2
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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}

$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – Yaddle May 27 '16 at 16:38
  • $\begingroup$ @Yaddle sorry it looks like I made a mistake in the triangle inequality $\endgroup$ – s.harp May 27 '16 at 16:54

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.