Is it true that for Hilbert-Schmidt operators $B$ and $C$ we have $|\langle B^*Ce,e\rangle|=\frac{1}{2}(\|Be\|^2+\|Ce\|^2)$ on unit vectors?

Question

On the Wikipedia page for trace-class, it is stated that if $A=B^*C$ for some Hilbert-Schmidt operators $B$ and $C$ then for any unit vector $e$, we have $|\langle Ae,e\rangle|=\frac{1}{2}(||Be||^2+||Ce||^2)$.

The reference given leads to an exercise in a book, so I tried proving it myself. I tried applying parallelogram identity and polarization identity, to no success.

Answer 1

Wikipedia references Conway's Course in Functional Analysis, which claims this in a hint to Exercise 20 on p.267. Nonetheless, this is false. Let $B$ be the orthoprojector on $e_1$, $C$ the orthoprojector on $e_2$ and $e=\frac1{\sqrt{2}}(e_1+e_2)$. Orthoprojectors to a one dimensional subspace are surely Hilbert-Schmidt, $e$ is a unit vector, and $Be=\frac1{\sqrt{2}}e_1$, $Ce=\frac1{\sqrt{2}}e_2$. Moreover, $$|\langle Ae,e\rangle|=|\langle B^*Ce,e\rangle|=|\langle Ce,Be\rangle|=\frac12|\langle e_1,e_2\rangle|=0.$$ But $\|Be\|^2+\|Ce\|^2=1\neq0$.