# Hahn-Banach for $L^1$-Norm

Let $L \colon H^1_0(\Omega) \to \mathbb{R}$ be a linear and continuous functional with \begin{align*} |L(v)| \leq C\| \nabla v\|_{L^1(\Omega)} := C\int_\Omega |\nabla v(x)|_{\mathbb{R}^n}dx \quad \forall v\in H^1_0(\Omega). \end{align*} How does Hahn-Banach's theorem yield a unique function $\lambda \in L^\infty(\Omega)^3$ such that \begin{align*} |\lambda(x)| &\leq C \text{ for a.e. } x\in \Omega,\\ L(v) &= C\int_\Omega \lambda \cdot\nabla v \,dx \quad \forall v\in H^1_0(\Omega) \end{align*} holds?

## 1 Answer

The technique is a standard "factoring through" trick:

1. Just define a map $T$ in the closure of the range of $\nabla \subset L^2(\mathbb{R})^{\oplus 3}$ by sending $\nabla u \mapsto L(u)$. You can extend that to the whole space by putting $0$ in the orthogonal complement (since you are in a Sobolev space there are orth. complements).
2. Prove than the map $T$ is bounded in $L^1(\mathbb{R})^{\oplus 3}$ and so it is in $L^\infty(\mathbb{R})^{\oplus 3}$.
3. Then use that $L = T \circ \nabla$ by the definition of $T$.