# Non-negative definite self adjoint operator

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?

• 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". – Omnomnomnom May 1 at 10:42
• Another approach here is to consider the positive definite map that $G$ induces on the quotient space $H/\ker G$ – Omnomnomnom May 1 at 10:45
• ah ok I see. Thank you! – ShaqAttack1337 May 1 at 10:56

## 1 Answer

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 $$|(Gf,g)|^2 \le (Gf,f)(Gg,g).$$