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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 $A$.

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Every continuous bilinear (or sesquilinear) form $a$ on $V$ induces a continuous (anti)linear map $A: V\to V', u\mapsto a(u,\cdot)$. Your assumption boils down to $A$ being bijective: "Every $l\in V'$ is of the form $a(u,\cdot)$" is surjectivity and that the solution $u$ being unique is injectivity. A bijective operator between Banach spaces is invertible, so that $A$ is an isomorphism. The constant $\alpha$ you're looking for is the operator norm of $A^{-1}$. For that we have to prove that $a$ is positive or negative definite.

That's where the subspaces come into the picture: If the problem $\forall v\in U: a(u,v)=l(v)$ has a unique solution for every $U$ und $l$, then in particular we can look at a functional that maps a fixed $v\neq 0 $ to $1$ and at the subspace $U=span(v)$. The solution we get is a multiple of $v$ and shows that $a(v,v)$ cannot be zero. Since $v$ was arbitrary, $a$ is either positive or negative definite.

Note that just like in the Lax-Milgram theorem itself, symmetry was not necessary.

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  • $\begingroup$ Thanks for your comments. However, how to find the constant $\alpha$? $\endgroup$
    – Simon Pun
    Commented Jan 25, 2018 at 16:20
  • $\begingroup$ As I said: It's the operator norm of $A^{-1}$. $\endgroup$
    – Johannes Hahn
    Commented Jan 25, 2018 at 20:04
  • $\begingroup$ If A is invertible, then I can conlude $\lVert u \rVert_V \leq \lVert A^{-1} \rVert \cdot \lVert a(u,\cdot) \rVert = \lVert A^{-1} \rVert \sup_{v \neq 0} \lvert \frac{a(u,v)}{\lVert v \rVert_V} \rvert$, but this supremum is larger than $\lvert a(u,u) \rvert$, so I cannot follow the inequality in the question. Am I missing something? $\endgroup$
    – tim-kt
    Commented Jun 4 at 12:33

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