Is this following bilinear form coercive?

Question

First of all I want to mention that this is homework, so don't spoil it and reveal all the answer. just some guidenss :)

Let $H$ be a Hilbert space, $T:H\rightarrow H$ a bounded linear operator for which there exists an $r>0$, such that for all $x\in H: \|Tx\|\geq r\|x\|$.

Define $B:H\rightarrow H$, a bilinear form, by $B(x,y)=\langle Tx,y\rangle$. Prove or disprove that $B$ is coercive.

What I did so far:

1. my intuition says it is false.
2. I managed to prove that $B$ is bounded
3. I realised that $T$ is an injection, therefore if $H$ is finite-dimensional $T$ is invertible, and my guts say that it is true iff $T$ is invertible (use the Lax-Milgram theorem and the fact that any inner-product is coercive). Therefore I tried looking into infinite dimentional Hilbert spaces, with a non invertible $T$ that would answer to the conditions yet make the bilinear form not coercive. I have't even found a $T$ that follows the conditions (let alone disporoves the claim).

I would appriciate help and assistence in order to make my mind more organized.

Thanks!

Answer 1

The answer is NO.

Simpest possible example $H=\mathbb R^2$ and $$ T=\left( \begin{matrix} 0& 1 \\ -1 & 0\end{matrix} \right) $$ Then, for $x=(x_1,x_2)$, we have that $Tx=(x_2,-x_1)$, and hence $$ \|Tx\|=\|x\|, $$ while $$ \langle Tx,x\rangle=0. $$

Answer 2

Yiorgos S. Smyrlis and Davide Giraudo have already explained that the result does not hold whenever the operator is indefinite. However, it is fruitful to think about how the concept could be extended to the indefinite case.

One might consider the following generalization that accounts for the possiblity of negative eigenvalues. Replace $$\inf_x \frac{B(x,x)}{||x||^2} = c > 0$$ with $$\inf_x \sup_{t \in \{+1, -1 \}} t\frac{B(x,x)}{||x||^2} = c > 0.$$

Now we could make this look a little nicer by $sup$'ing over vectors instead of a discrete set: $$\inf_x \sup_y \frac{B(x,y)}{||x||||y||} = c > 0,$$

and generalize this to operators $B:X \times Y \rightarrow \mathbb{R}$ instead of just $B : X^2 \rightarrow \mathbb{R}$, and we have the famous LBB inf-sup condition, which is fundamental for proving invertibility of operators defined by bilinear forms in general.