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.
Best Answer
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.