Show in Hilbert space: $\|T(h)\| \geq C \|h\|$ for all $h \in (\operatorname{Ker}T)^\perp$ for closed range operator

Question

I need your help in the following task:

Let $H$ be a Hilbert space and let $T \in L(H,H)$ have a closed range.

Show that there exists $C>0$ such that $\|T(h)\| \geq C \|h\|$ for all $h \in (\operatorname{Ker}T)^\perp$.

I just collected some properties since I don't know how to solve the problem.

Let $K = \operatorname{Ker}T$. I know that $K$ is a closed subspace of $H$ since $T$ is linear and bounded, hence continuous. $K$ can be written as $K = T^{-1}(\{0\})$, so it is a preimage of the closed set $\{0\}$ under a continuous function, hence it is closed.

Since $K$ is closed subspace, I can apply the projection theorem and I get that every $x \in H$ can be uniquely written as $x=x_K + x_{K^\perp}$ with $x_K \in K$ and $x_{K^\perp} \in K^\perp$.

If I now take an arbitrary $h \in K^\perp$ it can be written as $h = x - x_K$ where $x \in H$ and $x_K \in K$. Moreover I get $T(h)=T(x-x_k)=T(x)-T(x_K) = T(x)$ because $x_K \in K$.

In addition I thought about the basic definition of a bounded operator, which states that there is some constant $D>0$ such that $\|T(x)\| \leq D |\|x\|$ for all $x \in H$.

This would give $\|T(h)\| = \|T(x)\| \leq D \|x\| $, but this doesn't help me.

I also thought about using the operator norm of $T$ which would give the inequality $\|T(x)\| \leq \|T\|_{op}||x|| $

Moreover I don't know how to include the assumption that $T$ has closed range.

I hope you can give me some good hints or help me rearranging the facts above to get a smooth proof.

Answer 1

Hints: Define $S: H|Ker (T) \to T(H)$ by $S(x+Ker (T))=Tx$. Verify that this is bijective continuous linear map. By Open Mapping Theorem its inverse is continuous. Now verify that the sated inequality holds with $C=\|S^{-1}\|$.

Answer 2

Since the image of $T$ is closed, the restriction of $T$ to the orthogonal of $Ker(T)$ that we call $U$ is a continuous bounded map between Banach spaces. The open mapping theorem implies that it is invertible. For every $x\in U, \|T^{-1}(T(y))\|=\|y\|\leq D\|T(y)\|$. Take $D={1\over C}$.

Answer 3

Define the map $$ \tilde{T}:H/\mathrm{Ker}(T) \to \mathrm{Range}(T), [h]\mapsto \tilde{T}([h])=T(h). $$ Because $T$ has closed range, then $\tilde{T}$ is a continuous bijection from the Banach space $H/\mathrm{Ker}(T)$ to the Banach space $\mathrm{Range}(T)$. The norm on the second space is simply the restriction of the norm from $H$, whilst on the first space we are considering $$ \| [h]\|_{H/\mathrm{Ker}(T)} = \inf_{k\in \mathrm{Ker}(T)} \|h+k\|. $$ In particular, if $h\in \mathrm{Ker}(T)^\perp$ we have $$ \|h+k\|^2 = \|h\|^2+\|k\|^2 + 2\langle h,k\rangle = \|h\|^2+\|k\|^2, $$ which means that $$ \| [h]\|_{H/\mathrm{Ker}(T)} = \|h\|. $$ By the open mapping theorem, $\tilde{T}$ has a continuous inverse.

For every $h\in \mathrm{Ker}(T)^\perp$ we have $$ \|h\|_H = \|[h]\|_{H/\mathrm{Ker}(T)} = \|\tilde{T}^{-1}( \tilde{T}(h))\|_{H/\mathrm{Ker}(T)} \leq C \|T(h)\|_{\mathrm{Range}(T)} = C \|T(h)\|. $$