# Hardy Space Cancellation Condition

I have been reading the chapter on Hardy spaces in Stein's Harmonic Analysis book, and I am having a lot of trouble figuring something out.

The setting here is $\mathbb{R}^n.$ Let $f \in L^q$ be compactly supported, and let $\phi$ be a Schwartz function. As usual, define $\phi_t(x) = t^{-n} \phi(t^{-1} x).$ Define the maximal function $M_{\phi}f(x)$ to be $\sup_{t > 0} | \phi_t * f(x)|.$ Stein claims that if we assume that $\int f = 0$, then $M_{\phi}(f)$ is less than or equal to $c | x|^{-n-1}$ for large $x$. He says that the smoothness of $\phi$ and cancellation condition on $f$ are very important here.

I have not been able to figure out why this is true. I assume you have to integrate by parts and then use the fact that the gradient of $\phi$ is decreasing really quickly, but I can't seem to get the details to work out correctly. Can someone please explain why we have this decay bound on $M_{\phi}(f)$?

Let $B$ be the support of $f$. Pick $c_{b} \in B$. By the cancellation condition on $f$, we have $$M_{\phi}(f)(y)=\sup_{t>0} \left|\int_{y+B} \left(\frac{1}{t^n}\phi\left(\frac{x}{t}\right)-\frac{1}{t^n}\phi\left(\frac{y+b_{0}}{t}\right)\right)f(y-x)\, dx\right|$$

Using the mean value theorem, we get

$$\frac{1}{t^n}\phi\left(\frac{x}{t}\right)-\frac{1}{t^n}\phi\left(\frac{y+b_{0}}{t}\right)=\frac{1}{t^{n+1}}\phi'\left(\frac{x_{b}}{t}\right)$$ for some $x_{b} \in y+B$. Since $\phi$ is Schwartz function, we have

$$|\phi'(x)| \leq \frac{C}{1+|x|^{n+1}}$$

Now we can prove what we wanted.

$$\begin{split} |M_{\phi}(f)(y)| &= \sup_{t>0}\left|\int_{y+B}\frac{1}{t^{n+1}}\phi'\left(\frac{x_{b}}{t}\right)f(y-x) \, dx\right| \\ & \leq \|f\|_{L^q}\sup_{t>0}\left(\int_{y+B}\left|\frac{1}{t^{n+1}}\phi\left(\frac{x_{b}}{t}\right)\right|^{p} \, dx\right)^{1/p} \\ & \leq \|f\|_{L^q}\sup_{t>0}\left(\int_{y+B}\left|\frac{1}{t^{n+1}+x_{b}^{n+1}}\right|^p \, dx\right)^{1/p} \\ & =\|f\|_{L^q} \left(\int_{y+B}\frac{1}{x_{b}^{np+p}}\, dx\right)^{1/p} \\ &\approx \|f\|_{L^{q}} \,m(B)^{1/p}\, \frac{1}{(y+c_{b})^{n+1}} \approx \frac{C}{y^{n+1}} \end{split}$$

• @ Guillermo - what is $b_0$? Jul 12 '14 at 11:03
• @ Guillermo - I don't understand why we integrate over $y+B$, shouldn't we be integrating over $B(y, t)$ and is $b_0\in\mathbb{R}^n$ chosen so that $|y+b_0|\geq t$? Jul 14 '14 at 7:11
• @ Guillermo - I also don't follow the mean value theorem line; if $x_b/t$ lies in the line segment between the points $x/t$ and $(y+b_0)/t$, then shouldn't the RHS be something like $$\frac{1}{t^{n+1}}(y+b_0-x)\cdot\phi'(x_b/t)?$$ Jul 15 '14 at 7:47
• @Nirav you are right. RHS should be like that. But it doesn't matter because $y+b_{0}-x$ is in $B$, the bounded range. Jul 15 '14 at 7:53
• @Nirav $c_{b}$ is any point in $B$ and I assumed that WLOG $B$ is convex. Jul 15 '14 at 7:57