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.

  • $\begingroup$ You need some kind of compactness statement for this on the domain of definition. Then search for 'Dini' $\endgroup$
    – Thomas
    Mar 30, 2014 at 9:15
  • $\begingroup$ @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. $\endgroup$ Mar 30, 2014 at 11:16

1 Answer 1


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.

...Which happens to be false, as a single counterexample is enough to prove. Such as the following one.

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 $0$.

  • $\begingroup$ If it helps the OP (or anyone reading this) with visualizing the sequence, it is very similar to sequence given in answers to this post. In those answers graphs of those functions are depicted, too. $\endgroup$ Mar 30, 2014 at 11:14
  • $\begingroup$ @MartinSleziak Thanks. $\endgroup$
    – Did
    Mar 30, 2014 at 11:21

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