For each $n \in \Bbb N$, let $$f_{n}(x) = \begin{cases}nx, &0\leq x\leq \frac{1}{n}\\ 1, &\frac{1}{n}<x\leq1\end{cases}$$ Find the limit function of the sequence $f_{n}$. Show that the convergence of the sequence is not uniform.

I am having difficulty in writing a formal proof.

As $n$ tends to $\infty$, we have $x=0$, so $f(x)= 0$. So limit function is $f(x)= 0$.

Next step is to show that the convergence is not uniform. Since for $x=1$ the condition $|f_{n}(x) - f(x)|< \epsilon$ is not satisfied if we take $\epsilon$ less than $1$, so the convergence is not uniform.

Is this approach correct?

Please guide me to write a formal proof for the first part.


3 Answers 3


limit function $f$

For $x=0$ holds $x \le \frac1n$ for all $n\in\mathbb N$. So $f_n(0) = 0$ for all $n\in\mathbb N$. Thus $f(0)=0$.

For every $x>0$ exists a $N\in\mathbb N$ such that $x>\frac1n$ for all $n>N$ (Archimedean property). That is $f_n(x) = 1$ for all $n>N$. Thus $f(x)=1$.

So $$f(x) = \begin{cases}0 & x=0\\1 & 0<x\le 1.\end{cases}$$

no uniform convergence

Proof by contradiction: let be $0<ε<\frac12$. Assume there is a $N\in\mathbb N$ such that for all $x\in[0,1]$ and all $n\ge N$: $$\vert f_n(x)-f(x)\vert < ε.$$ Choose $x=\frac{1}{2N}$. Then $f_N(x) = \frac12$. So $$\vert f_N(x)-f(x)\vert = \vert \frac12 - 1\vert = \frac12 \not< ε $$ what is a contradiction.

  • $\begingroup$ pH 74. Well done. $\endgroup$ Mar 5, 2020 at 9:50
  • $\begingroup$ Thanks @PeterSzilas! I did not do such proofs for about 10 years. So it was a nice refreshment for myself… $\endgroup$
    – pH 74
    Mar 5, 2020 at 10:25
  • $\begingroup$ pH74. I struggle a bit too:) $\endgroup$ Mar 5, 2020 at 10:48


For $x >0$:

Pick $n_0$ (Archimedean principle) s.t for $n \ge n_0$: $1/n <x$, then $f_n(x)=1$.i.e. $\lim_{n \rightarrow \infty}f_n(x)=1$ for $x>0$.

For $x=0$: $\lim_{n \rightarrow \infty}f_n(0)=0.$.

$f_n$ are continuos on $[0,1]$.

If $f_n$ were uniformly convergent the limit function $f$ would be continuos. Hence?

  • $\begingroup$ $f$ is discontinuous, so convergence must be non-uniform. $\endgroup$
    – Mathaddict
    Mar 5, 2020 at 9:15
  • 1
    $\begingroup$ Mathsaddict.Yes, if f not continuos, convergence is not uniform. $\endgroup$ Mar 5, 2020 at 9:21

$\lim f_n(0)=0$ is clear. For $x>0$, $|f_n(x)-1|=0<\epsilon $ if $n >\frac 1 x$. So $f_n(x) \to f(x)$ for every $x$ where $f(0)=0$ and $f(x) =1$ for $x >0$.

If $ 0<\epsilon <1/2$ then $|f_n(\frac 1 {n^{2}}) -f(\frac 1 {n^{2}})|>\epsilon$ for every $n\geq 2$. Hence the convergence is not uniform.

  • $\begingroup$ Kavi.Second last line $f(1/n^2)$. $\endgroup$ Mar 5, 2020 at 9:24
  • $\begingroup$ @PeterSzilas Yes, thank you for pointing out. $\endgroup$ Mar 5, 2020 at 9:30

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