# Coercive vs positive definite for a symmetric bilinear form

I was wondering once again what the difference in concept of positive definite and coercive was. Of course, if you have a bilinear form $a(.,.)$, that is coercive, positive definiteness follows directly from the definition: $$a(u,u)\geq\alpha\left\|u\right\|^2>0\text{ for }u\neq 0$$ since $\alpha>0$. Hope, I'm not wrong so far. On the other hand, I do have neither a proof nor an example that either proves or disproves the other direction. Is anyone here to clarify this?

Thanks in advance!

• The two definitions coincide whenever the unit ball is compact, namely in finite-dimensional spaces. For a counter-example in an infinite-dimensional space, see the answer below. – Siminore Jul 24 '13 at 17:19
• @Siminore. Can you please prove that the two definitions coincide in finite dimensional spaces or give references of where I could find this? Thanks a lot! – Adriaan Nov 18 '14 at 13:28

## 1 Answer

On $L^2([0,1])$, consider $a(u,v)=\int_0^1 tu(t)\bar v(t)\,dt$. It is positive definite but not coercive.