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Let $T$ be an operator on $S(G)$ where $G$ is the line $R$ or the circle $T$, and $S(G)$ denotes the Schwartz space of functions on $G$.

We can ask if the operator T is bounded (as an operator from $S \cap L^2$ to $S \cap L^2$) in a variety of ways. These different forms of boundedness include weak type, restricted weak type, strong type, and restricted strong type. By strong type we mean that there is an inequality of the form $||Tf||_{2} \leq C ||f||_{2}$ and by weak type we mean that the operator satisfies a well-known distributional inequality. Placing the word 'restricted' before one of these names indicates that we only require the inequality to hold for functions that are the characteristic functions of sets. We'll denote these types of boundedness as S, W, RW, RS.

The following implications are easy to verify $S -> RS$, $S->W->RW$. More subtly, if we assume that $T$ commutes with translations, it is known that $S=W=RW$ (in the $L^2$ -> $L^2$ setting).

Now take a sequence of such operators, say $T_{n}$, each of which commute with translations, and form an associated maximal function $Mf(x) = sup_{n} |T_{n}f(x)|$.

Question: Is it the case that either $S=W$ or $W=RW$ hold in this setting?

Recall that Stein's maximal principle states that the function $Mf$ is defined almost everywhere (for all $f \in L^2$) if and only if the operator $M$ is weakly bounded.

Notice that the condition that $T_{n}$ commute with translations is equivalent to saying that $T_{n}$ is given by convolution. As evidence that the above statement might be true, recall a theorem of Moon which states that if $T_{n}$ is a sequence of operators given by convolution with integrable kernels then the associated maximal function (as an operator from $L^1$ to $L^1$) satisfies RW=W.

An application of a positive answer would be the following. Assume S=W and take $T_{n}$ to be the n-th partial summation operator (of the Fourier expansion of $f$). The above assertion implies that Carleson's theorem is equivalent to the seemingly stronger Carleson-Hunt inequality $||Mf||_2 < ||f||_2$. By Stein's maximal principle Carleson's theorem is equivalent to a weak $L^2$ bound on the maximal function, using the assertion $S=W$ implies that the weak type inequality implies the strong type inequality (Typically to get the strong type inequalities in this context one needs to prove that the maximal operator is weakly bounded on $L^p$ and $L^q$ with p >2 and q<2 and then interpolate). Also, if $RW=W$ one would only have to deal with characteristic functions in many questions regarding almost everywhere convergence.

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Certainly not $S=W$. Take the operators $T_g$ that convolve with all $L^2$-functions $g$ supported on some interval $I$ and satisfying $\|g\|_{L^2}\le |I|^{-1/2}$ (well, formally you requested a sequence but $L^2$ is separable, so just choose a countable dense set). Then the maximal operator you introduced is merely $\sqrt{M(|f|^2)}$ where $M$ is the usual maximal function. So, it is bounded from $L^2$ to weak $L^2$ only. Probably $RW=W$ is false also but it is too late here now for my brain to work properly.

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This is late, but you may still be interested. I recently learned that RW = W, that is restricted weak = weak for $L^1(\mathbb{R}^d)$. In 1974, Moon proved

Theorem (Moon '74): If $T_n f := f * g_n$ with $g_n \in L^1(\mathbb{R}^d)$ is a sequence of convolution operators and $M$ is their maximal function, then for $q \geq 1$, $M$ is restricted weak type (1,q) if and only if $M$ is weak type (1,q).

However, the statement is not true for $p > 1$. Even more can be proved. Stein and Weiss showed that there exist an operator say $T$ such that $T$ is restricted weak type (p,q), but not weak type (p,q). An example of such an operator is $Tf(x) = x^{-1/q} \int_0^{\infty} y^{-1/p'} f(y) dy$. Here $p'$ is the conjugate exponent of $p$.

See Moon's paper for these results. I discovered this in a paper of Akcoglu, Baxter, Bellow and Jones which, surpisingly says that Moon's theorem is not true if you replace $\mathbb{R}$ with $\mathbb{Z}$!

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Thanks for the comments. However, I don't think they address any of the questions asked. Above, Fedja showed that $S \neq W$. The remaining question is to determine if $RW=W$ for maximal convolution operators on $L^2$. Moon's theorem applies to $L^1$ only and the Stein and Weiss example isn't even a convolution operator! Indeed it is known (for convolution operators) that $W\neq S$ for $p <2$ but $W=S$ for $p=2$. It's a long standing open problem to decided what happens when $p>2$. – Mark Lewko Aug 8 '13 at 19:24
You're right; the operator is not a convolution operator. I missed your comment about Moon's theorem in the original question - sorry for the repeat. Do you have a reference for S = W = RW in the $L^2$ case (for a single convolution operator)? – K Hughes Aug 12 '13 at 17:43
I believe this was first proved by Cowling ( I learned about the result from a paper of Ash ( – Mark Lewko Aug 12 '13 at 21:43
Thank you for the references! – K Hughes Aug 22 '13 at 10:24
I do not understand @fedja argument; in particular, I disagree that the maximal operator Mark Lewko introduced is $\sqrt{M(f^2)}$. If we take $g_n$ to the characteristic function of the interval $(-n,n)$, then we have something like the usual Hardy--Littlewood maximal operator, say $M$ with $||g_n||_2 \leq (\sqrt{2n})^{-1/2}$, but not $\sqrt{M(f^2)}$. What am I missing? – K Hughes Aug 22 '13 at 10:35

Jim Wright pointed out to me that Stein's spherical maximal function fails to be weak-type at the endpoint $L^{\frac{d}{d-1}}(\mathbb{R}^d)$ for $d \geq 3$. This can be seen in Stein's counterexample, provided in his paper The example is $f(x) := |x|^{1-d} \cdot [\log (1/|x|)]^{-1}$ truncated into the ball $|x| \leq 1/2$. However, Bourgain proved that the spherical maximal function is restricted weak-type at the endpoint $L^{\frac{d}{d-1}}(\mathbb{R}^d)$ for $d \geq 3$ in his paper Estimations de Certaines Fonctions Maximales. I apologize for not being able to provide a link here as I could not find one. This shows that your question can fail for some $p$ in $(1,2)$. By studying different hypersurfaces, I think you can show that any rational $p$ in $(1,2)$ will not satisfy $RW = W$.

As for the question of $RW=W$ when $p=2$, the spherical maximal function does not apply as Seeger--Tao--Wright proved that the circular maximal function (= spherical maximal function in $\mathbb{R}^2$) is not restricted weak-type on $L^2(\mathbb{R}^2)$. See Proposition 1.5 in

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A small technical point: the spherical maximal function is defined over a family of averages parametrized by $n>0$; this is equivalent to, by scaling and truncation arguments, averages over the family parametrized by $n \in \mathbb{N}$. – K Hughes Aug 1 '15 at 13:13

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