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

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

1. $$\mu(Lf)=0$$ 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 $$\mathcal E(f_n)\xrightarrow{n\to\infty}0$$?

• Concerning question 1: If $\Gamma(g) > 0$, you can your choice of $\lambda$. Otherwise, (4) is linear in $\lambda$ and, hence, it implies $\Gamma(f,g) = 0$. – gerw Feb 15 at 7:30
• Concerning question 2: I think you can simply use Young's inequality for the second addend on the right hand side and then bring $E(f_n)/2$ to the left-hand side. – gerw Feb 15 at 7:32

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 $$a(x_n,x_n)\to 0$$.