Let $\Omega$ be a closed domain with smooth boundary in $\mathbb{R}^n$. Let $H^1_0(\Omega)$ be the closure of compactly supported smooth functions under the norm $\|u\|_1 = \int_\Omega u^2 + |\nabla u|^2\ dx$ and let $H^1(\Omega)$ be the closure of smooth, continuous functions under the same norm.

Any $H^1$ function which has nonvanishing trace cannot be approximated by any sequence of functions in $H^1_0$. So $H^1_0$ is a closed subspace of the Hilbert space $(H^1, \|\cdot\|_1)$, hence has an orthogonal complement.

What is a generating set of the orthogonal complement of $H^1_0$ in $H^1$?

Motivation is to get my hands on some concrete examples, rather than to just appeal to theorems that establish the existence of a right inverse to a trace operator.

Of course if anyone has references, I'm happy to follow them up. I've skimmed through Gilbarg-Trudinger and Evans and found nothing, but maybe I'm looking in the wrong place.

  • $\begingroup$ For $H^1([0,1])$, say $\langle f,g \rangle_{H^1([0,1])} = f(0)\overline{g(0)}+f(1)\overline{g(1)}+\langle f',g' \rangle_{L^2([0,1])}$. $H^1_0([0,1]) = \{ f \in H^1([0,1]),f(0) = f(1)=0\}=\{ f \in H^1([0,1]),f(0) = f(1)=\langle f',1 \rangle_{L^2([0,1])}=0\}$ so the orthogonal complement is $\{a+bx\}$. When changing the inner product (for say $(f,g ) = \langle f,g \rangle_{L^2([0,1])}+\langle f',g' \rangle_{L^2([0,1])}$) the orthogonal complement becomes $\{a \phi_1+b \phi_2\}$ $\endgroup$ – reuns Nov 20 '18 at 1:54
  • $\begingroup$ @reuns surely there is a condition on $a,b$? With $f(x) = \begin{cases}x, & x < \frac{1}{2}\\ 1-x, & x \geq \frac{1}{2}\end{cases}$ and $g(x) = a + bx$ I compute $\langle f, g\rangle_{H^1([0,1])} = \frac{1}{4}a + \frac{1}{8}b$ which is not always zero $\endgroup$ – Neal Nov 20 '18 at 2:04
  • $\begingroup$ That's why I gave two different inner products, only for the first one the orthogonal complement has an obvious basis, for the other inner product, I'd say we should substract to $1$ and $x$ their projection on the trigonometric orthogonal basis of $H^1_0$ $\endgroup$ – reuns Nov 20 '18 at 2:13
  • $\begingroup$ @reuns Ah, I see, thank you. Your comments make good sense. (I would note that I specified the inner product in the question.) $\endgroup$ – Neal Nov 20 '18 at 13:33

As reuns has stated in the comments, the answer depends on the inner product you choose in $H^1(\Omega)$. Let us fix the most common choice $$ (u,v)_{H^1(\Omega)} = \int_\Omega \nabla u \cdot \nabla v + u \, v \, \mathrm{d}x.$$

In this case, the orthogonal complement of $H_0^1(\Omega)$ consists precisely of the (weak) solutions $u \in H^1(\Omega)$ of $$ -\Delta u + u = 0$$ (without B.C.). Indeed, the weak formulation of this PDE is $$(u,v)_{H^1(\Omega)} = 0 \quad\forall v \in H_0^1(\Omega).$$

For different inner products, you get different PDEs.

  • $\begingroup$ Ah hah, this makes perfect sense. Does this PDE have a name? (Aside from "unit-eigenfunction equation with no boundary condition") $\endgroup$ – Neal Nov 24 '18 at 1:11
  • 1
    $\begingroup$ I don't know if there is a special name of this PDE. $\endgroup$ – gerw Nov 24 '18 at 6:03

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