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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 $Y$.

Then

  • $u(X)=Y$
  • $u$ 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!

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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 an Fréchet space, so it is closed in $W$.

We don't know yet that $u(V)$ is a Fréchet space in the topology induced by $W$. We know that it is a Fréchet space in the topology transported from $V$ via $u$, but that could theoretically be strictly finer than the topology induced by $W$.

However. If we can show that $u$ is an open mapping to $u(V)$ with its subspace topology induced by $W$, then we know that $u$ is a homeomorphism from $V$ to $u(V)$, and then we know that $u(V)$ is a Fréchet space, hence closed.

Let $F \subset W$ be an algebraic complement of $u(V)$. By assumption, $F$ is finite-dimensional, hence $F$ is a closed subspace. Now consider the space $X = W/F$. Since $F$ is a closed subspace of $W$, $X$ is a Fréchet space, and

$$V \xrightarrow{u} W \xrightarrow{\pi} X$$

is a continuous bijection between Fréchet spaces, hence a topological isomorphism, in particular open.

Now, for an open $U \subset V$, the set $u(U) = u(V) \cap \pi^{-1}((\pi\circ u)(U))$ is open in $u(V)$, since it is the intersection of $u(V)$ with the open (in $W$) set $\pi^{-1}((\pi\circ u)(U))$. So $u \colon V \to W$ is an embedding, hence $u(V)$ is a Fréchet space, hence closed.

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  • $\begingroup$ Thank you, but do you mean that in the open mapping theorem , the topology of $Y$ is changed? $\endgroup$ – Cao Nov 28 '13 at 17:21
  • $\begingroup$ No, the topology of $Y$ is what it is. The point is that here we don't know a priori that the open mapping theorem is applicable. There are two possibilities, either $u(V)$ is closed in $W$ [the above shows that that is the case], then everything is settled, $u(V)$ is a Fréchet space, $u$ is a homeomorphism from $V$ to $u(V)$, all fine; or $u(V)$ isn't closed, then we can't apply the open mapping theorem, then $u(V)$ is not of the second category in $\overline{u(V)}$. $\endgroup$ – Daniel Fischer Nov 28 '13 at 18:26
  • $\begingroup$ Thank you, I have not had a good understanding of the baire category. Thank you! $\endgroup$ – Cao Nov 28 '13 at 18:35

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