Uniform or pointwise convergence of a sequence of functions

Question

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?

Answer 1

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 $x$.

Answer 2

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 $n\in\mathbb N$, $\int_0^1f_n(x)\,\mathrm dx=\frac12$.