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.

  • 2
    $\begingroup$ There might be a typo in $z=1$, taking $k=1$ would fit the question better in my understanding. To know your background to answer the question efficiently it would be nice to know if you familiar with the formulation of the problem in (bi)linear forms, i. e. find $u$ s. t. $a(u, v) = f(v) \ \forall v \in X$. $\endgroup$
    – Korf
    Commented Jan 18 at 10:17
  • $\begingroup$ @Korf My apologies for the typo. You are completely correct, I meant $k = 1$. Thanks for pointing that out. And yes I am very familiar with bilinear forms. $\endgroup$
    – k12345
    Commented Jan 18 at 17:39
  • $\begingroup$ @Korf but I'm still a little unclear of how that points to the dual space so if you could please elaborate I would be very grateful to you. $\endgroup$
    – k12345
    Commented Jan 23 at 1:37

1 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.

  • 1
    $\begingroup$ This was an absolutely beautiful answer. Thank you so much for writing it all out so clearly. I'll award you the bounty as soon as it lets me (says I have to wait 35 minutes). Thank you again! $\endgroup$
    – k12345
    Commented Jan 24 at 19:30

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