1
$\begingroup$

Let $g_n, \phi \in \dot{H}^1(\mathbb R^d)= \{f: \nabla f \in L^{2}(\mathbb R^d) \}.$

Assume that $\int_{\mathbb R^d} \nabla g_n \cdot \nabla h \ dx \to \int_{\mathbb R^d} \nabla \phi \cdot \nabla h \ dx$ as $n\to \infty$ for all $h\in \dot{H}^1.$ (In other words, $g_n$ converges to $\phi$ weakly in $\dot{H}^1$.)

Question: Can we expect to find $\psi \in \mathcal{S}(\mathbb R^d)$ (Schwartz Space) so that $$\lim_{n\to \infty} \int_{\mathbb R^d} g_{n}(x) \psi(x) dx = \int_{\mathbb R^d} \phi(x) \psi (x)dx?$$

Side Note: Let $\phi$ be a radial bump function supported on $\{ \xi: |\xi|\leq 2 \}$ which equals to 1 on $\{\xi: |\xi|\leq 1\}.$ Put $h(\xi)= \phi(\xi)- \phi(2\xi),$ and $h_{1}(\xi)= h(\xi/2).$ Take $\psi(x)= (h_{1})^{\vee}(x)$ (Inverse Fourier transform of $h_1$) (Is this the right choice?)

$\endgroup$
  • $\begingroup$ What about $\psi \equiv 0$? $\endgroup$ – PhoemueX Mar 1 '17 at 13:28
  • $\begingroup$ @PhoemueX: Thanks: Yes for that it is true. But in my purpose, I need $\psi$ as the convolution kernel of Little-wood Paley projection operator. (So I have defined $\psi$ accordingly) Can you give any suggestion or remark? $\endgroup$ – abcd Mar 1 '17 at 14:33
  • $\begingroup$ @PhoemueX: Specifically, I am trying to understand the proof of Inverse Strichatz estimates Proposition 3.2, p.242 (Chapter 3). The above fact has been used in it (see p.242). Where this has been used. $\endgroup$ – abcd Mar 1 '17 at 14:36
1
$\begingroup$

One issue is your precise definition of $\dot {H}^1$. You need to factor out the constants, i.e., it is a subspace of $\mathcal {S}'/\Bbb {C} $.

Anyway, you have $g_n \to \phi $ weakly, i.e., you only need to verify that $g \mapsto \int g \psi $ is a linear bounded functional on $\dot {H}^1$. But now, if $\psi$ is a Schwartz function such that $\hat {\psi} $ is supported away from $0$, then $$ \left|\int g \psi \right|=| \langle g, \overline {\psi} \rangle_{L^2}| = |\langle \hat {g} \hat {\overline {\psi}}\rangle_{L^2}| \leq \int |\hat {\nabla g}(\xi)| \cdot |\overline { \hat{\psi}(-\xi)} |/(2\pi |\xi|) d\xi $$ can be estimated by a constant multiple of $\|\hat {\nabla g}\|_{L^2} = \|g\|_{\dot {H}^1} $.

Above, I repeatedly used Plancherel''s Theorem. Also, the last line might be slightly different depending on the normalisation that your are using for the Fourier transform (give or take a factor of $2\pi $).

$\endgroup$
  • $\begingroup$ Thanks. Please would you tell me how should I factor $\dot{H}^1$ by constants? How should I show it is a subspace of $\mathcal{S}'/ \mathbb C$? Any reference I should take look? (If it is long?) $\endgroup$ – abcd Mar 2 '17 at 10:29
  • 1
    $\begingroup$ @abcd: Mainly, I was just being picky. You defined $\dot{H}^1 = \{f : \nabla f \in L^2(\Bbb{R}^d)\}$, but you never stated where $f$ comes from (is $f \in L^2$, is $f$ a tempered distribution?). Since the norm on $\dot{H}^1$ is $\|f\|_{\dot{H}^1} = \| \nabla f\|_{L^2}$, you need to exclude functions/distributions $f \not \equiv 0$ with $\nabla f \equiv 0$. These are precisely the constants. This problem is discussed in some detail in "Grafakos, Modern Fourier Analysis, Section 1.1.1". In fact, he considers the space $\mathcal{S}' / \mathcal{P}$, i.e., modulo polynomials. $\endgroup$ – PhoemueX Mar 2 '17 at 17:20
  • 1
    $\begingroup$ @abcd: You dont need to show that $\dot{H}^1$ is a subspace of $\mathcal{S}' / \Bbb{C}$, you just need to define it properly :) $\endgroup$ – PhoemueX Mar 2 '17 at 17:21
  • $\begingroup$ Fine, Thanks a lot. $\endgroup$ – abcd Mar 3 '17 at 6:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.