# Bound operator norm

Consider the Hilbert space $\ell^2(\mathbb{N})$ and let $P:\ell^2(\mathbb{N})\to\ell^2(\mathbb{N})$ be an orthogonal projection. Let $(\delta_i)_{i\in\mathbb{N}}$ denote the canonical orthonormal basis of $\ell^2(\mathbb{N})$ and define for $C\subset\mathbb{N}$ $$Q_C=\sum_{k\in C}Q_k,$$ where $Q_kx=\langle\delta_k,x\rangle\delta_k$. Put differently, $Q_C$ is the orthogonal projection on $\ell^2(C)=\overline{\mathrm{span}\{\delta_k\}_{k\in C}}$.

Now I want to prove: $$\|Q_CP(\mathrm{Id}-Q_C)PQ_C\|\leq \frac14.$$

My attemp: By the min-max theorem, $\|Q_CP(\mathrm{Id}-Q_C)PQ_C\|=\sup_x\langle x,(Q_CP(\mathrm{Id}-Q_C)PQ_C)x\rangle$ for $\|x\|=1$. By the outermost $Q_C$'s, we may restrict from the beginning to $x\in\overline{\mathrm{span}\{\delta_k\}_{k\in C}}$. In summary, $$\|Q_CP(\mathrm{Id}-Q_C)PQ_C\|=\sup_{\substack{x\in\ell^2(C)\\\|x\|=1}}\langle x,P(\mathrm{Id}-Q_C)Px\rangle=\sup_{\substack{x\in\ell^2(C)\\\|x\|=1}}\left[\|Px\|^2-\sum_{k\in C}|\langle Px,\delta_k\rangle|^2\right].$$ However, this approach does not look very fruitful...

The way I would see your inequality is as $$\|(I-Q_C)PQ_C\|\leq\frac12,$$ since $$\|Q_CP(I-Q_C)PQ_C\|=\|Q_CP(I-Q_C)^*(I-Q_C)PQ_C\|=\|(I-Q_C)PQ_C\|^2.$$ And this is basically saying that the off diagonal entries of a projection cannot contribute more than $1/2$ to the norm. In spirit, this happens because the equality $P^2=P$ gives, for every diagonal entry, $$P_{kk}-P_{kk}^2=\sum_{j\ne k} |P_{kj}|^2.$$ To make this into a proof, we write $P$ in terms of the decomposition $H=Q_CH\oplus (I-Q_C)H$, $$P=\begin{bmatrix} A&B\\ B^*&C\end{bmatrix},$$ with $A,C\geq0$ (since $P$ is positive). If we now look at the equality $P^2=P$ at the $1,1$ entry, we get $$A^2+BB^*=A.$$ So $BB^*=A-A^2$. As $A$ is a positive contraction (from $P$ being a projection), we have $$\sigma(A-A^2)\subset\{\lambda-\lambda^2:\ \lambda\in[0,1]\}.$$ For all such $\lambda$ we have $\lambda-\lambda^2\leq1/4$, so $$\|B\|^2=\|BB^*\|=\|A-A^2\|\leq\frac14,$$ so $\|B\|\leq1/2$. Thus $$\|(I-Q_C)PQ_C\|=\|B\|\leq\frac12.$$
If $M$ and $N$ are subspaces in generic position in a Hilbert space $H$, with respective projections $\tilde{P}$ and $Q$, then there exists a Hilbert space $K$, and there exist positive contractions $S$ and $C$ on $K$, with $S^2 + C^2 = 1$ and $\ker S=\ker C=0$, such that $\tilde{P}$ and $Q$ are unitarily equivalent to $$\begin{pmatrix}1& 0\\0&0\end{pmatrix},\qquad \begin{pmatrix}C^2&CS\\CS&S^2\end{pmatrix}$$ respectively.
What's not written in the theorem but follows by the proof is that $K=\mathrm{ran}P$ and that $\tilde{P}$ and $Q$ admit the same unitary $U:H\to K\oplus K$. Moreover, $C$ and $S$ commute. Then using Halmos' theorem with $\tilde{P}=Q_C$, $Q=P$, we see by a little computation $$\|Q_CP(\mathrm{Id}-Q_C)PQ_C\|=\left\|\begin{pmatrix}C^2S^2&0\\0&0\end{pmatrix}\right\|=\|C^2-C^4\|.$$ An easy exercise in spectral calculus shows $\sigma(C^2-C^4)\subset [0,1/4]$ from which the result follows.