# 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.