# What is the orthogonal complement of $H^1_0$ in $H^1$?

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.

• 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 $\{a \phi_1+b \phi_2\}$ – reuns Nov 20 '18 at 1:54
• @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 – Neal Nov 20 '18 at 2:04
• 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 $H^1_0$ – reuns Nov 20 '18 at 2:13
• @reuns Ah, I see, thank you. Your comments make good sense. (I would note that I specified the inner product in the question.) – Neal Nov 20 '18 at 13:33

## 1 Answer

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.

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