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There is an argument that disturbs me somewhat in Rabinowitz : Minimax methods in critical point theory. p.94

We are trying to prove that for certain functions, the Palais - Smale condition can be simplified. Specifically, let $\Omega$ be a bounded domain in $\mathbb{R}^n$, with smooth boundary. Let $p$ be a function satisfying ($p$1) $p(x,\xi) \in C(\bar{\Omega}\times\mathbb{R},\mathbb{R})$
($p$2) There are constants $A,B>0$ with $|p(x,\xi)| \leq A+B|\xi|^s$, where $0\leq s < \frac{n+2}{n-2}, n\geq3$.

Claim : Let $I$ be the functional corresponding to $p$ defined by $$I(u) = \int_{\Omega}\frac{1}{2}|\nabla u|^2 - P(x,u) dx$$ where $P(x,\xi) = \int_0^{\xi}p(x,t)dt$. Then, if $(u_m)$ is a bounded sequence in $E = W_0^{1,2}(\Omega)$, such that $I'(u_m)\rightarrow 0$ , then $(u_m)$ has a convergent subsequence.

Beginning of the proof: Let $D : E\rightarrow E^{*}$ be the duality map between $E$ and its dual. Then, for $u,\phi \in E$, $$(Du)\phi = \int_{\Omega}\nabla u \dot\ \nabla \phi dx$$ Thus, $D^{-1}I'(u) = u - D^{-1}J'(u)$. And it continues from there.

Here, $J'(u)$ is the derivative of the second term in $I$. I don't understand what it means by the duality map. In Evans and Brezis for example, they are very adamant that we do not identify this space with its dual. So where is this very specific form of the map coming from? And why can we simply invert it like this? Also, it claims later that $D^{-1}$ is continuous. Why is this??


@5PM Now that I look at this though, you seem to already explicitly know the formula for pairing. You state that $\langle -\Delta f,f\rangle _{E^{*},E} = \int{|\nabla f|^2}$. Why is this? How do we know what the pairing formula is? For instance, I worked out the $L^p$ example you provided, but checking the two conditions worked because I knew that the dual of $L^p$ is $L^q$. So, $\langle -\Delta f,f\rangle_{E^{*},E} =?$ You would need to also know explicitly the norm on the dual, to show that $||-\Delta f||_{E^{*}} = ||f||_E (= (\int{|\nabla f|^2})^{\frac{1}{2}})$

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up vote 9 down vote accepted

The author does not identify $E^*$ with $E$: that's why there is a duality map and not the identity map. If $E$ is a nice (uniformly convex and uniformly smooth) real Banach space, there is a well-defined map $D:E\to E^*$ such that $\|Dx\|_{X^*}=\|x\|_X$ and $\langle Dx,x\rangle_{E^*,E}=\|x\|^2$. In other words, $Dx$ is the (unique) norming functional for $x$, scaled to have the same norm as $x$. Uniform smoothness implies that $D$ is uniformly continuous on bounded sets, while uniform convexity implies the same for the inverse $D^{-1}$. In fact, $D^{-1}$ is nothing but the duality map for $E^*$.

An example: if $E=L^p$, $1<p<\infty$, then $Df = |f|^{p-2}f$. You may want to check that this works.

When $E$ is a Hilbert space, $D$ is a linear map, and therefore an isometry: $\|Dx-Dy\|_{E^*}=\|x-y\|_E$. This takes care of all continuity questions.

When $E=W^{1,2}_0(\Omega)$ with the homogeneous norm $\|f\|_E^2=\int_\Omega |\nabla f|^2$, the duality map $D:E\to E^*=W^{-1,2}(\Omega)$ is our old friend Laplacian, $-\Delta$. Indeed, $$\langle -\Delta f,f\rangle_{E^*,E} = \int |\nabla f|^2$$ If they use the norm with $\|f\|_{L^2}$ term thrown in, then $D$ is probably not as explicit.

In any case, the statement $(Du)\varphi = \int_\Omega \nabla u\cdot \nabla \varphi$ simply repeats that $D$ is a duality map.

Answer to the edit by @Euler: yes, we must know what the pairing is in order to find the duality map, because the definition of duality map involves pairing. By the way: if we didn't know the pairing between $E^*$ and $E$, why would we call $E^*$ the dual of $E$?

The elements of $E^*$ are bounded linear functionals on $E$; as soon as the norm on $E$ is known, the norm on $E^*$ is known too, by the definition of a norm of linear functional. But since this notion of $E^*$ is kind of abstract, we want to identify of the elements of $E^*$ with some functions, or at least distributions. In the case of $W^{1,2}_0(\Omega)$ this can be done as follows: a distribution $f$ belongs to $W^{-1,2}(\Omega)$ if there exists a constant $C$ such that $$|f(\varphi)|\le C \|\varphi\|_{W_0^{1,2}} \text{ for all }\varphi \in C_c^\infty(\Omega)$$ The smallest such $C$ is the $W^{-1,2}$ norm of $f$. This is still kind of abstract, but at least we can do something: for example, check that the distributional Laplacian of a function $u\in W_0^{1,2}(\Omega)$ belongs to $W^{-1,2}$: $$ |(\Delta u)(f)| = \left|\int u \Delta f \right| = \left|\int \nabla u \cdot \nabla f \right| \le \|\nabla u\|_{L^2} \|\nabla f\|_{L^2} \tag{1} $$ where the first step is the definition of distributional Laplacian, the second step is integration by parts, and the third one is Hölder's inequality.

Assuming that $W^{1,2}_0$ is equipped with the homogeneous norm, (1) shows that $\|\Delta u\|_{W^{-1,2}}\le \| u\|_{W^{1,2}_0}$. In fact, equality holds because $f$ could be arbitrarily close to $u$, turning Hölder's inequality into near-equality.

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Ah, I get it. So in cases where we don't want to 'identify' a space with it's dual, as in the case $H_0^1 \subset L^2 \subset H^{-1}$, we still have the canonical isometry (for a Hilbert space). Actually that's very simple, thank you. – Euler....IS_ALIVE Jan 21 '13 at 0:05
I tried to post a comment, but it wasn't parsing properly, so I edited the question above. – Euler....IS_ALIVE Jan 21 '13 at 1:50
+1. Nice answer, clearing up a point that often causes confusion. – Nate Eldredge Jan 21 '13 at 3:43
Yes this is fantastic. I don't understand why it seems like all my PDE books just basically skipped over this. Thank you so much. – Euler....IS_ALIVE Jan 21 '13 at 3:55

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