Does pointwise convergence against a continuous function imply uniform convergence?

Solution 1:

Your example, $$ f_n(x) = \begin{cases} e^{1-1/({1-((n+1) x-1)^2})} & 0<x<\frac{2}{n+1} \\ 0 & \text{otherwise,} \end{cases} $$ behaves something like so:

I think it's crystal clear that this is a continuous (even differentiable, I think) sequence of functions that converges pointwise to zero. The convergence clearly can't be uniform, since there's always a point where the value of $f_n$ is $1$.

The example I had in mind looks something like this:

Note that the graph consists of just straight line segments, so it's quite simple to write down a piecewise formula for this. In fact:

$$ f_n(x) = \begin{cases} (n+1) x & 0<x<\frac{1}{n+1} \\ 2-(n+1) x & \frac{1}{n+1}<x<\frac{2}{n+1} \\ 0 & \text{otherwise}. \end{cases} $$

I emphasize, though, that (in my approach, at least) the picture comes first.

Solution 2:

That's not necessary true, standard example is $f_n(x) = x^n$ for $x\in[0,1)$ and $f_n(x) = 0$ otherwise. They converge point-wise to $f\equiv 0$ but not uniformly. What you have instead is Dini's theorem:

If $(f_n)$ are continuous functions, and they converge to a continuous function $f$ point-wise and monotonically on a compact set, then they converge uniformly to that function.