Dilation of Convex Set in Vector Space is Convex

Theorem

Let $\Bbb F \in \set {\R, \C}$.

Let $X$ be a vector space over $\Bbb F$.

Let $C \subseteq X$ be a convex subset of $X$.

Let $\alpha \in \Bbb F$.

Then $\alpha C$ is convex.

Proof

Consider first the case $\alpha = 0$.

We then have $\alpha C = \set { {\mathbf 0}_X}$.

This is convex by Singleton is Convex Set.

Now consider the case $\alpha \ne 0$.

Let $u, v \in \alpha C$ and $t \in \closedint 0 1$.

Then there exists $x, y \in C$ such that $u = \alpha x$ and $v = \alpha y$.

Since $C$ is convex, we have:

$t x + \paren {1 - t} y \in C$

Then, we have:

$\alpha t x + \alpha \paren {1 - t} y \in \alpha C$

That is:

$t u + \paren {1 - t} v = t \paren {\alpha x} + \paren {1 - t} \paren {\alpha y} \in \alpha C$

Since $u, v \in \alpha C$ and $t \in \closedint 0 1$ were arbitrary, $\alpha C$ is convex.

$\blacksquare$