Let $(X, \mathcal A,\mu)$ be a measure space, and $0<p<\infty$
Definition: The weak $L^p-$space on $(X, \mathcal A,\mu)$ denoted $L^{p,\infty}(X, \mu)$ is defined as the set of all $\mu$-measurable functions $f$ such that: $$\|f\|_{L^{p,\infty}} = \sup\{ t\mu\left(\{x\in X: |f(x)|>t\}\right)^{1/p}: t>0\}<\infty.$$
One can easily check that the map $f\mapsto \|f\|_{L^{p,\infty}}$ is a quasi norm. namely we have the quasi-triangular inequality $$\|f+g\|_{L^{p,\infty}} \le \max(2,2^{1/p})\left(\|f\|_{L^{p,\infty}}+\|g\|_{L^{p,\infty}}\right)$$
Now we assume is a $\sigma-$finite measure space. Let $0<r<p<\infty $ we define $$|\|f|\|_{L^{p,\infty}} = \sup_{0<\mu(E)<\infty} \mu(E)^{-\frac1r+\frac1p} \left(\int_E |f|^rd\mu\right)^{1/r}$$ the surpemun is taken over all measurable subsets $E$ of $X$ of finite measure.
Question: prove that $|\|\cdot|\|_{L^{p,\infty}}$ and $\|\cdot\|_{L^{p,\infty}}$ are equivalent. Then conclude that $L^{p,\infty}$ is normable for $p>1$ and metrisable for $0<p\le 1.$
I was able to prove that, $$\|f\|_{L^{p,\infty}}\le |\|f|\|_{L^{p,\infty}}$$ which is easy from the definition. Now How can I prove that
$$|\|f|\|_{L^{p,\infty}}\le c_p \|f\|_{L^{p,\infty}}$$
This would answer the others questions it is not difficult to see that $$f\mapsto |\|f|\|_{L^{p,\infty}}$$ is a norm for $p>1.$
It suffices to show that there exists a constant $c$ such that for each $E$ of positive and finite measure, $$ \int_E\left\lvert f\right\rvert^r\leqslant c\left\lVert f \right\rVert_{\mathbb L^{p,\infty } }^p\mu\left(E\right)^{1-p/r}. $$ To this aim, use Fubini's thereom to get $$ \int_E\left\lvert f\right\rvert^r=\int_{\mathbb R }\mu\left(\left\{ x\mid \left\lvert f(x)\right\rvert^r\mathbf 1_E (x) \gt t \right\} \right) \mathrm dt $$ and observing that $$ \left\{ x\mid \left\lvert f(x)\right\rvert^r\mathbf 1_E (x) \gt t \right\} =\left\{ x\mid \left\lvert f(x)\right\rvert^r\gt t \right\} \cap E $$ we derive that $$ \int_E\left\lvert f\right\rvert^r\leqslant \int_{\mathbb R }\min\left\{ \mu\left(\left\{ x\mid \left\lvert f(x)\right\rvert^r \gt t \right\} \right),\mu\left(E\right) \right\} \mathrm dt. $$ Since $$\mu\left(\left\{ x\mid \left\lvert f(x)\right\rvert^r \gt t \right\} \right)=\mu\left(\left\{ x\mid \left\lvert f(x)\right\rvert \gt t^{1/r} \right\} \right)\leqslant t^{-p/r}\left\lVert f \right\rVert_{\mathbb L^{p,\infty } }^p,$$ we are able to conclude.