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I'll first type the problem.

Let $(X,\mu)$ be a measure space and let $s>0$.

(a) Let $f$ be a measurable function on $X$. Show that if $0 < p < q < \infty$, we have $$ \int_{|f| \leq s} |f|^q \leq \frac{q}{q-p} s^{q-p} \|f\|_{L^{p,\infty}}^p. $$

(b) Let $f_j, 1 \le j \le m$, be measurable functions on $X$ and let $0 < p < \infty$. Show that $$\bigg|\bigg|\max_{1 \leq j \leq m} |f_j|\bigg|\bigg|_{L^{p,\infty}}^p \leq \sum_{j=1}^m \| f_j\|_{L^{p,\infty}}^p. $$

(c) Conclude from part (b) that for $0 < p < 1$ we have $$\|f_1 + \dots + f_m \|_{L^{p,\infty}}^p \leq \frac{2-p}{1-p} \sum_{j=1}^m \|f_j\|_{L^{p,\infty}}^p. $$

Hint: Part (a): Use the distribution function. Part (c): First obtain the estimate $$d_{f_1+\dots+ f_m}(\alpha) ≤ \mu({| f_1+\dots+ f_m|>\alpha,\max| f_j|≤\alpha})+d_{\max_j| f_j|}(\alpha)$$ for all $\alpha > 0$ and then use part (b).

I have managed to solve part (a) and (b) and to prove the estimate of the hint given for part (c) but I haven't been able to solve part (c). What I have done is this

$$\| f_1+\dots+f_m\|_{L^{p,\infty}}^p = \sup_{\alpha> 0} \alpha^p d_{f_1+\dots+f_m}(\alpha). $$ Thus I need to bound the right hand side for any $\alpha>0$. We have, by the estimate given in the hint

\begin{align*}\alpha^p d_{f_1+\dots+ f_m}(\alpha) &\le \alpha^p\mu({| f_1+\dots+ f_m|>\alpha,\max| f_j|≤\alpha})+\alpha^pd_{\max j} | f_j|(\alpha) \\ &\leq \alpha^p\mu({| f_1+\dots+ f_m|>α,\max| f_j|≤\alpha}) + \bigg|\bigg|\max_{1 \leq j \leq m} |f_j|\bigg|\bigg|_{L^{p,\infty}}^p\\ & \leq \alpha^p\mu({| f_1+\dots+ f_m|>\alpha,\max| f_j|≤\alpha})+\sum_{j=1}^m \| f_j\|_{L^{p,\infty}}^p \end{align*}

So I need to estimate $$\alpha^p\mu({| f_1+\dots+ f_m|>\alpha,\max| f_j|\le\alpha})$$ I noticed that if in part (a) I plug $q = 1$ and $s =1$ for $g= \max |f_j|$ I would obtain the result but I can't see how to actually relate $\alpha^p\mu({| f_1+\dots+ f_m|>\alpha,\max| f_j|≤\alpha})$ to $\int_{|\max_j |f_j|| \leq 1} | \max |f_j||$ Any help is appreciated.

Alex Ortiz
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Franlezana
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1 Answers1

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Let $g = \max|f_j|$ and $E = \{\max|f_j|\le\alpha,|f_1+\dots+f_m|>\alpha\}$. You can see that \begin{align*} \alpha^p\mu(E) &\le \int_E|f_1+\dots+f_m|^p\\ &=\int_E|f_1+\dots+f_m|^1\cdot|f_1+\dots+f_m|^{p-1}\\ &\le\alpha^{p-1}\int_E\big(|f_1|+\dots+|f_m|\big), \end{align*} the third inequality holding by the triangle inequality, and since $p-1 < 0$ and $|f_1+\dots+f_m|>\alpha$ on $E$.

Note that on $E$, $|f_j|\le g \le \alpha$ for each $j$, so ultimately using the bound from part (a) in the inequality marked by $(*)$, $$ \alpha^p\mu(E)\le \alpha^{p-1}\sum_j\int_{|f_j|\le\alpha}|f_j| \stackrel{(*)}{\le} \alpha^{p-1}\bigg(\frac{1}{1-p}\alpha^{1-p}\sum_j\|f_j\|_{p,\infty}^p\bigg) = \frac{1}{1-p}\sum_j\|f_j\|_{p,\infty}^p. $$ Together with what's written in the OP, this finishes the proof of the claim of part (c).

Alex Ortiz
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