We have matrices $A,B\in\mathbb{R}^{n,m}$. Let $P\in\mathbb{R}^{n,n}$ and $Q\in\mathbb{R}^{m,m}$ be the orthogonal projection matrices in the column space of $A$ and the row space of $A$ (or column space of $A^{\top}$), respectively. I want to prove that $$ \|P(B-A)Q\|_*\leq \sqrt{\text{rank}(A)}\|B-A\|_F, $$ where $\|\cdot\|_*$ is the nuclear norm and $\|\cdot\|_F$ the Frobenius norm. Because $\|\cdot\|_*^2\leq \text{rank}(\cdot)\|\cdot\|_F^2$ and $\text{rank}(P(B-A)Q)\leq \text{rank}(A)$, then $$ \|P(B-A)Q\|_*\leq \sqrt{\text{rank}(A)}\|P(B-A)Q\|_F. $$ But, is it true that $\|P(B-A)Q\|_F\leq \|B-A\|_F$?
I calculated that $\|P(B-A)Q\|_F^2=\|PBQ\|_F^2+\|A\|_F^2-2\langle B,A\rangle$, but it's not clear to me that $\|PBQ\|_F^2\leq \|B\|_F^2$. Maybe one can use submultiplicity of Frobenius norm, but the F-norm of $P$ is $\sqrt{\text{rank}(A)}$ and not 1 as one could wish.