# Show that the divergence operator is onto

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.

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".

• Strictly speaking then, -div maps onto the $\textit {the dual}$ of $L^2/\mathbb {R}$ which the author identifies with itself?
– Fozz
Sep 12, 2017 at 23:26
• The dual of that quotient is naturally the space of functions with zero integral. The dual of quotient is a subspace of dual.
– user357151
Sep 13, 2017 at 1:24
• What you refer to as "that quotient" is precisely "the space of functions with zero integral" is it not? Then what you're saying is that the dual of the quotient is itself? And by "natural" do you mean in some other sense than that of the standard Riesz representation sense for Hilbert spaces?
– Fozz
Sep 13, 2017 at 1:38
• I added two paragraphs to the answer to make the point better.
– user357151
Sep 13, 2017 at 2:35
• Thanks for your clarification. But one more question: it seems that you view $L^2/\mathbb{R}$ to be distinct from but isomorphic to $V$ whereas I understood your $V$ to be precisely equal to that quotient space. Then strictly speaking how do you define that quotient? In what space does it live?
– Fozz
Sep 13, 2017 at 2:49