Question

I've seen this statement of Radstrom cancellation: if $A +C \subset B+C$ where $A,B$ are convex, $B$ is closed, and $C$ is bounded, then $A \subset B.$

Is it essential that $A$ be convex?

Answer 1

No. Actually, the original version of the lemma does not require the convexity of A.

Reference: Hans Rådström (note spelling): An embedding theorem for spaces of convex sets. Proc. Amer. Math. Soc. 3 (1952) 165–169, MR0045938. doi:10.2307/2032477.

Answer 2

Alex, your idea of using a separation theorem can be turned into a full proof as follows:

As you said, $A \not\subset B$ iff there is a $a \in A$ and vector $x$ such that $\sup_{b \in B} \langle b,x \rangle < \langle a,x \rangle$. But then,

\begin{align} \sup_{z \in B + C} \langle z, x \rangle &= \sup_{b \in B, c \in C} (\langle b, x \rangle + \langle c, x \rangle ) \\ &= \sup_{b \in B} \langle b, x \rangle + \sup_{c \in C} \langle c, x \rangle \\ &< \langle a, x \rangle + \sup_{c \in C} \langle c, x \rangle \\ &= \sup_{z \in a + C} \langle z, x \rangle \\ &\leq \sup_{z \in B + C} \langle z, x \rangle, \end{align}

a contradiction.

Answer 3

Hmm, I have an idea for a proof which works even when $A$ isn't convex:

$A \not\subset B$ iff there is a $a \in A$ and vector $x$ such that $\langle a, x\rangle >0$ and $\langle b, x \rangle \leq 0$ for all $b \in B$ (since $B$ is a closed convex set). Let $b \in B$ and $c \in C$ maximize $\langle b + c, x \rangle$, then $\langle a + c, x \rangle > \langle b + c, x \rangle$, so there is a point in $A + C$ that isn't in $B+C$.

This implies that if $A \not\subset B$ then $A + C \not\subset B+C$. Of course, there aren't really $b,c$ which maximize that quantity above, but I believe you could make this rigorous using approximation arguments.

Answer 4

Replace $A$ by its convex hull and use the result for convex $A$.

Edit: as observed in the comments, this is not enough.