therefore, $(1-t)P+tQ\in S$.

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