Sum of Convex Sets in Vector Space is Convex

Theorem

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

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

Let $A$ and $B$ be convex subsets of $X$.

Then:

$A + B$ is convex.

Corollary

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

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

Let $A$ and $B$ be convex subsets of $X$.

Let $\lambda, \mu \in \Bbb F$.

Then:

$\lambda A + \mu B$ is convex.

Proof

Let $x, y \in A + B$ and $t \in \closedint 0 1$.

Then there exists $a, a' \in A$ and $b, b' \in B$ such that:

$x = a + b$

and:

$y = a' + b'$

Then:

\(\ds t x + \paren {1 - t} y\)

\(=\)

\(\ds t a + t b + \paren {1 - t} a' + \paren {1 - t} b'\)

\(\ds \)

\(=\)

\(\ds \paren {t a + \paren {1 - t} a'} + \paren {t b + \paren {1 - t} b'}\)

Since $A$ and $B$ are convex, we have:

$t a + \paren {1 - t} a' \in A$

and:

$t b + \paren {1 - t} b' \in B$

so that:

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

So $A + B$ is convex.

$\blacksquare$