$H$ 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 $H$.

  • $\begingroup$ What is $T$? How is it related to $a$? $\endgroup$ – Davide Giraudo Nov 7 '12 at 22:49
  • $\begingroup$ There exists a linear continuous functional T on H, such that a(x,y)=(Tx,y) $\endgroup$ – Montaigne Nov 7 '12 at 22:51
  • 2
    $\begingroup$ No, $T$ is a linear operator, not a linear functional. A functional would take $H$ to $\mathbb R$, not to $H$. $\endgroup$ – Robert Israel Nov 7 '12 at 22:59
  • $\begingroup$ Thanks for the information, but a(x,y)=(Tx,y) still should hold. Is there any other mistake in the formulation? $\endgroup$ – Montaigne Nov 7 '12 at 23:02

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 $H$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.