The problem comes from Naylor's Linear Operator Theory Section 5.23 Problem 14

Let $T$ be a normal operator on Hilbert space $H$ and let $T = A + iB$ be the Cartesian decomposition of $T$. Show that $$\mathrm{max}(\left\lVert A\right\rVert^2,\left\lVert B\right\rVert^2) \leq\left\lVert T\right\rVert^2 \leq \left\lVert A\right\rVert^2+\left\lVert B\right\rVert^2$$

I've shown that Cartesian decomposition of $T$ is unique, $T^*=A-iB$, and that $A$ and $B$ commutes.

For the $\left\lVert T\right\rVert^2 \leq \left\lVert A\right\rVert^2+\left\lVert B\right\rVert^2$ part, I used the fact that $T$ is normal and use that $\left\lVert T\right\rVert^2 =\left\lVert T T^*\right\rVert = \left\lVert T^2 \right\rVert$.

However the $\mathrm{max}(\left\lVert A\right\rVert^2,\left\lVert B\right\rVert^2) \leq\left\lVert T\right\rVert^2$ part is giving me trouble. At first, I made a series of inequalities to build up to$\left\lVert T\right\rVert^2$, but no luck.


From the properties of the decomposition, one gets $$ \|Tx\|^2 =\|Ax\|^2 + \|Bx\|^2, $$ hence $$ \max( \|Ax\|^2, \|Bx\|^2 ) \le \|Tx\|^2 \le \|T\|^2 \|x\|^2. $$ Now take the supremum over $x$ with $\|x\|\le 1$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.