Show that the divergence operator is onto

Question

Came across this while reading Temam's text on Navier Stokes. Lemma 2.4 states that the divergence operator $\nabla\cdot$ maps $\bf{H}_0^1(\Omega)$ $\textit{onto}$ $L^2(\Omega)/\mathbb{R}$ where the latter is the subspace of $L^2(\Omega)$ of elements $f$ satisfying $\int_\Omega f=0$. To prove this statement we consider $A=\nabla$ (the gradient operator) defined on $L^2(\Omega)$ (mapping into $\bf{H}^{-1}(\Omega)$). Earlier, it'd been shown that the operator $A$ restricted to $L^2(\Omega)/\mathbb{R}$ is injective. So if we simply set $R(A):=A(L^2(\Omega)/\mathbb{R})\,(=A(L^2(\Omega))$, we know that $A$ is an isomorphism between $L^2(\Omega)/\mathbb{R}$ and $R(A)$.

So far, I don't think anything profound has been said. Now, the proof proceeds with the following statement (paraphrased): by transposition, the adjoint $A^*\in L(\bf{H_0^1(\Omega)}$ $,L^2(\Omega))$ is an isomorphism from the $\textit{orthogonal}$ of $R(A)$ $\textit{onto}$ $L^2(\Omega)/\mathbb{R}$.

My question is, what's the justification for this last statement?? Perhaps I'm a little rusty, but I'm guessing I'm failing to make the necessary manipulations to relate the ranges of $A$ and $A^*$ and the orthogonal spaces of them etc. etc.

Answer 1

The relevant fact is: if an operator $T:X\to Y$ between normed spaces is bounded from below, meaning $\exists c>0$ such that $\|Tx\|\ge c\|x\|$ for all $x\in X$, then its adjoint $T^*:Y^*\to X^*$ is surjective.

Proof: given any linear functional $f\in X^*$, define $g(y)=f(T^{-1}y)$ on the range of $T$. Then $g$ is bounded, so by the Hahn-Banach theorem it extends to a bounded linear functional on $Y$. By construction $T^*g=f$. $\quad\Box$

Since $A$ is known to be an isomorphism of $L^2(\Omega)/\mathbb{R}$ onto something, it's bounded from below. Hence, its adjoint $A^*$, which is $-\operatorname{div}$, is surjective onto the dual of $L^2(\Omega)/\mathbb{R}$.

Let's discuss this dual. The dual of $L^2(\Omega)$ is itself, via the pairing $(f,g)=\int_\Omega fg$. When we take quotient by the subspace of constant functions, the only remaining functionals are those that vanish on that subspace. Thus, the dual of $L^2(\Omega)/\mathbb{R}$ is the space $V=\{g\in L^2(\Omega): \int_\Omega g=0\}$, via the same pairing $(f,g) = \int_\Omega fg$. The final conclusion is that the divergence of an element of $H^1_0$ can be any $L^2$ function with zero integral.

It's also true that $V$ is isomorphic to the quotient $L^2(\Omega)/\mathbb{R}$ itself, since $V$ is the orthogonal complement of constant functions. But I don't find it helpful to bring in this isomorphism here, as it obscures rather than helps. It makes sense to think about the domain of $\nabla$ as the quotient by constant functions, since those are killed by the gradient. On the other hand, the range of divergence is a space of functions, not any kind of a quotient. Just because an isomorphism exists doesn't mean we have to consider it.

The claim that $A^*$ is an isomorphism of $R(A)^\perp$ onto something is incorrect; in fact, $R(A)^\perp$ is precisely the kernel of $A^*$. The author may have forgotten to say "quotient by" somewhere, or replace "the orthogonal of" by "the dual of".