The question is about the following proof.
I understand why we say $$d_f(\alpha)\leq \min\left(\frac{\|f\|_{p,\infty}^{p}}{\alpha^p},\frac{\|f\|_{q,\infty}^{q}}{\alpha^q}\right).$$
It is the consequence of the definition of the weak Lebesgue norm $$\|f\|_{p,\infty}=\sup { \lambda (D_f(\lambda))^{1/p}}$$ where $D_f$ is the distribution function.
QUESTION 1: Am I correct at this stage?
I understand how we get first equality. But I dont understand how we get the second equality
Also I dont understand how we get third and fourth equalities.
QUESTION 2: Can you give me some hints to handle this proof? I reread previous material in this book several times and did not find any hint.
Maybe it depends on the constant $B$ which we introduced. I dont know the intuition behind this constant also.
For Q1, as you pointed out, since $f$ belongs to $L^{p, \infty}$ as well as to $L^{q, \infty}$, the following two inequalities hold simultaneously \begin{align*} d_f(\alpha)\leq \frac{\|f\|_{p, \infty}^p}{\alpha^p},\\ d_f(\alpha)\leq \frac{\|f\|_{q, \infty}^q}{\alpha^q}, \end{align*} thus \begin{align*} d_f(\alpha)\leq \min\left(\frac{\|f\|_{p, \infty}^p}{\alpha^p}, \frac{\|f\|_{q, \infty}^q}{\alpha^q}\right). \end{align*}
Now Q2. The constant $B$ is chosen to split up the integral in a convenient way, with the aim to handle the $\min$ function appearing from the above inequality. You first use the layer-cake representation formula and then the inequality of the first step. Observe that when $0\leq \alpha\leq B$ a little bit of algebraic manipulation leads us to \begin{align*} \alpha^{q-p}\leq \frac{\|f\|_{q, \infty}^q}{\|f\|_{p, \infty}^p} \Longleftrightarrow \frac{\|f\|_{p, \infty}^p}{\alpha^p}\leq \frac{\|f\|_{q, \infty}^q}{\alpha^q} \end{align*} hence in such interval \begin{align*} \min\left(\frac{\|f\|_{p, \infty}^p}{\alpha^p}, \frac{\|f\|_{q, \infty}^q}{\alpha^q}\right)=\frac{\|f\|_{p, \infty}^p}{\alpha^p}. \end{align*} Similarly, when $B\leq \alpha$ it holds \begin{align*} \min\left(\frac{\|f\|_{p, \infty}^p}{\alpha^p}, \frac{\|f\|_{q, \infty}^q}{\alpha^q}\right)=\frac{\|f\|_{q, \infty}^q}{\alpha^q}. \end{align*} By spliting up the integral in these two intervals and using the above reasoning you get the equality $(5.15)/(1.1.25)$.