If $H$ Hilbert, $N\subset H$ closed subspace, $T\colon H\to H$ bounded, $\dim(\ker(T))<\infty$ and $T(H)$ closed, then $T|_{N}$ has same properties 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.
 A: 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 $T(x)=y$.
A: 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.
