Suppose that $a(\cdot,\cdot):V \times V \rightarrow \mathbb{R}$ is a symmetric, continuous bilinear form defined on the Hilbert space V.
Assume that, for any continuous linear functional on $l \in V’$ and for any closed subspace $U \subset V$, the variational equation $$ a(u,v) = l(v) \quad \forall v \in U$$ has one and only one solution $u\in U$.
Show that there is a constant $\alpha>0$ such that, either $a(v,v) \geq \alpha ||v||_V^2$ for all $v\in V$, or $a(v,v) \leq -\alpha ||v||_V^2$ for all $v\in V$.
I am thinking that the bilinear form $a(\cdot,\cdot)$ can be related to an self-adjoint operator $A: V \rightarrow V$. The solvability of the variational equation $a(u,v) = l(v)$ can be transformed to the following equation $Au = \tau(l)$ where $\tau: V' \rightarrow V$ is the F. Riesz's isometry. The latter equation can be related to the "eigenvalues" of the self-adjoint operator $A$.
Due to the assumptions, there is no $v\ne0$ such that $a(v,v)=0$. This implies that either $a(v,v)>0$ for all $v\ne0$ or $a(v,v)<0$ for all $v\ne 0$. Then one should be able to proof the claim using the spectral theorem in the multiplication form. I do not have time right now.
Answer to previous version which assumed unique solvability for the variational equation only on $V$.
Not true. Take $V=\mathbb R^2$. Define $$ a(u,v)=u^T \pmatrix{1&0\\0&-1}v. $$ Then $a$ is symmetric, invertible, etc. But $$ a(u,u)=0 $$ for $u=\pmatrix{1\\1}$. (The example works with obvious modifications also for $V=\mathbb C^2$.)