A classical fact about linear operators in Hilbert spaces

Question

Studying the formulations which arise in hybridized mixed methods (say, mixed finite element method + hybridization), I got stuck with a rigorous proof of the following simple fact:

Let $\varphi$ be a linear map from $W$ to $\mathbb{R}$ and let $V \subset W$ where $V, W$ are Hilbert spaces. Suppose, $$ \varphi(v) = 0 \quad \forall v \in V \subset W $$ and we know the characterization of $V$ as $$ w \in V \iff c(\lambda,w) = 0, \quad \forall \lambda \in \Lambda $$ where $\Lambda$ is another Hilbert space and $c(\cdot,\cdot)$ is some nice bilinear form defined on $\Lambda \times W$. Then, as it is claimed in the classical book Brezzi,Fortin "Mixed Finite Element methods ..." (chapter 5, section 5.1.2, "Interelement multipliers", p.179 in my edition) or Brezzi,Boffi,Fortin "Mixed finite element methods and applications" (chapter 7, section 7.2.2, "Interelement multipliers", p.428), there exists such $\mu \in \Lambda$ that $$ \varphi(v) = c(\mu, v) \quad \forall v \in W $$

I think, the statement follows from something like a closed range theorem, but I cannot write down the argument rigorously.

Can someone help me with that?

Answer 1

Take $W = \Lambda = l^2$ and $V = \{0\}$. Define $c$ by $c(\vec{a},\vec{b}) = \sum \frac{1}{n} a_nb_n$ and $\phi: W \to \mathbb{R}$ by $\phi(\vec{b}) = \sum \frac{1}{n}b_n$. There is no $\vec{a} \in \Lambda$ such that $\phi(\vec{b}) = c(\vec{a},\vec{b})$ for all $\vec{b} \in W$.