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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?

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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 $\mathcal{A}_0$.

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