prove closure $\overline{f(V)}$ is a neiborhood of origin in Topological vector space Let X,Y be topological vector space,Y is Baire space,let $f$ be a continuous linear map $f:X\to Y$,assume $f(X)$ has non empty interior.
Prove given any neiborhood $V$ of origin for $X$, $\overline{f(V)}$ is neiborhood of origin for $Y$
(I think my proof is almost done,but not very clear)
My attempt,first since $f(X)$ has non empty interior the map is onto,now the question is how to use the fact that $Y$ is Baire space?I know we need to construct something like $\bigcup U_n$ which each of them is closed and no interior.But I don't know how to construct this sequence $U_n$ (It may be an ascending sequence?)
each of them seems be the set $\overline{f(V)}$ which is closed,and assume no interior contains in them,then if countable union to covers $Y$.it must has empty interior,hence contradiction?
To construct the countable union,note that any $V$ exist balanced absorbing neiborhood $A$ contains in it if we can show for this $A$ we have $\overline{f(A)}$ is neiborhood of origin,then it will also holds for $\overline{f(V)}$ Since absorbing we know for this $A$ ,$\bigcup_n nA$ covers $X$ hence $f(\bigcup_n nA) = Y $
To complete the proof,we need to show to fact:

*

*if $\overline{f(A)}$ contains no interior then $n\overline{f(A)}$ also contains no interior,since they are homeomorphic.

*if zero is not interior for $\overline{f(A)}$ then all point can not be interior for it.

 A: You are right that since $f(X)$ (which is a linear subspace of $Y$) has no empty interior then $f(X)=Y$: indeed, let $y_0\in f(X)$ so that there exists an open neighborhood $V\subset f(X)$. If $y\in Y$ then we can take the points $y_0+\lambda y$, $\lambda\in\mathbb{C}$. As $\lambda\to0$ we have that $y_0+\lambda y\to y_0$, so there exists some non-zero $\lambda\in\mathbb{C}$ so that $y_0+\lambda y\in V$. But then $y_0+\lambda y\in f(X)$, so $y\in f(X)$.
We give a definition: If $Z$ is a vector space and $E\subset Z$ we say that $E$ is absorbent when $Z=\bigcup_{r>0}rE$. We say that $E$ is balanced when for any $x\in E$ and $\alpha\in\overline{\mathbb{D}}$ we have that $\alpha x\in E$.
Note that if $Z$ is a TVS and $E$ is balanced then $\text{int}(E)$ is also balanced: indeed, if $x\in\text{int}(E)$ then there exists an open $V\subset E$ so that $x\in V$. If $\alpha\in\mathbb{C}$ with $|\alpha|\leq1$, we have that $\alpha\cdot x\in\alpha\cdot V\subset E$ and $\alpha\cdot V$ is open, thus $\alpha\cdot x\in\text{int}(E)$.

Lemma: Let $Z$ be a Baire TVS. If $E\subset Z$ is absorbent, then $E$ is a Baire space with the subspace topology.

Proof of the lemma: Let $(F_n)$ be a sequence of closed sets such that $E\subset\bigcup_nF_n$. For $k\in\mathbb{N}$ we have that $kE\subset\bigcup_nkF_n$, therefore since $Z=\bigcup_kkE$ (because $E$ is absorbent) we have that $Z=\bigcup_{n,k}kF_n$. Since $Z$ is a Baire space, there exists a pair of integers $k,n$ so that $\text{int}(kF_n)\neq\emptyset$. But $kF_n$ is homeomorphic to $F_n$, so $\text{int}(F_n)\neq\emptyset$. This shows that $E$ is a Baire space.

Claim: Let $X$ be a TVS and let $V$ be a neighborhood of $0$ in $X$. Then $V$ contains a balanced neighborhood of $0$.

Proof of the claim: Indeed, since scalar multiplication is a continuous operation $m:\mathbb{C}\times X\to X$ we have that $m^{-1}(V)$ is open in $\mathbb{C}\times X$ and contains $(0,0)$ so there exists $r>0$ and a neighborhood $W\subset X$ of $0$ such that $D(0,r)\times W\subset m^{-1}(V)$, so $m(D(0,r)\times W)\subset V$. But $m(D(0,r)\times W)=\{\alpha\cdot x: |\alpha|<r, x\in W\}$ is open, contains $0$ and if $\beta\in\mathbb{C}$ satisfies $|\beta|\leq1$ then for any $\alpha\in D(0,r)$ and $x\in W$ we have that $\beta\cdot(\alpha\cdot x\in m(D(0,r)\times W)$, since $|\alpha\beta|\leq r$.
So let $V$ be a neighborhood of $0$ in $X$ and take $U$ a balanced neighborhood of $0$ so that $U\subset V$. Now let $y\in Y$. As we said, $f$ is onto, so $y=f(x)$ for some $x\in X$. Since $\lambda x\to0$ as $|\lambda|\to0$, we have that there exists some $\lambda\in\mathbb{C}$ so that $\lambda x\in U$. So $y=\frac{1}{\lambda}f(\lambda x)\in\frac{1}{\lambda}f(U)\subset\frac{1}{\lambda}\overline{f(U)}$ and therefore $\overline{f(U)}$ is absorbent in $Y$ and it is also balanced, since $U$ is balanced. By the lemma we have that $\overline{f(U)}$ is a Baire space, so $\text{int}(\overline{f(U)})\neq\emptyset$. But since $\overline{f(U)}$ is balanced, its interior is also balanced, thus $0\in\text{int}(\overline{f(U)})\subset\text{int}(\overline{f(V)})$ which is what we wanted to show.
