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?
Non-negative definite self adjoint operator
functional-analysisself-adjoint-operators
Related Solutions
It is clear that $|\langle Tx,x\rangle|\leq \|T\|$ for $\|x\|=1$. For the converse, it suffices to show that $|\langle Tx,y\rangle|\leq \alpha$ for all $\|x\|=\|y\|=1$, with $\alpha=\sup\bigl\{|\langle Tx,x\rangle|: \|x\|=1\bigr\}$. We can clearly assume $\langle Tx,y\rangle \in\mathbb R$. Then $$ \langle Tx,y\rangle = (\langle T(x+y),x+y\rangle - \langle T(x-y),x-y\rangle)/4. $$ But then $$ |\langle Tx,y\rangle|\leq\alpha(\|x+y\|^2+\|x-y\|^2)/4=\alpha, $$ by the parallelogram identity.
I have paraphrased this nice derivation from the book Essential Results of Functional Analysis, by Zimmer.
Let $H$ be the underlying Hilbert space, $I$ the identity operator on $H$, $V=\operatorname{Range}(T-I)$ and $\overline{V}$ the closure of $V$. It is a standard result of operators on Hilbert space that $V^\perp = \operatorname{Ker}((T-I)^*) = \operatorname{Ker}(T^*-I)$. Since $T^*$ is assumed to have no positive eigenvalues, $V^\perp = \operatorname{Ker}(T^*-I)=0$. Hence $V$ is dense in $H$.
For any $u\in\mathcal{D}(T)$, $$\|(T-I)u\|^2 = \langle Tu-u,Tu-u\rangle = \|Tu\|^2+\|u\|^2 - \langle Tu,u\rangle - \langle u,Tu\rangle = \|Tu\|^2+\|u\|^2 - 2\langle u,Tu\rangle$$ where $\langle Tu,u\rangle = \langle u,Tu\rangle$ by the symmetry of $T$. Since $\langle u,Tu\rangle \le 0$ by assumption, we conclude $$\|(T-I)u\| \ge \|u\|$$ It follows that not only is $(T-I)$ one-to-one, but it is bounded below. So considering the inverse $(T-I)^{-1}$ defined on $V$, it is a bounded linear operator. If $z\in\overline{V}$ there exists a sequence $z_n\in V$ such that $z_n\rightarrow z$. Then $z_n$ is Cauchy. Letting $y_n = (T-I)^{-1}(z_n)$, we have that $y_n$ is Cauchy because $(T-I)^{-1}$ is bounded. Let $y$ be the limit of $y_n$. then $y_n\rightarrow y$ and $z_n=(T-I)y_n \rightarrow z$. Since $T$ is closed, so is $(T-I)$. Therefore $z\in V$ and $z=(T-I)y$. Hence $\overline{V} = V$. We already showed that $V$ is dense, so $V=H$.
Now, $T$ is symmetric, so $T^*$ is an extension of $T$, so $(T^*-I) = (T-I)^*$ is an extension of $(T-I)$. But $V=\operatorname{Range}(T-I)$ is already all $H$, so either $(T^*-I)$ isn't one-to-one, or $(T^*-I) = (T-I)$. We know however from assumption that $\operatorname{Ker}(T^*-I) = 0$. So $(T^*-I)$ is one-to-one, and it follows that $(T^*-I) = (T-I)$. From this it follows that $T^*=T$.
Best 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 $$ |(Gf,g)|^2 \le (Gf,f)(Gg,g). $$