# Upper and Lower bounds on Normal Operator Norm

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 $$\|x\|\le 1$$.