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:
- my intuition says it is false.
- I managed to prove that $B$ is bounded
- 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!
Best Answer
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 $$ \langle Tx,x\rangle=0. $$