One problem with proving that $K(H)'=\{cI\}$

Question

I want to prove that the commutant of the space of compact operators is $\{cI\}$. Let $A\in K(H)'$. Then $Af \otimes\overline{f} (x)=f \otimes\overline{f}Ax$, because $f \otimes\overline{f}$ is rank one, therefore compact. The previous relation comes down to $<x,f>Af=<Ax,f>f$. How can I formally prove that now follows $Af=cf$? For now, $c=c(f)$.

Then I do the same for $g$, then for $f+g$, and then I can conclude that $c$ is really a constant, by taking $f$ and $g$ linearly independent.

Answer 1

For any $f$ such that $Af\ne0$, you have $$ Af=\frac{\langle Af,f\rangle}{\langle f,f\rangle}\,f. $$ So, as you say, $Af=c(f)f$ for any $f\in H$. Now fix an orthonormal basis $\{e_n\}$. Then $$ c(e_j+e_k)\,(e_j+e_k)=A(e_k+e_j)=Ae_j+Ae_k=c(e_j)e_j+c(e_k)e_k. $$ By the uniqueness of coefficients in a basis, $c(e_j+e_k)=c(e_j)$ and $c(e_j+e_k)=c(e_k)$, so $c(e_j)=c(e_k)$. Call this number $c$. Now, for any $f\in H$, $$ Af=\sum_nf_n\,Ae_n=c\,\sum_nf_ne_n=cf. $$