I am looking for a proof that weak coercivity (Garding inequality) and the condition $a(u,v) = 0, \forall v \in V \implies u = 0$ is equivalent (or implies) to the inf-sup condition and the condition $a(u,v) = 0, \forall u \in V \implies v = 0$ (Necas theorem).
Now, $V, H, V'$ is a Hilbert triplet, $V$ Hilbert space densely and compactly embedded in $H$, other Hilbert space ($V=H^1_0, H = L^2$ for example).
Weak coercivity (Garding inequality) is: $a(u,u) \ge \alpha ||u||_V^2- \lambda ||u_n||_H^2$, for $\alpha, \lambda > 0$.
Inf-sup condition is: $\inf_{v \in V} \sup_{u \in V} \frac{a(u,v)}{||u||_V ||v||_V} \ge \alpha > 0$.
Now I know that if the inf-sup condition was false then, there is a norm $1$ sequence $u_n$ such that
$a(u_n,v) \leq \frac{1}{n}||v||_V, \forall v \in V$.
By the boundedness of the sequence there is a weak limit in $V$, and strong in $H$, up to subsequences, $\bar{u}$ and so
$0 < \alpha \le a(u_n, u_n) + \lambda||u_n||_H^2 \le \frac{1}{n} + \lambda ||u_n||_H^2 \implies ||\bar{u}|| _H> 0$, by passing to the limit.
But also $a(\bar{u}, v) \le 0, \forall v \in V$.
I don't know how to conclude that $a(\bar{u}, v)= 0, \forall v \in V$, which will solve the issue,giving the contradiction under the assumption that therefore $\bar{u} = 0$.
Thanks for tips.