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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.

Taylor Rendon
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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.

Glitch
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