# Operator with norm

I got the following problem to solve:

Let $H$ Hilbert space and $T: H \to H$ a bounded positive operator, i.e. \begin{align*} \langle x, T x \rangle \geq 0 & & \text{for all } x \in H. \end{align*}

Show that $\Vert \cdot \Vert_T: T(H) \times T(H) \to \mathbb R_{\geq 0},\ (x,y) \mapsto \sqrt{\Vert y \Vert^2 + \langle x, T x \rangle}$ defines a norm on $T(H) \times T(H)$.

What I could show so far, is the homogeneity of the norm, i.e. $\Vert \lambda(x, y) \Vert_T = \lambda \Vert (x, y) \Vert_T$ for $\lambda \in \mathbb C$.

But I really struggle to show $\Vert (x, y) \Vert_T = 0 \Leftrightarrow (x, y) = 0$, as well as the triangle inequality. I got the hint, that for the first statement it helps that $\langle x, y \rangle = 0$ for all $x \in \mathrm{ker}(T)$ und $y \in T(H)$ (i showed that allready). For the proof of the triangle inequality I'm supposed to use the Cauchy Schwarz inequality, but i really don't see how.

I would be grateful for some help on the topic.

## 1 Answer

For hermitian $T$ you have $\ker(T) \cap \rm{im}(T) = \{0\}$, as $\langle x,x \rangle = 0 \iff x=0$, so if $x = Tz$ and $x \in \ker(T)$ you have $\langle x, x\rangle=\langle x, Tz\rangle=\langle Tx,z\rangle = 0$.

Now for a positive bounded operator $T$ on a Hilbert space you have a root $T^{1/2}$ that is also positive. Then $\langle x, Tx\rangle=\langle T^{1/2} x,T^{1/2}x\rangle$, which is $0$ only if $x \in \ker(T^{1/2})$. But since $(T^{1/2})^2=T$ you must have that $\ker(T^{1/2})\subset \ker(T)$ (actually $=$ holds), but if $x \in \rm{im}(T)$ and $x\neq0$ you have $x\notin\ker(T)\supset \ker(T^{1/2})$ and $\langle x,Tx\rangle >0$.

To see the triangle inequality use that

$$\langle (y_1,x_1),(y_2,x_2)\rangle_T := \langle y_1,y_2\rangle + \langle x_1,T x_2\rangle$$

defines a scalar product that induces the norm (positive definiteness is shown above, sesqui-linearity and conjugate symmetry are obvious). Using CS with this scalar product gives the triangle inequality on the norm:

\begin{align}\|a+b\|^2&=\langle a+b,a+b\rangle=\|a\|^2+\|b\|^2+\langle a, b\rangle + \langle b,a\rangle\\ &≤\|a\|^2+\|b\|^2+2\|a\|\|b\|=(\|a\|+\|b\|)^2\end{align}

• The idea with the square root is actually pretty nice. But i cant see that you used Cauchy Schwarz. Is there an easier way to see the triangle inequality by using Cauchy Schwarz and not using the square root? – Yaddle May 27 '16 at 16:38
• @Yaddle sorry it looks like I made a mistake in the triangle inequality – s.harp May 27 '16 at 16:54