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I am reading a paper(Thomas Bartsch, Zhi-Qiang Wang, Commun. in Partial Differential Equations, 20,1725-1741(1995)), and encounter the following problem in page 1731:

($f1$) $f \in C(R^N \times R, R)$ satisfies $f(x,u)=o(|u|)$ as $u \rightarrow 0$ uniformly in $x$.

($f2$) There are constants $a_1,a_2>0$ and $s\in (1,\frac{N+2}{N-2})$ for $N \geq 3$ and $s \in (1,+\infty)$ for $N=1,2$ such that $$ |f(x,u)|\leq a_1+a_2|u|^s. $$

$$\Psi(u)=\int_{R^N}F(x,u)~\mathrm{d}x,\qquad \text{ where } ~~F(x,u)=\int_0^uf(x,t) ~\mathrm{d}t$$

How to prove $\Psi \in C^1(L^p(R^N),R)$ under condtions ($f1$) and ($f2$), where $p=s+1, s \in (1,\frac{N+2}{N-2})$ ?

What is the definition of the duality map and what is the property of the map (in page 1731)?

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How do you see $f$ as a functional on $L^p(R^n)$ in the first place? –  Harald Hanche-Olsen Nov 6 '12 at 14:11
What paper are you reading? It is always helpful to tell where your problem is from. –  martini Nov 6 '12 at 14:16
Thomas Bartsch, Zhi-Qiang Wang, Commun. in Partial Differential Equations, 20,1725-1741(1995) –  mzll2008 Nov 6 '12 at 14:24
The duality map is, if I am note mistaken, the map $\phi \mapsto \langle\phi,\cdot\rangle$ from a Hilbert space to its dual space. By Riesz representation theorem this is an isomorphism. –  Willie Wong Nov 6 '12 at 15:54
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2 Answers

up vote 1 down vote accepted

Assuming that the authors intended, as I commented that $\Psi \in C^1(L^p(\mathbb{R}^N)\cap L^2(\mathbb{R}^N), \mathbb{R})$ instead, then I sketch the proof here for continuity. Differentiability should follow analogously.

Let $u\in L^p\cap L^2$ be fixed. Let $\epsilon > 0$. Chose $R > 0$ such that $\|u\|_{L^2(\mathbb{R}^N \setminus B_R)} < \epsilon / 1000$. Then for $\delta < \epsilon /1000$ we have that any $v$ such that $\|u-v\|_{L^2} < \delta$ also has $\|v\|_{L^2(\mathbb{R}^N \setminus B_R)} < \epsilon / 500$. Similarly, since $\|v\|_p$ is roughly that of $\|u\|_p$, by Chebyshev's inequality we can chose some $\lambda$ such that $u|_{\{|u| > \lambda\}}$ and $v|_{\{|v|> \lambda\}}$ are small in $L^p$. By the fact that $u-v$ is size $\delta$ in $L^2\cap L^p$ we have that $v|_{\{|u|> \lambda\}}$ and $u|_{\{|v| > \lambda\}}$ are also small.

We use the fact that $$ f(x,q) \leq \min( q , a_1 + a_2 |q|^s)$$ by assumption, this implies that $$ F(x,q) \leq \min(\frac12 q^2, a_1 |q| + a_2 |q|^{s+1}) $$ so that $$ \int_\Omega F(x,u) \mathrm{d}x \leq C\min\left( \|u\|_2^2 ,\|u\|_p^p\right) $$ for some fixed constant $C$ depending on $a_2, a_1$. Hence on the "non-compact" sets we have that the difference $$ \int_{B_R^C \cup \{ |u| \text{ or } |v| > \lambda\}} F(x,u) - F(x,v) \mathrm{d}x \leq C\epsilon $$

For the remainder of the spatial domain, we use that $|x| < R$ and $|u|,|v| < \lambda$. Thus we have that $f(x,u)$ is bounded. So $$ |F(x,u) - F(x,v)| < C|u-v| $$ Since we are integrating over a compact set the integral can be bounded by some power of the volume of the set (at most $|B_R|$) times $\|u-v\|_{L^p}$. And so continuity is established.

