# Is this following bilinear form coercive?

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!

• Do you mean by "coercive" that $B(x,x) \geq c \Vert x \Vert^2$? If yes, try to proof it directly. – crixstox Nov 16 '14 at 17:36
• Yes, I do mean coercive like that. I will try – user188400 Nov 16 '14 at 17:41
• Probably I am wrong, but the iff in (3) seems problematic... you essentially have no information on the direction of $Tx$, even in finite dimension. – Milly Nov 16 '14 at 17:44
• Sorry, my bad! I made a mistake! Probably not that directly ... – crixstox Nov 16 '14 at 18:01
• Your assumption is symmetric, that is, if it is satisfied by $T$, then $-T$ satisfies it. But with $T:=-I$ we get an example or non-coercive operator. – Davide Giraudo Nov 16 '14 at 18:02

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 $$\langle Tx,x\rangle=0.$$
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.