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.


  • $\begingroup$ Do you mean by "coercive" that $B(x,x) \geq c \Vert x \Vert^2$? If yes, try to proof it directly. $\endgroup$ – crixstox Nov 16 '14 at 17:36
  • $\begingroup$ Yes, I do mean coercive like that. I will try $\endgroup$ – user188400 Nov 16 '14 at 17:41
  • $\begingroup$ 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. $\endgroup$ – Milly Nov 16 '14 at 17:44
  • $\begingroup$ Sorry, my bad! I made a mistake! Probably not that directly ... $\endgroup$ – crixstox Nov 16 '14 at 18:01
  • 2
    $\begingroup$ 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. $\endgroup$ – Davide Giraudo Nov 16 '14 at 18:02

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

  • $\begingroup$ almost the simplest example. as @davide giraudo mentioned take any hilbert space with T=-I. But that was very nice as well. thanks, you indeed helped me :) $\endgroup$ – user188400 Nov 16 '14 at 19:07

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.


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