Mathematics Asked by Daron on October 14, 2020

Suppose we have two convex sets $A,B$. It is straightforward to show the sum $A+B = {a+b:a in A,bin B}$ is convex. For take any $c=a+b$ and $c’=a’+b’$ and consider the convex combination

$$xc+(1-x)c’ = x(a+b)+(1-x)(a’+b’)$$ $$= big(xa+(1-x)a’big) + big( xb+(1-x)b’big).$$

By convexity the two summand are in $A,B$ respectively. Hence the sum is in $A+B$ as required.

Now suppose $A$ is $alpha$-strongly convex and $B$ is $beta$-strongly convex. That means the for each $a,a’ in A$ and $x in [0,1]$ that the ball centred at $xa+(1-x)a’$ of radius $alpha x(1-x)$ is contained in $A$. Likewise for each $b,b’ in B$ and $x in [0,1]$ that the ball centred at $xb+(1-x)b’$ of radius $beta x(1-x)$ is contained in $B$.

Can we say anything about the strong-convexity parameter of $A+B$? By that I mean $inf{ gamma ge 0: A+B$ is $gamma$-strongly convex $}$.

I'm going for the largest strong convexity constant, since any smaller value also works.

If $A$ is $alpha$-strongly convex and $B$ is $beta$-strongly convex, then $A+B$ is $gamma:=max{alpha,beta}$-strongly convex.

**Proof:** Let $a_1,a_2in A$, let $b_1,b_2in B$, and let $etain[0,1]$. For notational simplicity, I will set $a:=eta a_1+(1-eta)a_2$ and $b:=eta b_1+(1-eta)b_2$. Now suppose that $z$ is in the ball centered at $eta(a_1+b_1) + (1-eta)(a_2+b_2)=a+b$ with radius $eta(1-eta)gamma$. It suffices to show that $zin A+B$. By the construction of $gamma$ and definition of $z$, either

begin{equation} |a+b-z|=|a - left(z - bright)| leq eta(1-eta)gamma=eta(1-eta)max{alpha,beta} =eta(1-eta)alpha, tag{1} end{equation} or, begin{equation} |a+b-z|=|b - left(z - aright)|leq eta(1-eta)beta. tag{2} end{equation}

If (1) holds, then $z-b$ is in the ball centered at $a$ with the proper radius. Since $A$ is strongly convex, this implies $z-bin A$ and hence $zin A+bsubset A+B$. Likewise, if (2) holds, then $z-a$ is in the ball centered at $b$ with the proper radius. Since $B$ is strongly convex, we find $z-ain B$ and hence $zin B+asubset A+B$.

Answered by Zim on October 14, 2020

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