Formally we have that $$\Psi'(u)[v] = \int f(x,u) v \mathrm{d}x $$ It is not that hard to verify that this is a bounded linear functional (by Holder inequality) and is equal to the derivative. It remains to show that this is continuous in $u$. The argument is largely the same as above, except when estimating the "compact" piece you need to use uniform continuity to control the part where $|u-v|$ is small in $L^\infty$ and to use uniform boundedness and Chebyshev to get that when $|u-v|$ is not small, its support is small.

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What is the Chebyshev's inequality? –  mzll2008 Nov 7 '12 at 1:16
@mzll2008 This is Chebyshev's inequality, which essentially states that an integrable function cannot be too large too often. –  Willie Wong Nov 7 '12 at 8:39
Thanks for you help. –  mzll2008 Nov 9 '12 at 0:07
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As stated it is not true. Are there more conditions?

Let $$ f(x,t) = \begin{cases} t^2 & |t| \leq 1 \\ t^4 & |t| \geq 1 \end{cases}$$

Clearly $f$ satisfies (f1) as $f(x,u) = O(|u|^2) = o(|u|)$ as $u\to 0$ uniformly in $x$, and it satisfies $f(2)$ with $a_1 = a_2 = 1$, $s = 4$. For $N = 3$ we have that $(N+2)/(N-2) = 5 > 4$, so it is okay for $N = 1,2,3$.

Let $$ u(x) = \frac{1}{(1 + |x|)^\frac14} $$ clearly $u(x) \in L^5(\mathbb{R})$. But since $u \leq 1$, we have that

$$ F(x,u) = \int_0^u f(x,t) \mathrm{d}t = \int_0^u t^2 \mathrm{d}t = \frac13 u^3 $$

But the integral $\int_{\mathbb{R}} F(x,u) \mathrm{d}x$ does not converge, so $\Psi$ is not even $C(L^5(\mathbb{R}),\mathbb{R})$, never mind $C^1$.

Similarly for $N = 2$ we can take $u = (1+|x|)^{-1/2}$, and for $N = 3$ we can take $u = (1+|x|)^{-3/4}$ and produce analogous counterexamples.

The problem is that the decay condition $f(x,u) = o(|u|)$ is not strong enough for anything higher than $L^2$ control. Basically, to get the continuity (and similarly differentiability) of the functional, what you need to do is first cut off at a large radius $R$ (chosen relative to the function $u$ at which point you want to show continuity). Outside the large radius $R$ you have that $u$ and all functions $L^p$ close to $u$ have $L^p$ norm at most $\epsilon$, and must decay sufficiently fast. If we have a uniform in $x$ decay rate for $f(x,u)$ as $u\to 0$ that is sufficiently strong, then the integral of $F(x,u)$ outside $R$ is bounded by $L^p$ of $u$ outside $R$ and hence is negligible. Then we can work within the compact region where $u$ is large. Next you excise the points where $u$ blows up. By a Chebyshev inequality like argument these sets have small measure and neither $u$ nor $v$ can concentrate on them. So by excising some set on which $\int F(x,u)$ is of size $\epsilon$, we can assume $u$ is bounded. Now by uniform continuity of a continuous function on a compact set, we can conclude that $\int F(x,u)$ on this remaining set is continuous in $L^p$.

In short:

  1. Show that "non-compact" parts only contribute $\epsilon$
  2. Cut off the non-compact parts, the remaining part is compact and we can upgrade continuity of $f$ to uniform continuity.

I suspect (not having the CPDE paper in front of me at the moment) that (f1) should in fact be replaced by something with better decay properties. But as stated (f1) and (f2) are not sufficient to derive the desired conclusion.

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Actually, looking at the paper, since they are interested in functions in $W^{1,2}$ you actually have that $f\in L^p \cap L^2$. I am pretty sure that the authors intend that $\Psi \in C^1(L^p(\mathbb{R}^N)\cap L^2(\mathbb{R}^N), \mathbb{R})$... –  Willie Wong Nov 6 '12 at 15:51
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