the additivity for space of weak $L^p$

Question

There is an exercise in my textbook with working by myself :

Let $f_1,f_2,...,f_N$ be in $L^{p,\infty}(X,\mu)$ which is the well-known space of weak $L^p(X,\mu)$

Prove that for $1\leq p< \infty$ we have $$\bigg\lVert\sum_{j=1}^{N} f_j\bigg\rVert_{L^{p,\infty}}\leq N\sum_{j=1}^{N}\lVert f_j\rVert_{L^{p,\infty}}$$ proof : Now we proceed by induction on N,and using $\omega_{|f_j|}(\alpha)=\bigg|\{x\in X:|f_j(x)|>\alpha\}\bigg|$

For each $\alpha>0$ and $p\in[1,\infty)$ we have

\begin{align} \alpha^p\omega_{|f_1+f_2|}(\alpha)\leq\alpha^p\bigg(\omega_{|f_1|}(\dfrac{\alpha}{2})+\omega_{|f_2|}(\dfrac{\alpha}{2})\bigg)\\ = \alpha^p\omega_{|f_1|}(\dfrac{\alpha}{2}) +\alpha^p\omega_{|f_2|}(\dfrac{\alpha}{2})\\ =2^p\alpha^p\omega_{|f_1|}(\alpha)+2^p\alpha^p\omega_{|f_2|} (\alpha) \end{align}

Hence,we have $$\alpha\bigg(\omega_{|f_1+f_2|}(\alpha)\bigg)^{\dfrac{1}{p}}\leq\ 2\bigg(\alpha^p\omega_{|f_1|}(\alpha)+\alpha^p\omega_{|f_2|}(\alpha)\bigg)^{\dfrac{1}{p}}\leq2\alpha\bigg(\omega_{|f_1|}(\alpha)\bigg)^{\dfrac{1}{p}}+ 2\alpha\bigg(\omega_{|f_2|}(\alpha)\bigg)^{\dfrac{1}{p}}$$ ,where the last inequality holds by $p\in[1,\infty).$

Therefore, $$\alpha\bigg(\omega_{|f_1+f_2|}(\alpha)\bigg)^{\dfrac{1}{p}}\leq 2\lVert f_1\rVert_{L^{p,\infty}}\,\,+2\lVert f_2\rVert_{L^{p,\infty}}$$ Then, $$\bigg\lVert f_1+f_2\bigg\rVert_{L^{p,\infty}}\,\leq 2\lVert f_1\rVert_{L^{p,\infty}}\,\,+2\lVert f_2\rVert_{L^{p,\infty}}\,=2\bigg(\lVert f_1\rVert_{L^{p,\infty}}\,\,+\lVert f_2\rVert_{L^{p,\infty}}\bigg)$$

,we completes our proof by induction on finite number N.

So, is there anybody can check my proof for valid ?Thanks for considering my request.

Answer 1

The expression $\alpha^{p}\omega_{|f_{1}|}(\alpha/2)=2^{p}\alpha^{p}\omega_{|f_{1}|}(\alpha)$ is not true in general, but the philosophy is closed.

Rather, one could see without problem that $\{|f_{1}+\cdots+f_{N}|>\alpha\}\subseteq\displaystyle\bigcup_{i=1}^{N}\{|f_{i}|>\alpha/N\}$, so \begin{align*} \omega_{|f_{1}+\cdots+f_{N}|}(\alpha)=\mu(\{|f_{1}+\cdots+f_{N}|>\alpha\})\leq\sum_{i=1}^{N}\mu\left(\left\{|f_{i}|>\frac{\alpha}{N}\right\}\right)=\sum_{i=1}^{N}\omega_{|f_{i}|}(\alpha/N), \end{align*} so $\alpha^{p}\omega_{|f_{1}+\cdots+f_{N}|}(\alpha)\leq N^{p}(\alpha/N)^{p}\displaystyle\sum_{i=1}^{N}\omega_{|f_{i}|}(\alpha/N)\leq N^{p}\sum_{i=1}^{N}\|f_{i}\|_{L^{P,\infty}}^{p}$.

So $\|f_{1}+\cdots+f_{N}\|_{L^{p,\infty}}^{p}\leq N^{p}\displaystyle\sum_{i=1}^{N}\|f_{i}\|_{L^{P,\infty}}^{p}$. Now make use of the inequality that \begin{align*} (a_{1}+\cdots+a_{N})^{1/p}\leq\left(a_{1}^{1/p}+\cdots+a_{N}^{1/p}\right),~~~~p\geq 1, \end{align*} we get \begin{align*} \left(N^{p}\displaystyle\sum_{i=1}^{N}\|f_{i}\|_{L^{P,\infty}}^{p}\right)^{1/p}\leq N\sum_{i=1}^{N}\|f_{i}\|_{L^{p,\infty}}. \end{align*}