1
$\begingroup$

Let $X$ Hilbert space and $B \in L(X)$ positive, self adjoint operator. Then is this a norm or seminorm? $$|x|_B=(\langle Bx,x\rangle)^{1/2}$$

In [Li, Xungjing, and Jiongmin Yong. Optimal control theory for infinite dimensional systems.1995] p.232 they claim it's a seminorm.

But since $B$ is positive I can define $B^{1/2}$ which is a positive self adjoint operator and then $$|x|_B=(\langle Bx,x\rangle )^{1/2}=(\langle B^{1/2}x,B^{1/2}x\rangle )^{1/2}=|B^{1/2}x|$$ which is zero only for $x=0$ by the positivity of $B^{1/2}$.

What am I missing?

The thing that come to mind is that of course $\langle Bx,x\rangle =0$ when $Bx$ is orthogonal to $x$ but this seems not to be the case thanks to the previous equality.

$\endgroup$
8
  • 1
    $\begingroup$ Positive operators need not be injective. $B=0$ is also a positive operator. $\endgroup$ Jul 15 '21 at 9:39
  • 3
    $\begingroup$ It depends if a positive operator is allowed to have a kernel. My definition allows it! And if I want a positive operator not to have a kernel, I would say a "positive definite operator", I guess. $\endgroup$
    – Plop
    Jul 15 '21 at 9:41
  • 2
    $\begingroup$ Yes, it's rather a terminology question, 'positive operator' here means positive semidefinite. $\endgroup$
    – Berci
    Jul 15 '21 at 9:43
  • $\begingroup$ Yeah thanks I was pretty sure that it didnt admit a kernel. But checking better it does $\endgroup$
    – carlos85
    Jul 15 '21 at 9:55
  • $\begingroup$ yes right @Berci with positive I immidiately meant positive semidefinite, but it's semidefinite what it was meant $\endgroup$
    – carlos85
    Jul 15 '21 at 10:03
1
+100
$\begingroup$

As already explained by @Plop and @Berci, it depends on whether $B$ is strictly positive definite or only positive semidefinite.

For a positive semidefinite $B$, you may have some $v \in X$, $ v \neq 0 $ with $Bv = 0$. Then also $ |x|_B = 0 $ and $ |\cdot |_B $ this is only a seminorm, not a norm.

For a strictly positive definite $ B $, also $ B^{1/2} $ is strictly positive definite (by spectral calculus), so $ x \neq 0 $ implies $ |B^{1/2}x| > 0 $ and $$ |x|_B = |B^{1/2}x| > 0, $$

so you indeed have a norm.

$\endgroup$

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.