UMD Analysis Qualifying Exam/Jan09 Real

Problem 3

Let $f \in L^1 (-\infty,\infty) \!\,$ and suppose $\alpha > 0 \!\,$. Set $f_n(x)=\frac{f(nx)}{n^{\alpha}} \!\,$ for $n=1,2,\ldots \!\,$. Prove that for almost every $x \in (-\infty, \infty) \!\,$, $\lim_{n \rightarrow \infty} f_n(x) =0 \!\,$

Solution 3

Change of variable

By change of variable (setting u=nx), we have

$\int |f_n(x)| dx = \int \frac{|f(u)|}{n^{\alpha +1}} du \quad \quad (*)\!\,$

Monotone Convergence Theorem

Define $u_n(x) = \sum_{i=1}^{n} |f_i(x)| \!\,$.

Then, $u_n \!\,$ is a nonnegative increasing function converging to $\sum_{i=1}^{\infty} |f_i(x)| \!\,$.

Hence, by Monotone Convergence Theorem and $(*) \!\,$

$\begin{align} \int \sum_{i=1}^{\infty} |f_i(x)| dx &= \sum_{i=1}^{\infty} \int |f_i(x)| dx \\ &= \sum_{i=1}^{\infty} \int \frac{|f(x)|}{i^{\alpha +1 }}dx \\ &= \left( \int |f(x)| dx\right) \left( \sum_{i=1}^{\infty} \frac{1}{i^{\alpha +1 }} \right) \\ &< \infty \end{align} \!\,$

where the last inequality follows because the series converges ($\alpha >0 \!\,$ ) and $f \in L^1 \!\,$

Conclusion

Since

$\int \sum_{i=1}^\infty |f_i(x)| dx < \infty \!\,$,

we have almost everywhere

$\sum_{i=1}^\infty |f_i(x)| < \infty \!\,$

This implies our desired conclusion:

$\lim_{i \rightarrow \infty} f_i(x) = 0 \quad \mbox{a.e.}\!\,$