# Uniform convergence

I need an example for sum $$\sum_{1}^{\infty } u_n(x)$$ which converge absolutely and uniformly in [a,b] while the sum $$\sum_{1}^{\infty } \left |u_n(x) \right |$$ does not converge uniformly in [a,b].

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Here is a related question. –  David Mitra Jan 31 '13 at 17:12
The example there is in an open inetrval while i want to find an example in a close interval... but thanks for the help –  user59640 Jan 31 '13 at 17:22
I believe $u_n=(-1)^n x(1-x)^n$, $0\le x\le 1$ works. For $0<x<1$, we have $\sum_{n=0}^\infty |u_n|= x({1\over 1-(1-x)} )=1$, while for $x=0$ and $x=1$, the sum of the absolute values is $0$. $\sum_{n=0}^\infty |u_n|$ is therefore convergent but not uniformly convergent on $[0,1]$. That $\sum_{n=0}^\infty u_n$ converges uniformly follows fromn the fact that $(x(1-x)^n)$ is a decreasing sequence of nonnegative functions that converge uniformly to the zero function. –  David Mitra Jan 31 '13 at 17:22

It would suffice to find a sequence of nonnegative functions $(u_n)$ and an interval $[a,b]$ such that

$\ \ \$1) $\sum\limits_{n=0}^\infty u_n$ is convergent on $[a,b]$,

$\ \ \$2) $\sum\limits_{n=0}^\infty u_n$ is not uniformly convergent on $[a,b]$,

$\ \ \$3) $(u_n)$ is a decreasing sequence of nonnegative functions.

and

$\ \ \$4) $(u_n)$ converges uniformly to the zero function on $[a,b]$.

For then the series $\sum\limits_{n=0}^\infty(-1)^n u_n(x)$ will converge uniformly on $[a,b]$.

Indeed, conditions 3) and 4) guarantee that the series $\sum\limits_{n=0}^\infty(-1)^n u_n(x)$ is Uniformly Cauchy on $[a,b]$. To see this, set $f_n(x)=(-1)^n u_n(x)$. Then for any $x\in[a,b]$ and for any two nonnegative integers $m$ and $n$ with $m\ge n$, we have by 3) that $$|f_n(x)+f_{n+1}(x)+\cdots+ f_m(x)|\le |f_n(x)|.$$ Now from condition 4) we may make $|f_n(x)|$ uniformly small over $[a,b]$ by taking $n$ sufficiently large. It follows that $\sum\limits_{n=0}^\infty(-1)^n u_n(x)$ is Uniformly Cauchy, thus uniformly convergent, on $[a,b]$.

One example of a sequence satisfying the conditions above is given in the comments.

Another would be given by setting $u_n(x)={x^2\over (1+x^2)^n}$ on the interval $[-1,1]$. (Note, using standard facts of Geometric series, that the pointwise limit of $(u_n)$ is $f(x)=\cases{1+x^2, &$x\ne0$\cr 0, &$x=0$}$. So $\sum\limits_{n=0}^\infty u_n$ converges on $[-1,1]$, but not uniformly there. Also note that by Bernoulli's inequality, for $x\ne0$ $${x^2\over(1+x^2)^n}\le {x^2\over 1+nx^2}={1\over n+(1/x^2)}\le {1\over n};$$ The above, together with the fact that $u_n(0)=0$ for all $n$, shows that condition 4) is satisfied.)

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