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$H$ is real Hilbert space. $a\colon H\times H \to \mathbb R$ is a bilinear form on $H$ with $\lvert a(x,y)\rvert \leq C\lVert x\rVert \lVert y\rVert$ and $a(x,x) \geq \alpha \lVert x\rVert^2$. I would like to know if one of the following properties holds and how to prove it, may you could help me with that.

1) $T(H)$ is dense in $H$ where $T$ is a linear operator on $H$ such that $a(x,y)=\langle Tx,y\rangle$.

2) $T$ is injective and $T(H)$ is complete. It should be possible to prove it with $\lVert Tx\rVert \geq \alpha \lVert x\rVert$.

3) $T$ is an isomorphism from $H$ to $H$.

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What is $T$? How is it related to $a$? – Davide Giraudo Nov 7 '12 at 22:49
There exists a linear continuous functional T on H, such that a(x,y)=(Tx,y) – Montaigne Nov 7 '12 at 22:51
No, $T$ is a linear operator, not a linear functional. A functional would take $H$ to $\mathbb R$, not to $H$. – Robert Israel Nov 7 '12 at 22:59
Thanks for the information, but a(x,y)=(Tx,y) still should hold. Is there any other mistake in the formulation? – Montaigne Nov 7 '12 at 23:02
up vote 3 down vote accepted

The bilinear form $a$ defines a linear operator $T: H \to H$ by $\langle Tx, y \rangle = a(x,y)$, i.e. $Tx$ is the vector corresponding to the linear functional $y \to a(x,y)$. Since $\langle Tx, y\rangle \le C \|x \| \|y\|$, $\|Tx\| \le C \|x\|$, which says $T$ is bounded with $\|T\| \le C$. Since $\|Tx\| \|x\| \ge \langle Tx, x \rangle \ge \alpha \|x\|^2$ we get that $T$ is injective and $\|Tx\| \ge \alpha \|x\|$. Now $\langle Tx, x \rangle = \langle x, T^* x \rangle = \langle T^*x, x \rangle$ so $T^*$ is also injective. Since $\text{Ker}(T^*) = (\text{Ran}(T))^\perp$, we conclude that $\text{Ran}(T)$ is dense. Moreover, if $T x_n$ is a Cauchy sequence, since $\|T x_n - T x_m\| \ge \alpha \|x_n - x_m\|$ we see that $x_n$ is a Cauchy sequence, and thus $T x_n \to T(\lim_n x_n)$. This says $\text{Ran}(T)$ is closed. Putting it all together, then, $T$ is an isomorphism of $H$ onto $H$.

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Thank you very much, you helped me a lot. – Montaigne Nov 7 '12 at 23:40

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