My problem is that I find it kind of hard to contrast between uniform and pointwise convergence. For example with this proof I'm not quite sure whether I have proven uniform or poitwise convergence:

$ f_n(x) := \left\{\begin{array}{ll} 2n^2x, & x\in [0,\frac{1}{2n}) \\ -2n^2x+2n, & x\in [\frac{1}{2n},\frac{1}{n})\\ 0,& x \in [\frac{1}{n},1]\end{array}\right. . $

Claim: $(f_n)$ converges pointwise/ uniformly (im not sure) to $f$, where $f(x):= 0$

Proof: Let $\epsilon > 0$

Case one (x = 0):

$\mid f_n(0) -f(0)\mid = \mid 0-0\mid = 0 < \epsilon $ (by definition)

Case two ($x \in (0,1]$):

Set $N \in \mathbb{N}: \frac{1}{N} < x$, now when $n \geq N \Rightarrow \frac{1}{n} < x$ then by definition of $f_n$

$\Rightarrow f_n(x) = 0 \Rightarrow \mid f_n(x)-f(x) \mid = \mid 0-0\mid = 0 < \epsilon $ (by definiton)

Thus the claim is proven q.e.d

I'm asking which type of convergence is this now and whichever it is how would you show the other type?

  • $\begingroup$ Your proof is for pointwise convergence since $N$ depends on $x$. Consider $\max_x |f_n(x)-f(x)|= n$ to show this is not uniform convergence $\endgroup$ – Henry May 8 at 8:03

The convergence is not uniform. Note that $f_n(\frac 1 {4n})=\frac n 2$. Hence whatever $m$ we choose we cannot have $|f_n(x)| <1$ for $n \geq m$ and for all $x$.

  • 1
    $\begingroup$ why this $1/4n$ and not simply $f(1/2n)=n$ ? $\endgroup$ – zwim May 8 at 7:52
  • $\begingroup$ but why does this say that we cannot have uniform convergence here $\endgroup$ – Fo Young Areal Lo May 8 at 7:54
  • $\begingroup$ I just wanted to use $2n^{2}x$ instead of $-2n^{2}x+2n$. @zwim $\endgroup$ – Kavi Rama Murthy May 8 at 7:54
  • $\begingroup$ @FoYoungArealLo Defintion of uniform convergence of $f_n$ to $0$: given $\epsilon >0$ there exists $m$ such that $|f_n(x)| <\epsilon$ for all $n \geq m$ and for all $x$. The inequality must hold even if we make $x$ dependent on $n$. That is exactly where uniform convergence differs from pointwise convergence. $\endgroup$ – Kavi Rama Murthy May 8 at 7:56

What you wrote is a proof of the fact that the sequence $(f_n)_{n\in\mathbb N}$ convergees pointwise to the null function. The convergence is not uniform because, otherwise, you would have$$\lim_{n\to\infty}\int_0^1f_n(x)\,\mathrm dx=\int_0^10\,\mathrm dx=0.$$But this doesn't occur, since, for each $n\in\mathbb N$, $\int_0^1f_n(x)\,\mathrm dx=\frac12$.

  • $\begingroup$ Could you state why it doesn't converge uniformly with another argumentation without integrals since in my calculus class we haven't defined them yet $\endgroup$ – Fo Young Areal Lo May 8 at 7:59
  • $\begingroup$ I could, of course, but then I would just be repeating the fine answer provided by Kavi Rama Murthy. $\endgroup$ – José Carlos Santos May 8 at 8:00

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