Convex Equality

Question

if X is a vector space over say F then aset is "A" is convex iff (s+T)A=sA+tA where s and t are positive scalars.

I have done that if the equality holds then the result is true and if A is convex then the find side inclusion is obvious I got struck in proving the other side inclusion.

Answer 1

Let $sa+tb \in sA+tA$, with $a,b \in A$.

Since $A$ is convex and $\frac{s}{s+t} + \frac{t}{s+t} =1$, you have $$\frac{s}{s+t}a + \frac{t}{s+t}b \in A$$ Hence $$sa+tb = (s+t)\left( \frac{s}{s+t}a + \frac{t}{s+t}b \right) \in (s+t)A$$

So you have $sA+tA \subseteq (s+t)A$. The opposite inclusion is trivial, since $$(s+t)a=sa+ta \in sA+tA$$