Holder inequality for variable Lebesgue spaces

63 Views Asked by At

I'm reading the proof on the book: http://www.crm.cat/en/Publications/Publications/2011/Pr1093.pdf

and I don't understand this sentence in page 24:

Combining the above terms, and using the fact that each is needed precisely when the $L^{\infty}$ norm of the corresponding characteristic function equals 1

Can someone explain me from where $||\chi_{\Omega}||_{L^{\infty}}$ appear ?

Thank you

1

There are 1 best solutions below

2
On

The proof proves and combines three bounds that can be stated as follows.

\begin{align} \int_{\Omega_\infty} |f(x)g(x)| \leq& \|f\|_{p(\cdot)}\|g\|_{p'(\cdot)} \\ \int_{\Omega_1} |f(x)g(x)| \leq& \|f\|_{p(\cdot)}\|g\|_{p'(\cdot)} \\ \int_{\Omega_*} |f(x)g(x)| \leq& \bigg(1 + \frac{1}{p_-} - \frac{1}{p_+}\bigg)\|f\|_{p(\cdot)}\|g\|_{p'(\cdot)} \end{align} A crude bound would come from noting that the left hand sides of these bounds sum up to $\int_\Omega |f(x)g(x)|$ and hence this quantity can be bounded by the sum of the right hand sides. The problem is that if one of $\Omega_\infty$, $\Omega_1$ or $\Omega_*$ has measure $0$ then the right hand side in the list of inequalities above is a terrible bound.

Since $\|\chi_\omega\|_\infty = 0$ if $\omega$ has $0$ measure and is $1$ otherwise, we can improve the inequalities above to deal with this issue. We obtain \begin{align} \int_{\Omega_\infty} |f(x)g(x)| \leq& \|f\|_{p(\cdot)}\|g\|_{p'(\cdot)} \|\chi_{\Omega_\infty}\|_\infty\\ \int_{\Omega_1} |f(x)g(x)| \leq& \|f\|_{p(\cdot)}\|g\|_{p'(\cdot)}\|\chi_{\Omega_1}\|_\infty \\ \int_{\Omega_*} |f(x)g(x)| \leq& \bigg(1 + \frac{1}{p_-} - \frac{1}{p_+}\bigg)\|f\|_{p(\cdot)}\|g\|_{p'(\cdot)}\|\chi_{\Omega_*}\|_\infty. \end{align} Summing up these inequalities gives the desired bound.