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I have a question about functional analysis and operator theory.

Definition

Let $(H,(\cdot,\cdot)_{H})$ be a real Hilbert space and $D$ be a dense subspace of $H$. Let $(\mathcal{E},D)$ be a positive definite closed symmetric bilinear form on $H$ i.e.

$\mathcal{E}(f,f) \geq 0 \text{ for all }f \in D$, $ \mathcal{E}(f,g)=\mathcal{E}(g,f) \text{ for all } f,g \in D$ and

$ \langle f,g \rangle:=\mathcal{E}(f,g)+(f,g)_{H} $ is a inner-product on $D$ and $D$ is a Hilbert space w.r.t. this inner-product.

By Schwarz's inequality and the trivial inequality $\mathcal{E}(f,f) \leq \langle f,f\rangle$, we can see $\left|\mathcal{E}(f,g) \right| \leq \langle f,f\rangle^{1/2} \langle g,g \rangle^{1/2}$. This implies the map $D \ni g \mapsto \mathcal{E}(f,g) \in \mathbb{R}$ is continuous w.r.t. $\langle \cdot, \cdot \rangle^{1/2}$. By Riesz's theorem, there exists bounded linear operator $T$ on $D$ such that $\mathcal{E}(f,g)=\langle Tf,g \rangle$.

On the other hand, we can define linear operator $(L,D(L))$ on $(H,(\cdot,\cdot)_{H})$ associated with $(\mathcal{E},D)$ i.e. Let \begin{align*} D(L)=\left\{ f \in D : g \mapsto \mathcal{E}(f,g) \text{ is continuous on }D \text{ w.r.t. } (\cdot,\cdot)_{H}^{1/2} \right\}. \end{align*}

Then, we can find $F \in H$ such that $\mathcal{E}(f,g)=(F,g)_{H}$ for each $g \in D$ (Riesz's theorem) and we denote $Lf:=F$.

My Question

Identifying $D$ with its dual $D'$ we have that \begin{align*} H' \subset D' \cong D \subset H \text{ densely and continuously } \end{align*}

and $( \cdot,\cdot )_{H}$ restricted to $H' \times D$ coincides with $\langle \cdot,\cdot \rangle$. Then, it holds that \begin{align*} D(L)&=\left\{ f \in D : g \mapsto \mathcal{E}(f,g) \text{ is continuous on }D \text{ w.r.t. } (\cdot,\cdot)_{H}^{1/2} \right\} \\ &=\left\{ f \in D : g \mapsto \langle Tf,g \rangle \text{ is continuous on }D \text{ w.r.t. } (\cdot,\cdot)_{H}^{1/2} \right\} \\ &=\left\{ f \in D : g \mapsto (Tf,g)_{H} \text{ is continuous on }D \text{ w.r.t. } (\cdot,\cdot)_{H}^{1/2} \right\} \\ &=D . \end{align*}

But I think this relation $D(L)=D$ strange...

Did I make mistakes? Please let me know.

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    $\begingroup$ Your application of the Cauchy-Schwarz inequality is wrong, you only get $\mathcal{E}(f,g)^2\leq \mathcal{E}(f,f)\mathcal{E}(g,g)$. $\endgroup$ – MaoWao Jun 2 '16 at 13:36
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    $\begingroup$ By the Schwarz's inequality, $\left| \mathcal{E}(f,g) \right| \leq \mathcal{E}(f,f)^{1/2}\mathcal{E}(g,g)^{1/2}$. Note that $\mathcal{E}(f,f) \leq \langle f,f \rangle$. $\endgroup$ – sharpe Jun 2 '16 at 14:41
  • $\begingroup$ Sorry, I mixed up your notations of $\langle\cdot,\cdot\rangle$ and $(\cdot,\cdot)_H$. $\endgroup$ – MaoWao Jun 2 '16 at 20:28

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