The functionals $\|\cdot\|_{L^{p,q}}$ do not satisfy the triangle inequality. Given a measurable function $f$ on a measure space $(X,\mu)$ and $0<p,q\leq \infty$, define  $$\|f\|_{L^{p,q}}=\left\{
\begin{array}{ll}
\displaystyle{\left(\int_{0}^\infty\left(t^{1/p}f^*(t)\right)^q\,\frac{dt}{t}\right)^{1/q}}, & \mbox{si } q<\infty,  \\
\sup_{t>0}t^{1/p}f^*(t), & \mbox{si } q=\infty.  
\end{array}
\right.$$
Consider the functions $f(t)=t\quad$ and$\quad g(t)=1-t$ defined on $[0,1]$. My question is: How can I find $f^*$ and $g^*$? where \begin{align*} f^*: [0,\infty)&\longrightarrow [0,\infty)\\ t&\longmapsto f^*(t)=\inf\{s>0: d_f(s)\leq t\}\end{align*}
and $$d_f(s)=\mu\left(\{x: |f(x)|>s\}\right),\ \ s>0$$
denotes the distribution function.
The Loukas Grafakos-Classical Fourier Analysis book suggests that $f^*(\alpha)=g^*(\alpha)=(1-\alpha)\mathcal{X}_{[0,1]}(\alpha)$. Here $\mathcal{X}$ denotes the characteristic function.
 A: The measure intended is just standard Lebesgue measure on $[0,1]$, so
$$
d_f(s) = \mu(\{t \in [0,1] : t > s\}) = 
\begin{cases}
\mu((s,1]) & \text{if } 0 < s < 1 \\
\mu(\varnothing) & \text{if } s \ge 1 
\end{cases} 
\\
= 
\begin{cases}
1-s & \text{if } 0 < s < 1 \\
0 & \text{if } s \ge 1. 
\end{cases} 
$$
Similarly,
$$
d_g(s) = \mu(\{t \in [0,1] : 1-t > s\}) = 
\begin{cases}
\mu([0,1-s)) & \text{if } 0 < s < 1 \\
\mu(\varnothing) & \text{if } s \ge 1 
\end{cases} 
\\
= 
\begin{cases}
1-s & \text{if } 0 < s < 1 \\
0 & \text{if } s \ge 1, 
\end{cases} 
$$
and we arrive at the key point: $f$ and $g$ have the same distributions.  On the other hand, $f+g =1$, so
$$
d_{f+g}(s) = 
\begin{cases}
1 & \text{if } 0 < s < 1\\
0 & \text{if } s \ge 1.
\end{cases}
$$
To compute $f^\ast = g^\ast$, we observe that for $0 \le t \le 1$,
$$
1-s \le t \Leftrightarrow 1-t \le s
$$
while for $t >1$ we have $1-s \le t$ for all $s \in [0,1]$.  Thus,
$$
f^\ast(t) = g^\ast(t) = 
\begin{cases}
1-t & \text{if } 0 \le t \le 1 \\
0 & \text{for } t > 0,
\end{cases}
$$
and $(f+g)^\ast$ can be computed similarly.  Now you can plug into the definitions of the quasinorms to verify the assertion about the failure of the triangle inequality.
