show that the function is in $L^r$

60 Views Asked by At

Let $f$ be a measurable function and $1 \le p < r < q < \infty$. If there is a constant $C$ such that $$\mu ( \{ x : |f(x ) | > \lambda \} ) \le \frac { C }{ \lambda ^p + \lambda ^ q} $$ for every $ \lambda > 0$, show that $f \in L ^r $.


I know that if both $f \in L^p $ and $f \in L^q $ then $f \in L ^r$. But the condition provided seems not suffices to show that (I can only show that it's in weak $L^p$ or weak $L^q $). Any help is appreciated.

1

There are 1 best solutions below

0
On BEST ANSWER

Hint: $$\int_X f(x)^r\,d\mu(x) = \int_0^\infty r\lambda ^{r-1} \mu(\{x\in X:f(x)> \lambda\})\,d\lambda.$$