I'm reading Functional Analysis book of Walter Rudin, and there's one point in this book that I don't know why he states that. Here is the statement:
$f$ is a linear mapping from F-space $X$ into a topological vector space $Y$ such that for every neighborhood $W$ of $0$ in $Y$, there exist neighborhood $V$ of $0$ in $X$ such that $f(V) \subset \overline{W}$. Then $f$ is continuous.
If $f$ satisfy the condition $f(V) \subset W$, then it's trivially true due to the definition of continuity and linear property. But with the closure on the right side, I don't know why. Can someone help me clarify this. Thanks so much.
Using Theorem 1.11, If $\mathscr{B}$ is a local base for a topological vector space $X$, then every member of $\mathscr{B}$ contains the closure of some member of $\mathscr{B}$, we obtain that a topological vector space is regular/a $T_3$ space.
Given any neighbourhood $U$ of $0$ in $Y$, choose a neighbourhood $W$ of $0$ with $\overline{W}\subset U$ per 1.11. Then there exists a neighbourhood $V$ of $0$ in $X$ with $f(V) \subset \overline{W}\subset U$ by assumption, hence $f^{-1}(U)$ contains a neighbourhood of $0$ in $X$, that is, $f$ is continuous in $0$.
That holds for all topological vector spaces, $F$-space or not.