# Uniform or pointwise convergence of a sequence of functions

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?

• 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 – 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 $$x$$.
• why this $1/4n$ and not simply $f(1/2n)=n$ ? – zwim May 8 at 7:52
• I just wanted to use $2n^{2}x$ instead of $-2n^{2}x+2n$. @zwim – Kavi Rama Murthy May 8 at 7:54
• @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. – 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 $$n\in\mathbb N$$, $$\int_0^1f_n(x)\,\mathrm dx=\frac12$$.