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Let $A, B$ be two nonempty convex set in normed space $X$. We always have $$ \text{int}(A)\bigcap B\ne\emptyset\;\Longrightarrow\; 0\in\text{int}(A-B). $$ Indeed, suppose that $\text{int}(A)\bigcap B\ne\emptyset$. Let $u\in\text{int}(A)\bigcap B$. Then there exists $r>0$ such that $B(u;r)\subset A$. Then $0\in B(u;r)-\{u\}\subset A-B$. Hence $0\in\text{int}(A-B)$. The reverse implication is not true in general. Indeed, let $A=\{0\}, B=X$. Then $\text{int}(A)\bigcap B=\emptyset$ but $0\in X=\text{int}(A-B)$.

Question Sufficient conditions on $A, B$ to have $$ \text{int}(A)\bigcap B\ne\emptyset\;\Longleftrightarrow\; 0\in\text{int}(A-B). $$

My attemption We hope we can use this equivalence $$ \text{int}(A)\bigcap B\ne\emptyset\;\Longleftrightarrow\; 0\in\text{int}(A)-B. $$ to solve my problem.

Thank you for all kind help and comments.

Blind
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1 Answers1

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If convex set $B$ contains more than one point, there is a convex set $A$ with empty interior such that $0 \in \text{int}(A-B)$. Namely, if $b_1, b_2 \in B$, let $f$ be a continuous linear functional such that $f(b_1) \ne f(b_2)$, and let $A = \{x: f(x) = (f(b_1) + f(b_2))/2$.

This shows that your conditions would have to involve $A$ as well as $B$.

Robert Israel
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