I have a sequence of functions $f_n:[0,2] \to \mathbb R$ defined by

$$ f_n(x) = \begin{cases} n^3x^2 & 0<x<1/n\\ n^3(x-\tfrac{2}{n})^2 & 1/n\le x<2/n \\ 0 & \text{otherwise} \end{cases} $$

I needed to determine the pointwise limit for the function and to determine if $(f_n)$ converges uniformly to $f$. So what I did was:

If $x=0$, $\lim_{n\to \infty} f_n(0)=0$ for all $n$.

If $x>0$, let $\dfrac{2}{N}<x$, then $f_n(x)=0$ for all $n>N$.

So the pointwise limit of $f(x)$ is $0$.

But how would I determine if $(f_n)$ converges uniformly? Does it have something to do with the supremum? $\sup|f_n(x)-f(x)|$?


1 Answer 1


Your reasoning that your sequence $(f_n)$ converges pointwise to the zero function $f(x)=0$ on $[0,2]$ is correct.

Considering uniform convergence now, let's first recall the definition of uniform convergence of a sequence of functions:

The sequence $(g_n)$ converges uniformly to $g$ on $I$ if for every $\epsilon>0$ there is a positive integer $N$ such that $$\tag{1} |g_n(x)-g(x)|<\epsilon,\quad\text{for all } n\ge N\ \text{and}\ x\in I. $$

Note that $(1)$ must hold for each $n\ge N$ and every $x\in I$. (It does have something to do with the supremum $\sup\limits_{x\in I}|g_n(x)-g(x)|$: this must be be able to be made as small as desired by taking $n$ sufficiently large.)

Back to your sequence. Note that for any positive integer $n$, we have $$\tag{2}f_n(1/n)=n.$$

Now let $\epsilon=1 $ and let $N$ be any fixed positive integer. By $(2)$, we have $$|f_N(1/N) -f(1/N)|=f_N(1/N)=N\ge {1 }=\epsilon.$$ This shows that for $\epsilon=1$ there is no positive integer $N$ such that $(1)$ holds (with $g_n=f_n$ and $g=f$). Consequently, $(f_n)$ does not converge uniformly on $[0,2]$ (or in fact, on any interval containing $0$).

The graphs of the $f_n$ prove illuminating. The sequence resembles the "witch hat" sequence: it consists of spikes near $x=0$ whose widths approach zero but whose heights grow large. The graphs of the first few terms of the sequence are shown below:

enter image description here

  • $\begingroup$ Cheers david! May I ask, is what I did to determine the pointwise limit correct? $\endgroup$
    – Frank
    Feb 15, 2012 at 1:42
  • $\begingroup$ @Frank, yes, it's perfectly correct. $\endgroup$ Feb 15, 2012 at 1:59
  • $\begingroup$ okay, thanks again for your help. $\endgroup$
    – Frank
    Feb 15, 2012 at 2:02
  • $\begingroup$ @DavidMitra Where did you plot those? $\endgroup$
    – Pedro
    Feb 15, 2012 at 2:12
  • $\begingroup$ @Peter Using JSXgraph. $\endgroup$ Feb 15, 2012 at 2:13

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