Visualizing where uniform convergence fails but pointwise convergence holds

Question

I am well acquainted with the concepts of pointwise and uniform convergence, namely on the former one fixes a point $x$ and then investigates if a given sequence of functions converges. In the latter, one sees if all $x$ can converge "at once". A more intuitive way to frame the latter is one can construct an $\epsilon$-tube around $f$ that contains all functions $f_n$ for all $n \geq N$ for some $N$ (depending on $\epsilon$ of course).

A friend of mine is taking analysis for the first time asked for an analogous image, but instead where pointwise convergence holds yet uniform convergence fails and I fell short. I am hoping someone on here can come up with a better mental picture than I can.

Answer 1

A canonical example, mentioned in the comments, is $f_n(x)=x^n$ on $[0,1]$. Converges pointwise to $0$ for $x<1$ and to $1$ for $x=1$. The convergence cannot possibly be uniform because a uniform limit of continuous functions is continuous.

A more illustrative example, in my opinion, is a "travelling bump". Let $f_n=1_{[n,n+1]}$. That is, $$ f_n(x)=\begin{cases}1,&\ n\leq x\leq n+1\\[0.3cm] 0,&\ \text{otherwise}\end{cases} $$ Then $f_n\to0$ pointwise, since for any $x$ we will eventually have $x<n$ for $n$ large enough. Meanwhile $\max\{|f_n(x)|: x \}=1$ for all $n$; that is $\operatorname{dist}(f_n,0)=1$ for all $n$, so the convergence is not uniform. Of course the example can be easily modified to have $f_n$ continuous or even smooth.

Answer 2

A classic example is a sigmoid function:

$$f(x;n)= \frac{1}{1+e^{-nx}}$$

This function converges to a Heaviside step function as $n \to \infty$ pointwise but not uniformly, since it has a fixed point at $f(0)=\frac12$ — a permanent discontinuity of $\frac12$ from the limiting value at $n=\infty$

Answer 3

Consider $A=[0,1)$ and $f_{n}=x^n$, then $f_{n}$ converges pointwise to zero but not uniformly.

Every point in $[0,1)$ eventually goes to $0$, for instance focus on the grey point, it has an $x$ value very close to $1$, but it still goes to $0$ eventually, the same is true for any point, however together the sequence of functions is never completely less than even $0.5$, and so while the sequence converges pointwise to $0$ (as depicted by the grey dot) it does not converge uniformly(since it's always 'stuck' to the point $(1,1)$).