8
$\begingroup$

In Stein and Shakarchi's Functional Analysis (Princeton Lectures in Analysis Vol. 4), the authors claim in Section 2 Exercise 17 that the function $$f(x):=\dfrac{\chi_{|x|\leq 1/2}}{x(\log|x|)^{2}}$$ does not belong to the real Hardy space $H^{1}(\mathbb{R})$. Specifically, the authors write

"Consider the function $f$ defined by $f(x)=1/(x((\log x)^{2})$ for $0<x\leq 1/2$ and $f(x)=0$ if $x>1/2$, and extended to $x<0$ by $f(x)=-f(-x)$. Then $f$ is integrable on $\mathbb{R}$, with $\int f=0$, hence $f$ is a multiple of a 1-atom in the terminology of Section 5.2 Verify that $M(f)\geq c/(|x|\log|x|)$ for $|x|\leq 1/2$, hence $M(f)\notin L^{1}$, thus by Theorem 6.1 we know that $f\notin H_{r}^{1}$."

$H_{r}^{1}$ is their notation for the (atomic) real Hardy space. Theorem 6.1 refers to the $L^{1}$ boundedness of the maximal convolution operator $M(f)(x):=\sup_{t>0}|\Phi_{t}\ast f(x)|$ on $H_{r}^{1}$, where $\Phi$ is $C^{1}$ and compactly supported.

This seems false. Decompose $f$ as $$\sum_{j=2}^{\infty}f_{j},\quad f_{j}:=f\chi_{2^{-j}\leq |x| < 2^{-j+1}}$$ Then $$\|f_{j}\|_{L^{\infty}}\leq \left(2^{-j}(\log|2^{-j+1}|)^{2}\right)^{-1}=\dfrac{2^{j}}{(j-1)^{2}(\log 2)^{2}}\leq c2^{-j}j^{-2}$$ By odd symmetry $\int f_{j}=0$ for all $j$. Since $|\left\{2^{-j}\leq x<2^{-j+1}\right\}|=2^{-j+1}$, we can define $\infty$-atoms $a_{j}$ by $$a_{j}(x):=(2c)^{-1}j^{2}f_{j}$$ and write $$f=\sum_{j}2cj^{-2}a_{j},$$ which belongs to $H^{1}(\mathbb{R})$. In fact, it seems that $f$ is precisely an example that the subspace of compactly supported $L\log L$ functions is properly contained in $H^{1}(\mathbb{R})$. Indeed, for $c>0$ sufficiently small \begin{align*} \int_{-1/2}^{1/2}|f(x)|\log^{+}|f(x)|dx&=\int_{-c}^{c}\dfrac{1}{|x|(-\log|x|)}dx=\infty \end{align*} One can also see that $f\notin L\log L$, as the Hardy-Littlewood maximal function of $f$ is not integrable on a neighborhood of the origin.

Returning to my original assertion, am I being silly here? Or is this indeed an error in the text.

$\endgroup$
2
  • $\begingroup$ Maybe try math overflow? $\endgroup$
    – Rescy_
    Commented Nov 7, 2015 at 22:38
  • 2
    $\begingroup$ @Rescy_: I don't think it's appropriate for Math Overflow, as it's not exactly a research-level question. Either I'm making an error in my offered atomic decomposition, or I am not. I'm sure there are users on this site who can help me out. $\endgroup$ Commented Nov 7, 2015 at 22:41

1 Answer 1

4
+50
$\begingroup$

You are sane. Your argument with the atomic decomposition is of course irrefutable, and $f\in H^1$. For an independent way to see this, we can use the fact that $H^1(\mathbb R)$ functions can also be characterized as the real parts of the boundary values of holomorphic functions $F$ on $\mathbb C^+$ with $\sup_{y>0} \int_{-\infty}^{\infty} |F(x+iy)|\, dx <\infty$. Now I would really like to take (and in the first version, I did take) $F(z)=1/(z\log^2 z)$ (with, let's say, $\textrm{Im}\, \log z\in (0,\pi)$ for $z\in\mathbb C^+$). This doesn't quite work though because of the pole at $z=1$. We can fix this by modifying as follows: $$ F(z) = \left( \frac{1}{z} - \alpha \frac{1+i}{z+i} - (1-\alpha) \frac{1+2i}{z+2i} \right) \frac{1}{\log^2 z} , $$ where we choose $\alpha$ so that $F$ becomes continuous at $z=1$ ($\alpha=2+i$ is what I got). Then $F\in H^1$, and near $x=0$, $F(x)$ behaves in essentially the same way as $f(x)$, so from the characterization of $H^1$ via $Mf$ it now follows that $f\in H^1$ also.

Stein's and Shakarchi's claim that $(Mf)(x)\gtrsim 1/|x\log x|$ is simply incorrect. I suspect what they had in mind was to take $t=x$ if $\Phi$ is supported by $[-1,1]$, say, so that $$ |(\Phi_x * f)(x)| = \frac{1}{x} \int_0^2 \frac{\Phi(1-u)\, du}{u\log^2 ux} , $$ and this would be $\gtrsim 1/|x\log x|$ if I could ignore the $\Phi(1-u)$, but of course this is small where the rest of the integrand is large, so I'm getting extra small factors. This isn't quite a direct proof that $Mf\in L^1$ (though we know this from your argument), but with more care one should be able to obtain one from this.

$\endgroup$
0

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .