# $\langle \nabla g_n, \nabla h \rangle \to \langle \nabla \phi, \nabla h \rangle \implies \langle g_n, \psi \rangle \to \langle \phi, \psi\rangle$?

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?)

• What about $\psi \equiv 0$? – PhoemueX Mar 1 '17 at 13:28
• @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? – abcd Mar 1 '17 at 14:33
• @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. – abcd Mar 1 '17 at 14:36

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 $2\pi$).
• 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?) – abcd Mar 2 '17 at 10:29
• @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. – PhoemueX Mar 2 '17 at 17:20
• @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 :) – PhoemueX Mar 2 '17 at 17:21