Why Applying A Continuous Linear Elliptic Differential Operator Gives A Functional and Clarifying Dual Space Definition

Question

I'm reading a paper on Multiscale FEM Methods and I just need a bit of help better visualizing how to interpret the dual space in the PDE setup.

We are given a continuous, linear, elliptic differential operator $O: [H^1(\Omega)]^k \to X_0^*$ for $k=1 ,2 ,3$. We are considering the problem $$ Ou=f\quad\text{ in }X^*_0 $$ where $X^* _0$ is defined as the dual space of $X_0$, defined as $$ X_0 = \{ x \in [H^1(\Omega)]^k : x=0 \text{ on the Dirichlet boundary, $x = 0$ on the Multiscale FEM border} \}. $$

We take $u \in X = \{ x \in [H^1(\Omega)]^k : x=0 \mbox { on the Dirichlet boundary} \}$.

In order to visualize this I took $k=1$ and the operator $O$ to be the Laplacian: furthermore I took $u = \sin(\pi x)$ on $\Omega = [0,2]$. Then we know that $u$ is zero on the boundaries and $u \in H^1(\Omega)$ since $$ \|u\|_{H^1(\Omega)} = 1+\pi^2 < \infty. $$ However, when we apply the differential operator I get $$ f = \pi^2 \sin(\pi x). $$ This function does map input values to the field of real numbers but it is not a linear function. this goes against my understanding of what it means to be an element of the dual space.

So my questions are:

1. Am I misunderstanding something about the definition of a dual space? (I thought you just had to map elements from the space to a field via a linear function).
2. Am I misunderstanding something else fundamental to the setup of this problem?

Sorry if this is a silly question and thank you very much for any insights.

Answer 1

As I do not know how much you know of functional analysis and its application to PDEs, I have written quite a long answer. If you are only interested in your two questions, you can only read points 1 and 4.

1. Definition of the dual space

You are right that the (topological) dual space of a $\mathbb{R}$-vector space $X$ is just the space of all linear functionals from $X$ to $\mathbb{R}$. However, in the setting of functional analysis, one is a bit more strict: $$X^* := \{\varphi : X\longrightarrow\mathbb{R}\ |\ \varphi \text{ is linear and bounded}\}$$

Boundedness for an operator $T:X\longrightarrow Y$ between two normed vector spaces $(X,\|\cdot\|_X)$ and $(Y,\|\cdot\|_Y)$ means that there exists some $C>0$ so that $$\|Tx\|_Y \leq C\|x\|_X$$ for all $x\in X$. If $T$ is linear, then boundedness is equivalent to continuouity. So one could equivalently define $X^*$ as $$X^* :=\{\varphi :X\longrightarrow\mathbb{R} \ | \ \varphi \text{ is linear and continuous}\}.$$

2. Sobolev Spaces

As you know, the Sobolev spaces $W^{\ell,p}(\Omega)$ provide a usefull generalization to the spaces of $\ell$-differentiable functions $C^\ell(\Omega)$. For generall $\ell\in\mathbb{N}$ and $1\leq p\leq \infty$, they are Banach spaces (i.e. complete normed vector spaces), and in the special case $p=2$ they are even Hilbert spaces, i.e. come with a natural scalar product (which induces a complete norm) defined by $$\langle u,v\rangle = \int\limits_\Omega u(x) v(x) dx$$ and are thus often denoted by $H^\ell(\Omega)$. This is a very usefull property, in great part due to the Theorem of Riesz-Fréchet

On a Hilbert space $(H,\langle\cdot,\cdot\rangle)$, there exists an isomorphism $H\overset{\sim}{\longrightarrow} H^*$ given by $v\longmapsto \langle v,\cdot\rangle$.

This theorem is generalized by the Lemma of Lax-Milgram

Given a bilinear form $B :H\times H\longrightarrow\mathbb{R}$ on a Hilbert space $(H,\langle\cdot,\cdot\rangle)$, there exists a unique continuous linear operator $T:H\longrightarrow H$, so that $$B(u,v)=\langle Tu,v\rangle\ \forall u,v\in H.$$

3. Usefullness for PDEs

Now, let $D : H^2(\Omega)\longrightarrow H^0(\Omega)=L^2(\Omega)$ be an elliptic differential operator, e.g. the Laplacian $\Delta$.

Then, after checking that $B = \langle D\cdot,\cdot\rangle$ actually defines a continuous bilinear form on $H^2(\Omega)$, one can re-write any PDE of the form $Du = f$ for $f\in L^2(\Omega)$ as its weak formulation $$ B(u,v)=\langle f,v\rangle \ \forall v\in H^2(\Omega)\ (*)$$

Thus,the Lemma of Lax-Milgram yields the existence of a unique such $u\in H^2(\Omega)$.

Your formulation boils down to the following: By Riez-Fréchet, we can see $Du$ and $f$ both as elements in the dual space $(H^2)^*$ and thus can express the weak formulation $(*)$ as $$Du =f \text{ in }(H^2(\Omega))^*.$$

4. Your Example After applying the Laplacian to your chosen $u\in X$, you get $Du = \pi^2\sin\pi x \in L^2(\Omega)$. The induced element of the dual space $(L^2(\Omega))^*$ thus is $$\langle Du,v\rangle = \pi^2\langle\sin\pi x,v\rangle = \pi^2\int\limits_\Omega \sin(\pi x) v(x) dx,$$ which of course is linear and bounded/ continuous.