Here LCTVS stands for locally convex topological vector space. I'm getting no idea how to approach the problem. I have tried using Minkowski functional but couldn't get much. I know a result (Separation Theorem) which says-
$X$ be TVS and $A_1,A_2\subseteq X$ be disjoint, nonempty, convex. Then
(a) If $A_1$ is open, $\exists f\in X^*$ and $\lambda\in\Bbb{R}$ such that $\text{Re}f(x)<\lambda\le \text{Re}f(y)\ \forall x\in A_1,y\in A_2$
(b) If $A_1$ compact, $A_2$ closed, $X$ locally convex, then $\exists f\in X^*$ and $\lambda_1,\lambda_2\in\Bbb{R}$ such that $\text{Re}f(x)<\lambda_1<\lambda_2< \text{Re}f(y)\ \forall x\in A_1,y\in A_2$
I don't know whether the above theorem help me anything here. Can anyone give me any idea or hint to solve the problem Thanks for help in advance.
For each $a\in A$ with $a+K\subseteq B+K$ one has to show $a\in B$ and (after replacing $K$ by $a+K$ and $B$ by $B-a$) we may assume $a=0$. Assuming $0\notin B$ the sepration theorem gives $\varphi \in X^*$ with $\text{Re}\varphi(b)\ge \lambda >0$ for all $b\in B$. Fix $k_0\in K\subseteq K+B$ and choose $k_1\in K$ and $b_1\in B$ with $k_0=k_1+b_1$. Recursively, one gets $k_n\in K$ and $b_n\in B$ with $k_0=k_n+b_1+\cdots+b_n$. Since $K$ is compact and $\varphi$ is continuous, Re$\varphi(k)\ge c$ for some real $c$ which yields $\text{Re}\varphi(k_0)\ge c+n\lambda \to\infty$ which is obviously absurd.
We did not need compactness of $K$ but only boundedness. On the other hand, $K\neq\emptyset$ is obviously crucial (since $\emptyset+A=\emptyset$).