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Let $H$ be a Hilbert space (over $\mathbb{C}$) and $N\subset H$ a closed subspace. (So N is also Hilbert.) Suppose that $T\colon H\to H$ is a bounded linear operator such that $\dim(\ker(T))<\infty$ and $T(H)$ is closed. I want to prove that $T|_{N}\colon N\to H$ has same properties, i.e. $\dim(\ker(T|_{N}))<\infty$ and $T(N)$ is closed.

Since $\ker(T|_{N})=N\cap\ker(T)$, it is clear that $\dim(\ker(T|_{N}))<\infty$. However, I have no idea how to prove that $T|_{N}(N)=T(N)$ is closed in $H$.

Any suggestions are greatly appreciated.

Calculix
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2 Answers2

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Let $p:H\rightarrow H/KerT$ the quotient map,since $dim(KerT)$ is finite, $p(N)$ is closed. See thes second answer here: Is the closedness of the image of a Fredholm operator implied by the finiteness of the codimension of its image?

Consider the map $\bar T:H/Ker T\rightarrow Im(T)$ it is a bijective continuous map betwwen Banach spaces. The open mapping theorem implies that its inverse is continuous. Let $x_n$ be a sequence of elements of $N$ such that $y=lim(T(x_n))=lim(\bar T(p(x_n))$, we have $(\bar T)^{-1}(y))=lim(\bar T)^{-1}(p(x_n))=lim_n(p(x_n))$ since $p(N)$ is closed, we deduce that $\bar=(\bar T)^{-1}(y)\in p(N)$, this implies that there exists $x\in N$ with $p(x)=\bar x$ and $T(x)=y$.

Tsemo Aristide
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You can show this using sequences. Given a convergent sequence $\{ h_k \}_{k=1}^\infty\subseteq T[N]$ to an element $h_0\in H$, you want to show that $h_0\in T[N]$. Notice that since $T[H]$ is closed, $h_0\in T[H]$. You can write $T(n_k)=h_k$ where $\{n_k \}\subseteq N$. If it has a sub-sequence of elements in $\ker(T)$ then $h_0=0\in T[N]$.

Otherwise, look at $T\vert_W $ where $W\oplus \ker(T)=H$. Then $W$ is closed since $\dim\big( \ker(T) \big)<\infty$, and $T\vert_W:W\rightarrow T(H)$ is a bounded bijective linear map between Hilbert spaces, and has a bounded inverse by a corollary of the mapping theorem. This implies that the sequence $\{ n_k \}$ converges.

Keen-ameteur
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  • Thanks for your reply. Why do you need that $\dim(\ker(T))<\infty$? Isn't it always true that $H=\ker(T)^{\perp}\oplus\ker(T)$? (Since $T$ is continuous, $\ker(T)$ is closed.) And the orthogonal complement of any set (for example $W:=\ker(T)^{\perp}$) is always closed. – Calculix Dec 01 '19 at 13:55
  • I had the Banach space case in mind where being complemented by a closed subspace is harder. Now that I look at it again, I tend to agree with what you say about it being unnecessary. – Keen-ameteur Dec 01 '19 at 14:06
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    It need not be the case that the sequence ${n_k}$ converges. We have $n_k = (T\lvert_W)^{-1}(h_k) + z_k$ for some $z_k \in \ker T$, and ${z_k}$ needn't even be bounded. And $\dim \ker T < \infty$ is necessary even for Hilbert spaces, if $\ker T$ and $\operatorname{im} T$ are both infinite-dimensional then it may be that $T(N)$ is not closed for some closed $N$. – Daniel Fischer Dec 01 '19 at 14:32
  • You're right. I'm wondering if there is something to salvage from this, or is it better to erase this answer. – Keen-ameteur Dec 01 '19 at 14:45
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    It's certainly salvageable. I prefer to look at it via a quotient space, but one can translate it into language avoiding quotients. Look at $X = H/\ker T$ and the induced $\tilde{T}\colon X \to T(H)$, which is an isomorphism (in the category of topological vector spaces). The crucial point is that $\pi(N)$ – where $\pi \colon H \to X$ is the canonical projection – is a closed subspace of $X$, because $\ker T$ is finite-dimensional (thus $\pi^{-1}(\pi(N)) = N + \ker T$ is closed). Hence $T(N) = \tilde{T}(\pi(N))$ is closed in $T(H)$. ($H$ can be Banach in this, needn't be Hilbert.) – Daniel Fischer Dec 01 '19 at 15:40