Weak $L^{p}$ spaces are quasi-normed?

Question

Let $(X,\mathcal{B}, \mu)$ be a measure space. Then for $0< p < \infty$ by definition

$L^{p,\infty}(X,\mathcal{B}, \mu)$ is the class of all measureable functions $f$ such that

\begin{eqnarray*} \|f\|_{p,\infty} &:=& \text{inf}\{c > 0: d_{f}(\alpha)\leq \frac{c^{p}}{\alpha^{p}}\text{ for all }\alpha > 0\}\\ &=& \text{sup}\{\gamma d_{f}(\gamma)^{\frac{1}{p}}:\gamma > 0\} \end{eqnarray*} where $$d_{f}(\alpha) = \mu(\{x\in X:|f(x)| > \alpha\})$$

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I'm trying to verify that $\|\cdot\|_{p,\infty}$ is a quasi-norm on $L^{p,\infty}$.

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The non-trivial thing to check is that for all $f,g\in L^{p,\infty}(X,\mathcal{B}, \mu)$, we have $\|f + g\|_{p,\infty} \leq c_{p}(\|f\|_{p,\infty} + \|g\|_{p,\infty})$, where $c_{p} = \text{max}\{2,2^{\frac{1}{p}}\}$.

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For $1\leq p < \infty$, I was able to show that $\|f + g\|_{p,\infty} \leq 2(\|f\|_{p,\infty} + \|g\|_{p,\infty})$ using the supremum definition.

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For $0 < p < 1$, I need to show that $\|f + g\|_{p,\infty} \leq 2^{\frac{1}{p}}(\|f\|_{p,\infty} + \|g\|_{p,\infty})$, but I'm stuck.

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The idea is supposed to be to use the following property of $d_{f}$:

$$d_{f + g}(\alpha + \beta)\leq d_{f}(\alpha) + d_{g}(\beta)$$

which implies in particular that

$$d_{f + g}(\alpha)\leq d_{f}\left(\frac{\alpha}{2}\right) + d_{g}\left(\frac{\alpha}{2}\right)$$

Can anyone help me finish the proof? Thanks in advance!

Answer 1

Fix $\gamma>0$ and $0<p<1$. Since $t^{1/p}$ is an increasing function for $t>0$, then using the last inequality in your question we get $$ \left(\frac{d_{f+g}(\gamma)}{2}\right)^{1/p} \leq\left(\frac{d_{f}(\gamma/2)+d_{g}(\gamma/2)}{2}\right)^{1/p} $$ Since $t^{1/p}$ is a convex function we get $$\left(\frac{d_{f}(\gamma/2)+d_{g}(\gamma/2)}{2}\right)^{1/p} \leq\frac{d_{f}(\gamma/2)^{1/p}+d_{g}(\gamma/2)^{1/p}}{2} $$ Therefore $$ d_{f+g}(\gamma)^{1/p}\leq 2^{1/p}\left(\frac{1}{2}d_f\left(\frac{\gamma}{2}\right)^{1/p}+\frac{1}{2}d_g\left(\frac{\gamma}{2}\right)^{1/p}\right) $$ Multiplying by $\gamma$ we obtain $$ \gamma d_{f+g}(\gamma)^{1/p} \leq 2^{1/p}\left(\frac{\gamma}{2}d_f\left(\frac{\gamma}{2}\right)^{1/p}+\frac{\gamma}{2}d_g\left(\frac{\gamma}{2}\right)^{1/p}\right) \leq 2^{1/p}(\Vert f\Vert_{p,\infty}+\Vert g\Vert_{p,\infty}) $$ Tking supremum over $\gamma>0$ in the last inequality we get $$ \Vert f+g\Vert_{p,\infty}\leq 2^{1/p}(\Vert f\Vert_{p,\infty}+\Vert g\Vert_{p,\infty}) $$