# Show $\alpha C + \beta C = (\alpha + \beta) C$ for a convex set $C$ and nonnegative $\alpha, \beta$.

Let $C\subset\mathcal{H}$ be a convex subset of a real Hilbert space $\mathcal{H}$ and let $\alpha,\beta$ be two nonnegative real numbers. I am asked to show that $\alpha C + \beta C = (\alpha + \beta) C$ where we have defined $\alpha C = \{\alpha x : x\in C\}$ and $C + D = \{x+y : x\in C, y\in D\}$.

It's clear to me that this property does not hold for nonconvex sets $C$, for example for the subset of $\mathbb{R}^2$ given by $C=\{(-1,-1),(1,1)\}$ it is trivial to show that $C+C \neq 2C$ by direct calculation.

So far I have tried to directly show subset inclusion in both directions.

Let $x\in (\alpha + \beta) C$. Then, $x=(\alpha + \beta)y$ for some vector $y\in C$. However, $(\alpha + \beta)y = \alpha y + \beta y \in \alpha C + \beta C\implies (\alpha + \beta)C \subset \alpha C + \beta C$.

In the other direction, I have the following, let $x\in \alpha C + \beta C$. Then, $x = \alpha y + \beta z$ with $y,z\in C$. I cannot see how convexity comes into play here? A hint is preferred to an actual answer.

Given $x\in\alpha C+\beta C$, the only points of $C$ you can immediately put your fingers on are $y$ and $z$ with $x=\alpha y+\beta z$.
Then the only additional points that convexity gives you are $ty+(1-t)z$ for $0\le t\le 1$.
The only points in $(\alpha+\beta)C$ that such a point gives you, are $(\alpha+\beta)(ty+(1-t)z)$ for $0\le t\le 1$.
Alas, you'd rather have $x$ ...
• Ok I see now, we know that $\frac{\alpha}{\alpha+\beta}\in]0,1[$ and $\frac{\beta}{\alpha+\beta}\in]0,1[$ as well. Furthermore we know $\frac{\alpha}{\alpha+\beta} + \frac{\beta}{\alpha+\beta} =1$ so we can write $\alpha y + \beta z = (\alpha + \beta)(\frac{\alpha}{\alpha+\beta}y + \frac{\beta}{\alpha+\beta}z)\in (\alpha+\beta)C$ – Tony Oct 13 '17 at 16:08