How can we show that this nonnegative symmetric bilinear form is closable?

Question

Let

• $(E,\mathcal E)$ be a measurable space
• $\mu$ be a measure on $(E,\mathcal E)$ and $$\mu f:=\int f\:{\rm d}\mu$$ for Borel measurable $f:\mathbb R\to\mathbb R$ with $f\ge0$ or $\mu|f|<\infty$
• $\mathcal A_0$ be a subspace of $\left\{f:E\to\mathbb R\mid f\text{ is bounded and }\mathcal E\text{-measurable}\right\}$ closed under multiplication and dense in $L^p(\mu)$ for all $p\ge1$
• $\Gamma$ be a bilinear symmetric operator on $\mathcal A_0$ and $$\Gamma(f):=\Gamma(f,f)\;\;\;\text{for }f\in\mathcal A_0$$

Assume

1. $\forall f\in A_0:\exists c\ge0:\forall g\in\mathcal A_0:\left|\mu\Gamma(f,g)\right|\le c\left\|g\right\|_{L^2(\mu)}$
2. $\mu\Gamma(f^2,g)+2\langle\Gamma(f),g\rangle_{L^2(\mu)}=2\mu\Gamma(fg,f)$ for all $f,g\in\mathcal A_0$
3. $\Gamma(f)\ge0$ for all $f\in\mathcal A_0$

By 1., there is a unique linear symmetric operator $(\mathcal A_0,L)$ on $L^2(\mu)$ with $$\langle Lf,g\rangle_{L^2(\mu)}=-\mu\Gamma(f,g)\;\;\;\text{for all }f,g\in\mathcal A_0.\tag1$$ By 2., $$L(fg)=2\Gamma(f,g)+fLg+gLf\;\;\;\text{for all }f,g\in\mathcal A_0.\tag2$$

Question 1: How can we show that $${\Gamma(f,g)}^2\le\Gamma(f)\Gamma(g)\tag3$$ for all $f,g\in\mathcal A_0$? By 3., $$0\le\Gamma(f+\lambda g)=\Gamma(f)+2\lambda\Gamma(f,g)+\lambda^2\Gamma(g)\;\;\;\text{for all }\lambda\in\mathbb R.\tag4$$ If $\Gamma$ would be positive definite, then we could choose $$\lambda:=-\frac{\Gamma(f,g)}{\Gamma(g)}$$ and conclude. Are we able to prove positive definiteness of $\Gamma$? If not, is there an other way to show $(3)$?

Let $$\mathcal E(f,g):=\mu\Gamma(f,g)\;\;\;\text{for }f,g\in\mathcal A_0.$$ By $(3)$, $${\mathcal E(f,g)}^2\le\mathcal E(f)\mathcal E(g)\;\;\;\text{for all }f,g\in\mathcal A_0.\tag5$$

Assume

4. $\mu(Lf)=0$ for all $f\in\mathcal A_0$

By 4., $$\mathcal E(f,g)=-\langle f,Lg\rangle_{L^2(\mu)}=-\langle Lf,g\rangle_{L^2(\mu)}\;\;\;\text{for all }f,g\in\mathcal A_0.\tag6$$

Question 2: How can we show that $\mathcal E$ is closable? Let $(f_n)_{n\in\mathbb N}$ be $\mathcal E$-Cauchy with $$\left\|f_n\right\|_{L^2(\mu)}\xrightarrow{n\to\infty}0.\tag7$$ By $(6)$ and $(5)$, $$0\le\mathcal E(f_n)\le\left\|Lf_m\right\|_{L^2(\mu)}\left\|f_n\right\|_{L^2(\mu)}+{\mathcal E(f_n-f_m)}^{\frac12}{\mathcal E(f_n)}^{\frac12}\;\;\;\text{for all }m,n\in\mathbb N.\tag8$$ My problem is that $\mathcal E(f_n)$ is occuring on the right-hand side of $(8)$. Why can we nevertheless conclude $\mathcal E(f_n)\xrightarrow{n\to\infty}0$?

Answer 1

Gerw basically answered your questions, but I think it's worth summarizing.

1. The Cauchy-Schwarz inequality $$ a(x,y)^2\leq a(x,x)a(y,y) $$ holds for every symmetric positive semidefinite bilinear form $a\colon V\times V\to \mathbb{R}$. You have already shown this in the case $a(y,y)>0$. If $a(y,y)=0$, then $$ a(x+\lambda y,x+\lambda y)=a(x,x)+2\lambda a(x,y) $$ is a nonnegative affine function of $\lambda$, hence $a(x,y)=0$.
2. If $T$ is a symmetric linear operator in $H$ such that $\langle Tx,x\rangle\geq 0$ for all $x\in D(T)$, then the bilinear form $$ a(x,y)=\langle Tx,y\rangle $$ is closable. Let $(x_n)$ be an $a$-Cauchy sequence such that $\|x_n\|_H\to 0$. With the same computation as in the question, one gets $$ a(x_n,x_n)\leq\|Tx_m\|_H\|x_n\|_H+a(x_n-x_m,x_n-x_m)^{1/2}a(x_n,x_n)^{1/2}. $$ The first term goes to zero as $n\to\infty$, while $a(x_n-x_m,x_n-x_m)$ is small for large $m,n$ and $a(x_n,x_n)$ is bounded (since $(x_n)$ is $a$-Cauchy). Thus $a(x_n,x_n)\to 0$.