Closed images

I recently faced a problem concerning linear operators on Hilbert spaces, and I proved a small result. It finally appeared that I didn’t need it, but I have decided to record it anyway. Since it is very simple, I have absolutely no doubt that it has already been proved, but in keeping with the spirit of this blog, I don’t really care. I will keep it short, so there will be no discussion or motivation.

Proposition

Let \mathcal{H}_1 and \mathcal{H}_2 be two Hilbert spaces. Let T:\mathcal{H}_1\to \mathcal{H}_2 be a linear bounded operator. If Im(T)=T(\mathcal{H}_1) is a closed subspace of \mathcal{H}_2 and if Ker(T) is finite dimensional, the image by T of any closed subspace of \mathcal{H}_1 is a closed subspace of \mathcal{H}_2.

The proof I found relies on the following Lemma.

Lemma

If E is a normed linear space, F a closed subspace of E and G a finite dimensional subspace of E, then F+G is a closed subspace of E.

Proof

Let us first note that we only have to deal with the case where F\cap G=\{0\}. Indeed, in the general case, we can choose G' a complementary vector space in G relative to F\cap G. The space G' is obviously finite dimensional and F+G'=F+G.

Let us  assume from now on that F\cap G=\{0\}. We denote by B_{G} the unit ball of G. It is a compact set for the topology given by the original norm on E. The application x \mapsto d(x,F) (for the distance deduced from the norm) is continuous and positive on B_G. Therefore, there exists \varepsilon>0 such that d(x,F)\ge \varepsilon for all x \in B_G. By homogeneity, this implies that d(x,F)\ge \varepsilon\|x\| for all x \in G.

Now let (x_n) be a sequence of vectors in F+G that converges to x\in E. There exist (y_n) and (z_n), sequences of vectors in F and G respectively, such that x_n=y_n+z_n. Now, according to the previous paragraph,
\displaystyle \|z_n\| \le \frac{1}{\varepsilon}d(z_n,F)\le \frac{1}{\varepsilon}\|y_n+z_n\|=\frac{1}{\varepsilon}\|x_n\|.
The sequence (z_n) is therefore bounded. Since G is finite dimensional, this implies the existence of z\in G and (z_{n_k}) a subsequence such that z_{n_k} \to z. And now we are finished: y_{n_k}=x_{n_k}-z_{n_k} \to x-z, and therefore, F being closed, x-z \in F. We obtain x = (x-z)+z \in F+G.

We can now prove the proposition. Let us consider \hat{T}: Ker(T)^{\perp} \to Im(T), the restriction of T. Since Ker(T)^{\perp} and Im(T) are closed subspaces of \mathcal{H}_1 and \mathcal{H}_2 respectively (this is not automatic for Im(T), we had to make the hypothesis), they are both Hilbert spaces. Furthermore, \hat{T} is continuous, and bijective by construction. According to Banach’s Closed Graph Theorem, it has a bounded inverse \hat{T}^{-1}. This implies, among other things, that the image of closed subsets of Ker(T)^{\perp} are closed subsets of  Im(T), and thus of \mathcal{H}_2.

Let us now consider F, a closed subspace of \mathcal{H}_1. It is easy to check that T(F)=T((F+Ker(T))\cap Ker(T)^{\perp}). According to the lemma, F+Ker(T) is closed and thus (F+Ker(T))\cap Ker(T)^{\perp} is a closed subspace of Ker(T)^{\perp}. According to what we have just seen, T((F+Ker(T))\cap Ker(T)^{\perp}) is closed, which conclude the proof.

I think that the proposition  could be extended to Banach spaces. It should be enough to replace the orthogonal complement by the quotient space, but I need to revise some Functional Analysis before writing it up. The lemma leads to an interesting question: when is the sum of two closed vector spaces a closed vector space. A mathoverflow discussion deals with this topic, but here again I need to think about it a little longer before writing something.

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