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 $N = G$.