Prove $|tr(PAP') - tr(PBP')| \leq ||A-B||_{\infty} ||P'||_1^2$

Question

How can I show the following identity?

$|\text{tr}(PAP') - \text{tr}(PBP')| \leq ||A-B||_{\infty} ||P'||_1^2$

Answer 1

You have $$ |\text{tr}(PAP') - \text{tr}(PBP')|=|\text{tr}(PAP'-PBP')|=|\text{tr}(P(A-B)P')|=|\text{tr}((A-B)P'P)|. $$ Now you apply the well-known Hölder inequality $$ |\text{tr}(XY)|\leq \|X\|_\infty\,\|Y\|_1. $$

If you don't see it, here are more details. From the above, you get $$ |\text{tr}(PAP') - \text{tr}(PBP')|\leq\|A-B\|_\infty\,\|P'P\|_1. $$ With $\{\mu_j\}$ the singular values of $P$, you have $$\|P'P\|_1=\sum_j\mu_j^2\leq\left(\sum_j\mu_j\right)^2=\|P\|_1^2.$$