I have been trying some questions on uniform convergence.Got stuck in one of those questions which says that

For a positive real number p, let (f$_n$) is a sequence of functions defined on [$0,1$] by

$$f_n(x) = \begin{cases} n^{p+1}x, \text{if 0 $\le$ $x$ $\lt$ $\frac{1}{n}$}\\ \frac{1}{x^p}, \text{if $\frac{1}{n}$ $\le$ $x$ $\le$ 1} \end{cases}$$

I have found its point-wise limit given by

$$f(x)= \begin{cases} 0, \text{if $x$ = $0$}\\ \frac{1}{x^p}, \text{if 0 $\lt$ $x$ $\le$ $1$} \end{cases}$$

I am stuck in proving whether its uniformly convergent or not.

I take any $\epsilon$ $\gt$ $0$.Now I need to know that does there exist a natural number m such that $\lvert f_n(x)-f(x)\rvert$ $\lt$ $\epsilon$ for all n $\geq$m and for all $x$ in [$0,1$]?



1 Answer 1


$|f_n(\frac 1 {n^{p+1}}) -\frac 1 {(\frac 1 {n^{p+1}})^{p}}|=|1-n^{p(p+1)}| \to \infty$. Hence $\sup_x |f_n(x)-f(x)|$ does not tend to $0$ and the convergence is not uniform

  • $\begingroup$ Just for variety, and no better than the answer given, another approach is to observe that each $f_n$ is continuous on $[0,1]$ while $f$ is not. Uniform convergence of continuous functions has a continuous limit (if that theorem is available to you), so the convergence cannot be uniform. $\endgroup$
    – WA Don
    Jun 4, 2020 at 14:23
  • $\begingroup$ @WADon thank you $\endgroup$
    – Gitika
    Jun 4, 2020 at 15:08
  • $\begingroup$ @Kavi Rama Murthy thanks a lot $\endgroup$
    – Gitika
    Jun 4, 2020 at 15:09

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