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In one of the steps for the proof of Theorem 2.3.20, Grafakos shows that you can pass a continuous, linear functional through an integral by considering the convergence (in the topology of the Schwartz space on $\mathbb{R}^n$) of the Riemann sum of the inner integral. In particular, Grafakos shows that

$$ \sum_{m=1}^{(2N^2)^n} x^{\alpha} \partial_x^{\beta} \widetilde{\varphi}(x - y_m) \psi(y_m) |Q_m| \to \int_{\mathbb{R}^n} x^{\alpha} \partial_x^{\beta} \widetilde{\varphi}(x - y) \psi(y) dy $$

as $N \to \infty$ (in $L^{\infty}$), where we consider a partition on $[-N, N]^n$ into $(2N^2)^n$ cubes $Q_m$ of side length $1/N$ and with $y_m$ as the center of each cube $Q_m$.

Now, for the actual question. In the proof, Grafakos states that

$$ x^{\alpha} \partial_x^{\beta} \widetilde{\varphi}(x - y_m) \psi(y_m) |Q_m| - \int_{Q_m} x^{\alpha} \partial_x^{\beta} \widetilde{\varphi}(x - y) \psi(y) dy = \int_{Q_m} x^{\alpha} (y - y_m) \nabla(\partial_x^{\beta} \widetilde{\varphi}(x - \cdot) \psi)(\xi) dy $$

for some $\xi = y + \theta(y_m - y)$, where $\theta \in [0, 1]$.

I'm stuck on Grafakos' derivation of the estimation above. In particular, I am not sure on where the gradient operator comes from (I suspect it's an integration by parts argument, but then I'm not sure where $\xi$ comes from).

If anyone could shed some light on the thought process behind the estimation, I would greatly appreciate it.

opaladozen
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1 Answers1

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I finally figured out that Grafakos uses the mean-value theorem, which is stated in Appendix I.2. I really wished that he would at least give a hint that he's using it...

For the future reader, I'll elaborate on the idea (while being a little bit loose with details and not worrying about making the signs the same as Grafakos' estimate). Assuming $g$ is a sufficiently nice function, Appendix I.2 of Grafakos states $$ g(t_1) - g(t_2) = \nabla f((1 - c)t_2 + ct_1) \cdot (t_1 - t_2) $$ for some $c \in (0, 1)$.

Next, let $g(t) = (\partial_x^{\beta} \widetilde{\varphi}(x - \cdot) \psi)(t)$ with the variable subsitutions $t_1 = y_m$ and $t_2 = y$. Then, from above, we have $$ \partial_x^{\beta} \widetilde{\varphi}(x - y_m) \psi(y_m) - \partial_x^{\beta} \widetilde{\varphi}(x - y) \psi(y) = (y_m - y) \cdot \nabla((\partial_x^{\beta} \widetilde{\varphi}(x - \cdot) \psi)((1 - \theta)y + \theta y_m)) $$ for some $\theta \in (0, 1)$. Now, observe that $(1 - \theta)y + \theta y_m = y + \theta(y_m - y)$. Then, integrate both sides over $Q_m$ with respect to $y$ and the rest of the proof in Grafakos' book should follow with no issues.

opaladozen
  • 39