# multiplying a uniformly convergent sequence of functions by $x^{-1/2}$

Given that $\sqrt{x}f_n(x) \rightarrow g$ uniformly where each $f_n$ is integrable.

I would like to show $\lim\limits_{n\rightarrow \infty}\int_0^1 f_n(x)dx$ exists.

So I have tried to show that $f_n$ converges uniformly. But I have not succeeded, is there something else I could try?

-
Is the convergence uniform on $\left[0,1\right]$? Did you try to show that the sequence $\{I_n\}$ where $I_n=\int_0^1f_n(x)dx$ is a Cauchy sequence? –  Davide Giraudo Oct 18 '11 at 18:38
@ Davide yes it's uniform on [0,1] and No I haven't tried that. –  user9352 Oct 18 '11 at 18:57
Since we talk about integrable in connection with a uniform limit I guess we look at Riemann integrable? –  AD. Oct 18 '11 at 19:29
@AD. yeah (I tried to convey that with the dx )but I'd be interested in looking at Lebesgue as well I just haven't thought about it yet. –  user9352 Oct 18 '11 at 20:38

Try this:

$|\int_0^1 f_n(x) dx - \int_0^1 f_m(x) dx | \leq \int_0^1 \frac{1}{\sqrt{x}}|\sqrt{x}f_n(x) - \sqrt{x}f_m(x)| dx$

and use the fact that $\sqrt{x}f_n(x) \to g(x)$ uniformly to get that the sequence is Cauchy.

-
In other words, using $U(f)(x)=\sqrt{x}f(x)$, $\|f_1-f_2\|_1\leqslant2\|U(f_1)-U(f_2)\|_\infty$. –  Did Oct 21 '11 at 6:04
It is interesting that even though $f_n$ may not converge to a Riemann integrable function, $\displaystyle\lim_{n\to\infty}\int_0^1f_n(x)\;\mathrm{d}x$ exists. (+1) For example, let $g(x)=\sqrt[3]{x}$ and $$f_n(x)=\left\{\begin{array}{}\frac{1}{\sqrt[6]{x}}&\text{for}&x>\frac{1}{n}\\0& \text{for} &x\le\frac{1}{n}\end{array}\right.$$ –  robjohn Oct 21 '11 at 7:55
@robjohn In your example, $f_n\to x^{-1/6}$ which is Lebesgue (and Riemann) integrable. In general, the argument I gave above can be used to show that $f_n$ is Cauchy in $L^1(0,1)$ and so converges to an integrable function. –  Jeff Oct 25 '11 at 5:50
And I think you mean $g(x)=x^{1/2-s}$. –  Jeff Oct 25 '11 at 5:56
@Jeff: $x^{-1/6}$ is Lebesgue integrable, but is not Riemann integrable on $[0,1]$ since it is not bounded (all upper sums are infinite). –  robjohn Oct 25 '11 at 9:35

Since $\sqrt{x}f_n(x)$ converges uniformly to $g(x)$, we have that that for any $\epsilon>0$, there is an $N>0$ so that for $m,n\ge N$, $|\sqrt{x}(f_m(x)-f_n(x))|\le\epsilon$ for all $x\in[0,1]$. Thus, $|f_n(x)|\le|f_m(x)|+\frac{\epsilon}{\sqrt{x}}$. Thus, for $n\ge N$, $f_n$ is dominated by the sum of two integrable functions. Since $f_n(x)\to \frac{g(x)}{\sqrt{x}}$ pointwise, we have that $\displaystyle\lim_{n\to\infty}\int_0^1 f_n(x)\;\mathrm{d}x$ exists by Dominated Convergence.

Furthermore, $\displaystyle\lim_{n\to\infty}\int_0^1 f_n(x)\;\mathrm{d}x=\int_0^1 \frac{g(x)}{\sqrt{x}}\mathrm{d}x$ and $\displaystyle\left\|\frac{g(x)}{\sqrt{x}}\right\|_{L^1[0,1]}\le\left\|f_m\right\|_{L^1[0,1]}+2\epsilon$

-
This would be for the Lebesgue integral. –  robjohn Oct 21 '11 at 7:20