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Show that $f$ is belongs to weak $L^{1}(E)$ and $f$ is bounded on $E$. Then $f$ is $L^{p}(E)$ for each $1<p<\infty$ ,where $E$ is a subset of $\mathbb{R^{n}}.$

Here is my proof :

Now we take the notation that $\lVert f\rVert_{L^{1,\infty}}\,$ to be the space of weak $L^{1}(E)$

Find a positive number $K$ such that $|f(x)|\leq K$ for every $x\in E$ since $f$ is bounded on $E$.

In addition, $f\in L^{1,\infty}(E)$ that we can set for each $\alpha>0,$ $$\omega(\alpha)\leq \frac{\lVert f\rVert_{L^{1,\infty}}}{\alpha}$$

,where $\omega(\alpha)$ is the distribution function of $f$.

We now estimate the $L^{p}$ norm of $f$ when $p$ between $1$ and $\infty$ as below :

\begin{align} \lVert f\rVert_{L^{p}(E)}^{p}&=p\int_{0}^{\infty}\alpha^{p-1}\omega(\alpha)\, d\alpha\\ &=p\int_{0}^{\infty}\alpha^{p-1}\chi_{\{\alpha\,:\,0<\alpha\leq K\}}\omega(\alpha)\, d\alpha\\ &=p\int_{0}^{K}\alpha^{p-1}\chi_{\{\alpha\,:\,0<\alpha\leq K\}}\omega(\alpha)\, d\alpha\,+p\int_{K}^{\infty}\alpha^{p-1}\chi_{\{\alpha\,:\,0<\alpha\leq K\}}\omega(\alpha)\,d\alpha\\ &=p\int_{0}^{K}\alpha^{p-1}\omega(\alpha)\,d\alpha\,+0\\ &\leq p\int_{0}^{K}\alpha^{(p-1)-1}\lVert f\rVert_{L^{1,\infty}}\,\,d\alpha\\ &=p\bigg(\frac{1}{p-1}\alpha^{p-1}|_{0}^{K}\bigg)\lVert f\rVert_{L^{1,\infty}}\\ &=\frac{p}{p-1}K^{p-1}\lVert f\rVert_{L^{1,\infty}} \end{align}

Whence,

\begin{align} \lVert f\rVert_{L^{p}(E)}&\leq\bigg({\frac{p}{p-1}}\bigg)^{\frac{1}{p}}K^{\frac{p-1}{p}}\lVert f\rVert_{L^{1,\infty}}^{\frac{1}{p}}=\bigg({\frac{p}{p-1}}\bigg)^{\frac{1}{p}}K^{1-{\frac{1}{p}}}\lVert f\rVert_{L^{1,\infty}}^{\frac{1}{p}}\,<\infty \end{align}

Is there anyone has the time to check my working for validity ? any valuable suggestion and advice would be appreciated.Thanks for patient reading.

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I fail to see why $\omega(\alpha)\leq K$, but luckily the rest of your reasoning goes through. The correct way to write it should be $\omega(\alpha)=\omega(\alpha)\chi_{0\leq\alpha\leq K}$.

Anyway, the virtue of this proposition is that, there is less information could be told by just weak $L^{1}$, in particular, it is very wrong to say that weak $L^{1}$ can control $L^{p}$. Rather, if we impose additional condition of being $L^{\infty}$, then all other $L^{p}$ are controlled. So this brings the air that $L^{\infty}$ is just strong enough to turn the situation into this way.

So in the set theoretic way, this means $L^{1,\infty}\cap L^{\infty}\subseteq L^{p}$.

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