Convergence of a series of functions in L^p

Question

I found some problems in solving this exercise:

Find $p>1$ such that the series $\sum_n f_n$ converges in $L^p(\mathbb{R})$ (with lebesgue measure) where: \begin{equation} f_n(x)=\dfrac{1}{1+n^2\sqrt{x}}\chi_{[\exp(2n),2\exp(2n+2)]} \end{equation}

My idea was to calculate the $L^p$ norm of the previous function and find for which values of $p$ the series $\sum_{n=1}^{\infty}\|f_n\|_p$ is convergent, then knowing that $(L^p,\|\|_p)$ is a Banach space the series $\sum_n f_n$ has to be convergent in $L^p$. The problem is that the calculation on the $L^p$-norm of $f_n$ seems to be too much complicate.

Can someone help me?

Answer 1

Instead of trying to calculate the $L^p$ norm, try to find an upper bound and lower bound.

I find $$ \frac{e^{2n}(2e^2-1)}{(1+n^2\sqrt{2}e^{n+1})^p}\leq \|f\|_p\leq \frac{e^{2n}(2e^2-1)}{(1+n^2e^{n})^p}\leq \frac{e^{2n}(2e^2-1)}{n^{2p}e^{np}} $$

And it is enough to conclude on the convergence or divergence of the serie of the norm (and since the fonction are positive, of the potential divergence of the serie of fonction)

Answer 2

We can write

\begin{align} f_n(x)&=\dfrac{1}{1+n^2\sqrt{x}}\chi_{[\exp(2n),2\exp(2n+2)]}\\[0.3cm] &=\dfrac{1}{1+n^2\sqrt{x}}\chi_{[\exp(2n),\exp(2n+2)]}+\dfrac{1}{1+n^2\sqrt{x}}\chi_{[\exp(2n+2),2\exp(2n+2)]}. \end{align} Because everything is positive, the convergence of $\sum f_n$ is equivalent to the convergence of the two series on the right. The advantage is the the series from the right have terms with disjoint support. So \begin{align} \bigg\|\sum_{n=m}^k\dfrac{1}{1+n^2\sqrt{x}}\chi_{[\exp(2n),2\exp(2n+2)]}\bigg\|_p^p &=\int_0^\infty\sum_{n=m}^k\dfrac{1}{(1+n^2\sqrt{x})^p}\chi_{[\exp(2n),2\exp(2n+2)]}\,dx\\[0.3cm] &=\sum_{n=m}^k\int_{e^{2n}}^{e^{2n+2}}\dfrac{1}{(1+n^2\sqrt{x})^p}\,dx\\[0.3cm] \end{align} Using the obvious lower and upper bounds for the integrand, $$ \sum_{n=m}^k \dfrac{e^{2n}(e^2-1)}{(1+n^2e^{n+1})^p}\,dx\leq\sum_{n=m}^k\int_{e^{2n}}^{e^{2n+2}}\dfrac{1}{(1+n^2\sqrt{x})^p}\,dx\leq\sum_{n=m}^k \dfrac{e^{2n}(e^2-1)}{(1+n^2e^{n+1})^p}\,dx. $$ The other series behaves similarly. Now you can simply study the convergence of the numeric series.