# Let $W$ be convex set show that there exists convex set $V$ such that $W = V + V$

Let $$W$$ be the convex set, show that there exists a convex set $$V$$ such that $$W = V + V$$.

My attempt was to take $$V = \frac{1}{2}W$$.

Firstly, I have shown that such $$V$$ is indeed convex.

Proof

Let $$\frac{1}{2}w_1, \frac{1}{2}w_2 \in V$$, $$\lambda \in [0,1]$$.

Then we have

$$\lambda\cdot(\frac{1}{2}w_1) + (1-\lambda)\cdot(\frac{1}{2}w_2) = \frac{1}{2}\cdot(\lambda w_1 + (1-\lambda)w_2)$$

By the convexity of $$W$$ we receive that the bracket on the right-hand side again belongs to $$W$$. Hence $$V$$ is convex.

Secondly, I have to show that $$V+V = W$$. I have divided it into two inclusions.

$$V+V \subset W$$

It is true due to the convexity of $$W$$ with $$\lambda = \frac{1}{2}$$.

Second inclusion is also trivial, because each $$w \in W$$ can be written as $$\frac{1}{2}w + \frac{1}{2}w$$.

Is my reasoning correct?

• Yes that looks correct. Oct 22 '20 at 15:40