# Completion of a subspace of a Hilbert space with respect to a positive semi-definite symmetric bilinear form

Let

• $$H$$ be a $$\mathbb R$$-Hilbert space
• $$\mathcal E$$ be a positive semi-definite symmetric bilinear form on a dense subspace $$\mathcal A_0$$ of $$H$$ and $$\mathcal E(f):=\mathcal E(f,f)\;\;\;\text{for }f\in\mathcal A_0$$

Now, let $$\langle f,g\rangle_{\mathcal E}:=\langle f,g\rangle_H+\mathcal E(f,g)\;\;\;\text{for }f,g\in\mathcal A_0,$$ $$\left\|\;\cdot\;\right\|_{\mathcal E}$$ denote the norm induced by $$\langle\;\cdot\;,\;\cdot\;\rangle_{\mathcal E}$$ and $$\mathcal D(\mathcal E):=\left\{f\in H\mid\exists(f_n)_{n\in\mathbb N}\subseteq\mathcal A_0:\left\|f_n-f\right\|_H\xrightarrow{n\to\infty}0\text{ and }\mathcal E(f_m-f_n)\xrightarrow{m,\:n\:\to\:\infty}0\right\}.$$

How can we show that there is a unique extension of $$\mathcal E$$ to a positive semi-definite symmetric bilinear form on $$\mathcal D(\mathcal E)$$ and a unique extension of $$\langle\;\cdot\;,\;\cdot\;\rangle_{\mathcal E}$$ to an inner product on $$\mathcal D(E)$$?

If $$(f_n)_{n\in\mathbb N}\subseteq\mathcal A_0$$ and $$f\in H$$ with $$\left\|f_n-f\right\|_H\xrightarrow{n\to\infty}0\tag1$$ and $$\mathcal E(f_m-f_n)\xrightarrow{m,\:n\:\to\:\infty}0,\tag2$$ then $$(f_n)_{n\in\mathbb N}$$ is $$\left\|\;\cdot\;\right\|_{\mathcal E}$$-Cauchy and hence $$\left\|f_n\right\|_{\mathcal E}\xrightarrow{n\to\infty}\rho\tag3$$ for some $$\rho\in\mathbb R$$. In order to conclude that $$\left\|f\right\|_{\mathcal E}:=\rho$$ is well-defined, we need to show that $$\rho$$ doesn't depend on $$(f_n)_{n\in\mathbb N}$$. How can we do that?

You cannot, because it is not true. Take for example $$H=L^2([0,1])$$, $$\mathcal{A}_0=C([0,1])$$ and $$\mathcal{E}(f)=|f(0)|^2$$. The sequences $$(f_n)$$ and $$(g_n)$$ given by $$f_n(x)=0$$ and $$g_n(x)=(1-nx)_+$$ for $$x\in [0,1]$$ both converge to $$0$$ in $$L^2$$, are Cauchy w.r.t. $$\|\cdot\|_{\mathcal{E}}$$, yet $$\mathcal{E}(f_n)=0$$, while $$\mathcal{E}(g_n)=1$$.
The condition you need is called closability, which basically means that everything works as you expect. Put differently, you need that $$\mathcal{E}$$ (as a function of one argument) is lower semicontinuous on $$\mathcal{A}_0$$.