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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?

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    $\begingroup$ Yes that looks correct. $\endgroup$
    – runway44
    Oct 22 '20 at 15:40

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