Estimate for weak $L^{1}$ norm

Question

Let the weak $L^{1}$ norm on $f$ be defined by $\|f\|_{\mathrm{WL}^{1}} = \sup_{t > 0}t D_{f}(t)$ where $D_{f}(t) = \mu(\{x \in \mathbb{R}: |f(x)| > t\})$ and $\mu$ is the standard Lebesgue measure. Consider the $N!$ functions on $[0, 1]$ defined by $$f_{\sigma} = \sum_{j = 1}^{N}\frac{N}{\sigma(j)}\chi_{[(j - 1)/N, j/n)}$$ where $\sigma$ is a permutation of $\{1, 2, \ldots, N\}$. Why is $$\left\lVert \sum_{\sigma \in S_{N}}f_{\sigma}\right\rVert_{\mathrm{WL}^{1}} = N!(1 + 1/2 + \cdots + 1/N)?$$

Answer 1

Rearranging terms in the sum $\sum_{\sigma} f_\sigma$, one sees that every interval $\left[\frac{j-1}{N}, \frac{j}{N}\right)$ appears exactly once in any summand $f_\sigma$, each time with weight $N/\sigma(j)$. So the sum $\sum_\sigma f_\sigma$ is equal to the constant function $\sum_\sigma \frac{N}{\sigma(j)}$.

It remains to compute the sum. When $\sigma$ runs over all permutations, $\sigma(j)$ takes all values $1, 2\ldots N$, and each value is repeated $(N-1)!$ times. Therefore $$\sum_{\sigma\in S_N} \frac{1}{\sigma(j)}=(N-1)!\left( 1+\frac{1}{2}+\dots+\frac{1}{N}\right).$$

To conclude one only needs to observe that the weak $L^1$ norm of a constant function $f\equiv C$ defined on an interval of unit length is exactly $\lvert C\rvert$.