# uniform convergence sequence

The sequence of function $\{f_n\}$ defined on $\mathbb{R}$, every function is decreasing function (if $x \geq y$ then $f_1(x)\geq f_1(y)$, $f_2(x)\geq f_2(y)$,.......) and sequences of function is decreasing function ($f_n\geq f_{n+1}$) and $\{f_n \}$ converges point wise.
Then $f_n$ converges to $f$ uniformly.

Example :

$f_n(x)=x+\frac{1}{n}$ , $n=1,2,3$..... defined on $(-\infty , 0]$. This sequence of function is uniformly convergence

My assumption is correct.

• You need some kind of compactness statement for this on the domain of definition. Then search for 'Dini' – Thomas Mar 30 '14 at 9:15
• @Thomas I'll just add that Dini's theorem does not have the assumption that the functions themselves are monotonous. (Which is an assumption in this question.) But from the example posted it Did's anwer, it seems that this additional assumption does not help. – Martin Sleziak Mar 30 '14 at 11:16

Perhaps one should first recall that one cannot prove that a statement holds simply exhibiting a case where it holds. Here, to exhibit a sequence $(f_n)$ such that all the hypotheses hold and such that $f_n\to f$ uniformly proves nothing about the validity of the general statement.
Assume that $f_n(x)=1$ if $x\leqslant0$, $f_n(x)=1-nx$ if $0\leqslant x\leqslant\frac1n$ and $f_n(x)=0$ if $x\geqslant\frac1n$. Then, as desired, each function $f_n$ is nonincreasing, the sequence $(f_n(x))$ is nonincreasing for every fixed $x$, and $f_n\to f$ pointwise with $f(x)=1$ if $x\leqslant0$ and $f(x)=0$ if $x\gt0$.
But the convergence $f_n\to f$ is not uniform since $f_n(\frac1{n^2})=1-\frac1n$ and $f(\frac1{n^2})=0$ hence the difference $f_n(\frac1{n^2})-f(\frac1{n^2})$ does not converge to $0$.