# Bounded projector in Hilbert space

### Solution:

So here is a completely different approach. By writing $$H=M\oplus M^\perp$$, we can see the elements of $$B(H)$$ as $$2\times 2$$ block matrices. The fact that $$P$$ is an idempotent with range $$M$$ gives us $$P=\begin{bmatrix} I&B\\0&0\end{bmatrix}.$$ The operator $$B:M^\perp\to M$$ is bounded, since $$P$$ is bounded. Of course, $$M=\left\{\begin{bmatrix} x\\0\end{bmatrix}:\ x\in M\right\}.$$Let us determine $$N$$. We know that $$N=(\ker P)^\perp$$. And $$\ker P=\left\{\begin{bmatrix} x\\y\end{bmatrix}:\ x+By=0 \right\}.$$ This allows us to write $$\ker P=\left\{\begin{bmatrix} -Bz\\z\end{bmatrix}:\ z\in M^\perp\right\}.$$ If $$\begin{bmatrix} x\\ y\end{bmatrix}\in (\ker P)^\perp$$ we have, for all $$z\in M^\perp$$ $$0=\left\langle\begin{bmatrix} x\\ y\end{bmatrix},\begin{bmatrix} -Bz\\ z\end{bmatrix} \right\rangle=\langle x,-Bz\rangle+\langle y,z\rangle=\langle -B^*x+y,z\rangle.$$ Then $$-B^*x+y\in M^{\perp\perp}=M$$ (here we use that $$M$$ is closed). Now, since $$B$$ maps $$M^\perp$$ to $$M$$, its adjoint $$B^*$$ maps $$M$$ to $$M^\perp$$. So $$-B^*x+y\in M^\perp$$. As $$M\cap M^\perp=\{0\}$$, we get that $$-B^*x+y=0$$. In other words, $$N=(\ker P)^\perp=\left\{\begin{bmatrix} x\\ B^*x\end{bmatrix}:\ x\in M\right\}.$$ Now, given $$m\in M$$ and $$n\in N$$, we have $$m=\begin{bmatrix} x\\ 0\end{bmatrix},\ \ \ \ n=\begin{bmatrix} z\\ B^*z\end{bmatrix}$$ for some $$x,z\in M$$. Then \begin{align} \sup_{n\in N\setminus\{0\}}\frac{|\langle m,n\rangle|}{ \|n\|} &=\sup_{z\in M}\frac{|\langle x,z\rangle|}{\sqrt{\|z\|^2+\|B^*z\|^2}}\\[0.3cm] &\geq\sup_{z\in M}\frac{|\langle x,z\rangle|}{\sqrt{\|z\|^2+\|B^*\|^2\,\|z\|^2}}\\[0.3cm] &=\frac1{\sqrt{1+\|B\|^2}}\,\sup_{z\in M}\frac{|\langle x,z\rangle|}{\|z\|}\\[0.3cm] &=\frac{\|x\|}{\sqrt{1+\|B\|^2}} =\frac{\|m\|}{\sqrt{1+\|B\|^2}} \end{align} (since $$x\in M$$, the equality $$\sup_{z\in M}\frac{|\langle x,z\rangle|}{\|z\|}=\|x\|$$ is an easy consequence of Cauchy-Schwarz). So, for any nonzero $$m\in M$$, \begin{align} \sup_{n\in N\setminus\{0\}}\frac{|\langle m,n\rangle|}{\|m\|\, \|n\|} \geq\frac{1}{\sqrt{1+\|B\|^2}}=\frac1{\|P\|}. \end{align} To see this last equality, note that $$\|P\|^2=\|P^*\|=\|PP^*\|=\|I+BB^*\|=1+\|BB^*\|=1+\|B\|^2.$$

Let us first prove the following lemma.

Lemma: Assume $$F,G$$ are closed subspaces of a Hilbert space such that $$F\cap G^\perp = \{0\}$$. Then there is $$\epsilon >0$$ such that for every $$f \in F\setminus \{0\}$$, $$\sup_{g \in G\setminus \{0\}} \frac{\vert \langle f,g\rangle\vert}{\Vert f \Vert\cdot \Vert g \Vert}\geq \epsilon$$.

Proof: Consider the map $$\phi : F \rightarrow G^*$$ sending each $$f$$ to the linear form $$g \mapsto \langle f,g\rangle$$. The map $$\phi$$ is anti-linear, bounded by one, and injective by our assumption.

Now, $$G$$ is a Hilbert space. So let us denote by $$\flat$$ the anti-linear map $$G^* \rightarrow G$$ sending any linear form $$\lambda$$ on $$G$$ to the vector $$\flat(\lambda)$$ such that, for every $$v\in G$$, $$\lambda(v) = \langle \flat(\lambda),v\rangle$$. We can check that $$\flat \circ \phi = P_G$$, where $$P_G$$ is the (restriction on $$F$$ of the) orthogonal projector on $$G$$. Indeed, if $$f \in F$$, then for every $$g \in G$$, $$\langle f ,g \rangle = \langle P_G(f) + (f- P_G(f)),g\rangle = \langle P_G(f),g\rangle$$.

The projector $$P_G$$ has closed range, so, since $$\flat$$ is an (anti-linear) isomorphism, $$\phi$$ has a closed range too. So the co-restriction $$\phi : F \rightarrow \phi(F)$$ is a bounded, bijective anti-linear operator between Hilbert spaces, so by the Banach isomorphism theorem, its inverse is continuous. So $$\phi$$ is bounded below, that is, there is $$\epsilon > 0$$ such that for every $$f \in F$$, $$\Vert \phi(f)\Vert \geq \epsilon \Vert f \Vert$$. Now, $$\Vert \phi(f)\Vert = \sup_{g \in G\setminus \{0\}} \frac{\vert \langle f,g\rangle\vert}{\Vert g \Vert}$$ so we are done.

Now, in your setting, if you assume that $$P$$ is nonzero, then $$M \cap N^\perp = im(P) \cap \ker(P) = \{0\}$$, so you can just apply the lemma to $$M = F$$ and $$N = G$$.