Limits of integral series

Question

Suppose we have the series of functions:

\begin{equation} F(x)=\sum_{n=1}^{\infty} f_n(x) \end{equation}

where convergence is uniform.

Additionally, consider the partial functions of the series:

\begin{equation} F_m (x)=\sum_{n=1}^{m} f_n(x) \end{equation}

Suppose that:

\begin{equation} \int_{0}^{1} \frac{F_m(x)}{x^{3}} ,dx \end{equation}

forms a Cauchy sequence, hence convergent.

Can we then claim that:

\begin{equation} \int_{0}^{1} \frac{F_m(x)}{x^{3}} ,dx \to \int_{0}^{1} \frac{F(x)}{x^{3}} ,dx \end{equation}

as $m \to \infty$?

observation: \begin{equation} \int_{0}^{1} \frac{F(x)}{x^{3}} ,dx \end{equation}

exist

Answer 1

The answer is no with a simple counterxample: consider $G_n(x)=n\chi_{[0,1/n]}(x)$ and $F_n(x)=x^3G_n(x), 0 \le F_n(x) \le 1/n^2$

(so one can define $f_n=F_n-F_{n-1}$)

Then clearly $F_n \to 0$ uniformly and $\int_0^1G_n(x)dx=1 \to 1$, while $F=0$ so $\int_0^1 F(x)dx/x^3=0$

With a little smoothing one clearly can make everything continuous or differentiable as wished etc

Answer 2

No.

Indeed, make the substitution $x=u^{-1/2}$. That is, let $G_n(u):=F_n(u^{-1/2})$ and $G(u):=F(u^{-1/2})$, so that $\int_0^1\frac{F_n(x)}{x^3}\,dx=\frac12\,\int_1^\infty G_n$ and $\int_0^1\frac{F(x)}{x^3}\,dx=\frac12\,\int_1^\infty G$. So, we can restate the question as follows:

Let $(G_n)$ be a sequence of functions on $[1,\infty)$ converging uniformly to a function $G$ on $[1,\infty)$ so that $\int_1^\infty G_n$ converges (as $n\to\infty$) and $\int_1^\infty G$ exists. Does it then follow that $\int_1^\infty G_n\to\int_1^\infty G$?

To show that this is not true, let e.g. $$G_n:=\frac1n\,1_{[n,2n]}.$$ Then $G_n\to0=:G$ uniformly but $$\int_1^\infty G_n=1\not\to0=\int_1^\infty G.$$