Question about required inequality for proof that Hilbert Transform is weak-(1,1) I'm reading a couple of proofs that $H$, the Hilbert transform, is weak-$(1,1)$ and have struggled to justify an inequality that all the proofs mention can be proved without justification.
Given
$$
b(x) = \sum_j b_j(x)
$$
then they use the fact that
$$ |Hb(x)| \leq \sum_j |Hb_j(x)| \text{ almost everywhere}.$$
Interestingly, both proofs mention that this is immediate if the above sum is finite and can be proved otherwise using the fact that $\sum_j Hb_j$ converges to $Hb$ in $L^2$.  Indeed the finite case is clear, and the infinite case seems intuitive to me, but I'm having trouble using norm convergence to show the inequality!  Here's what I can do.
By the linearity of the transform we have that $H(\sum_{j=1}^n b_j(x)) = \sum_{j=1}^n Hb_j(x)$.  As stated above we can immediately have
$$
\left| \sum_{j=1}^n Hb_j(x) \right| \leq \sum_{j=1}^n |Hb_j(x)|,
$$
and so by the $L^2$ convergence we have
$$
\int |Hb(x)|^2dx \leq \int \sum_{j=1}^n |Hb_j(x)|^2 = \sum_{j=1}^n \int |Hb_j(x)|^2.
$$
This is true for all $n$, thus we have
$$
\int |Hb(x)|^2dx \leq \sum_j \int |Hb_j(x)|^2 = \int \sum_j |Hb_j(x)|^2.
$$
However I now seem to have reached an impasse, as there seems no reason to me why this inequality of integrals would imply an inequality of integrands.  I can't help but feel I've gone awry somewhere, or went in the wrong direction.  Any clarity is much appreciated!
 A: Here's an approach that still uses integrals, but in a slightly different way.
Suppose to the contrary that
$$| Hb(x) |
\leq
\sum_j |Hb_j (x)|
$$
does not hold almost everywhere. Then we can find a set $E$ of finite positive measure where the opposite (strict) inequality holds. Therefore the integral
$$
\int_E
\bigl(
|Hb(x)| - \sum_j |Hb_j (x)|
\bigr) \, dx
$$
has a positive value, call it $A$.
Now using Hölder's inequality, the fact that the series $\sum_j Hb_j$ converges to $Hb$ in $L^2$ implies that the series $\sum_j \chi_E Hb_j$ converges to $\chi_E Hb$ in $L^1$, where $\chi_E$ denotes the characteristic function of $E$.
On the other hand, for any $N$ we have
\begin{align*}
\left\|
\chi_E Hb - \sum_{j=1}^{N} \chi_E Hb_j
\right\|_{1}
&= \int
\biggl|
\chi_E(x) Hb(x)
- \sum_{j=1}^{N} \chi_E(x) Hb_j(x)
\biggr|
\, dx \\
&\geq \int_E
\biggl(
| Hb(x) |
- \sum_{j=1}^{N} | Hb_j(x) |
\biggr)
\, dx \\
&\geq \int_E
\biggl(
| Hb(x) |
- \sum_j | Hb_j(x) |
\biggr)
\, dx \\
&= A.
\end{align*}
This contradicts the $L^1$ convergence, implying the result.
