Consider a Hilbert space $\mathcal{H}$ and let $A$ be a bounded operator on $\mathcal{H}$, $B$ a possibly unbounded operator with domain $\mathcal{D}(B) \subset \mathcal{H}$ such that the composition $AB$ admits an extension to a bounded operator on $\mathcal{H}$. Furthermore, let $P$ be an orthogonal projection on $\mathcal{H}$. Does it follow that the composition $APB$ is also a bounded operator on $\mathcal{H}$?
The scenario I am considering is something of the following sort. Consider the composition of operators $(-\Delta + 1)^{-1}(-\Delta)$ acting on the Hilbert space $L^2(\mathbb{R}^n)$. Clearly the negative laplacian $-\Delta$ is not defined on all of $L^2(\mathbb{R}^n)$ (say I define it with domain $\mathcal{D}(-\Delta) = H^2(\mathbb{R}^n)$). However, using the Fourier transform, it is fairly easy to show that the composition $(-\Delta + 1)^{-1}(-\Delta)$ is a bounded operator (since $0 \leq \frac{\lvert{k}\rvert^2}{\lvert{k}\rvert^2 + 1} < 1$). Now consider an orthogonal projection $P$. Is it also true that the operator $(-\Delta + 1)^{-1}P(-\Delta)$ is bounded? As soon as I deal with a projection the Fourier Transform fails to be a good tool, so I'm not sure which tool to use to approach the problem.
Intuitively, I believe the general case to be true. Even though $B$ may be unbounded, as the composition $AB$ is bounded it means that $A$ can "deal" with any "roughness" that $B$ finds in the extension of $AB$ to the entire Hilbert space. If I further restrict the output of $B$ by replacing it with $PB$, I restrict the set of possible "bad" outputs that get fed into $A$, so the same should hold. If a counterexample holds, I would love to see it (I have not been able to come up with one myself). I would appreciate any discussion on this topic. Thanks for reading!
No, you need more restriction on $P$.
One way to defeat your intuition is to have nontrivial eigenspaces for large eigenvalues of $B$ and projecting them to some intermediate subspace not "nearly orthogonal" to $\ker A$ (or the small spectrum part of $A^*A$). So $PB$ masks and massages the "bad output" to trick $A$ into picking them up.
As an example, consider $\mathcal{H}=\ell_2$, $B$ the diagonal operator with $Be_{2n-1}=e_{2n-1}$ and $Be_{2n}=ne_{2n}$ for all $n\geq 1$. Let $A$ be the orthogonal projection onto $\ker(B-I)$, so $AB=A$ is clearly bounded. Now if $P$ the orthogonal projection onto $\overline{\operatorname{span}}\{e_1+e_2,e_3+e_4,\dots,e_{2n-1}+e_{2n},\dots\}$, we easily compute $$ APBe_{2n}=AP(ne_{2n})=nA\frac{e_{2n-1}+e_{2n}}2=\frac{n}2e_{2n-1} $$ showing $APB$ is unbounded.