Upper bound for norm of Hilbert space operator

Hint. We can be write $$ T = A + iB $$ where $A$ and $B$ are self-adjoint operators defined by $$ A := \frac 1 2 (T + T^+)\\ B := -\frac i 2 (T - T^+) $$

You can use the sesquilinear forms. Especially a square form.

So let's say that $f:H \times H \to \Bbb C$ is a square form that $f(x,x)=\langle Tx,x \rangle$ then you can say that $\| f \| \le \|T \| \le 2\|f \|$.

By sesquilinear map we mean a map $\phi:H\times H\to\Bbb C$ with properties:

1. The map is linear to the first variable: the map $x \to \phi(x,y):H \to \Bbb C$ is linear.
2. The map is antilinear to the second variable: the map $x \to \overline{ \phi(x,y)}:H \to \Bbb C$ is linear.

Also, $\|f\| := \sup\limits_{\|x\|,\|y\|\le 1} \langle Tx,y \rangle$.