I'm reading proofs that $H$, the Hilbert transform, is weak-$(1,1)$, so I'd like to show that there is a constant $C>0$ such that
$$| \{|Hf| > \lambda\}| \le \frac{C}{\lambda} \|f\|_{L^1}$$
for all $\lambda > 0$ and $f \in L^1([-\pi,\pi]).$
All the proofs start by assuming that $ \|f\|_{L^1} < \lambda$ , why it is enough to discuss this case only? Any justification please!
Observe that for the inequality in question is equivalent to $$ \left| \left\{ \left| H\left(\frac{f}{\|f\|_{L^1}}\right) \right| > \frac{\lambda}{ \|f\|_{L^1}} \right\} \right| \leq C \frac{\|f\|_{L^1} }{\lambda}. $$ for all non-trivial functions $f\in L^1$. Now for $\lambda \geq \|f\|_{L^1}$, the right-hand side becomes larger, while the left-hand side becomes smaller. For this reasons, it suffices to prove that it for $\|f\|_{L^1} < \lambda$. In fact, by setting, $g= f / \lambda$, where $f \in L^1$ and $\lambda >0$, the weak-type inequality is equivalent to establishing that
$$ \left| \left\{ \left| H(g) \right| > 1 \right\} \right| \leq C $$ for every $g \in L^1$ with $\|g\|_{L^1}=1$. In other words, it just a standard normalization using a scaling and homogeneity argument, which allows to "disregard" the parameter $\lambda >0$.