Here is the proposition: assume that $V$ and $W$ are both Fréchet spaces, $u$ is a continuous map from $V$ to $W$. If we suppose that $u$ is injective, and $dim_C$$W/u(V)$ is finite, then $u(V)$ is a closed subspace of $W$.
I thought out a proof, but did not use the condition that $dim_C$$W/u(V)$ is finite, so I hope anyone could help me to find out the reason. The following is my proof:
The reference is Functional Analysis written by Walter Rudin. In this book, there is the Theorem $1.27$ on Page $21$, which states that Suppose that $Y$ is a subspace of a topological vector space $X$, and $Y$ is an Fréchet space (in the topology induced from $X$). Then $Y$ is a closed subspace of $X$.
So I tried to prove that the image of $u$ is also a Fréchet space in $W$, then I can use the Theorem $1.27$ to prove the proposition in the beginning.
We notice that there is a generalized open mapping theorem in the same book, it is the Theorem $2.11$ on Page $48$. It supposes that
- $X$ is an Fréchet space.
- $Y$ is a topological vector space.
- $u$ from $X$ to $Y$ is continuous and linear.
- $u(X)$ is of the second category in $Y$.
- $u$ is an open mapping
- $Y$ is an Fréchet space.
Here "be of the second category" is the same as in the definition of Baire Category.
Now if we replace $X$ by $V$, replace $Y$ by the image of $u$, and view $u$ as a 1-1 map from $V$ to $u(V)$, then according to the open mapping theorem, $u(V)$ is a Fréchet space, so it is closed in $W$.
As I am not familiar with functional analysis,I do not know where I am wrong. Please tell me, thank you very much!