2
$\begingroup$

Let $G$ be a non-negative definite self adjoint operator on a Hilbert space $H$. I want to show that for all $f,g\in H$ we have $$(Gf,h)^2\leq (Gf,f)(Gh,h).$$ Can anyone help?

$\endgroup$
3
  • $\begingroup$ Case 1: suppose $(Gh,h) = 0$. Case 2: if $(Gh,h) \neq 0$, then we can divide by $(Gh,h)$ and so we can use the proof of the Cauchy Schwarz inequality, noting that $\langle f,h\rangle := (Gf,h)$ "acts like an inner product". $\endgroup$ Commented May 1, 2019 at 10:42
  • $\begingroup$ Another approach here is to consider the positive definite map that $G$ induces on the quotient space $H/\ker G$ $\endgroup$ Commented May 1, 2019 at 10:45
  • $\begingroup$ ah ok I see. Thank you! $\endgroup$
    – user525192
    Commented May 1, 2019 at 10:56

1 Answer 1

0
$\begingroup$

This holds whether $G$ is non-negative definite or non-negative semi-definite. For example, if it is non-negative semi-definite, then the following defines an inner product for every $\epsilon > 0$. $$ \langle f,g \rangle_{\epsilon} = (Gf,g)+\epsilon (f,g) $$ Consequently, the Cauchy-Schwarz inequality holds: $$ |\langle f,g\rangle_{\epsilon}|^2\le \langle f,f\rangle_{\epsilon}\langle g,g\rangle_{\epsilon} $$ Letting $\epsilon \downarrow 0$ gives $$ |(Gf,g)|^2 \le (Gf,f)(Gg,g). $$

$\endgroup$

You must log in to answer this question.