Uniform limit not being equal to pointwise limit?

Question

Let $X=C[0,1]$ the space of continuous functions on $[0,1]$ with the metric $d(f,g)=\int_{0}^{1}\left |g(t)-f(t) \right |dt$.

For the function $f_n(t)=e^{-nt}$ I can see that it will converge uniformly to $f\equiv 0$ as $\int_{0}^{1} e^{-nt} dt =\frac{1-e^{-nt}}{n}\rightarrow 0$

The problem is that $\lim_{n\rightarrow \infty }f_n(0)=1 $ but $f(0)=0$. How come there is this point $t=0$ where the sequence of functions at $0$ does not converge to the limit function at $0$?

Thank you

Answer 1

I think you've mixed up different notions of convergence.

• $f_n$ does not converge uniformly to zero. It doesn't even converge pointwise to zero because, as you've noted, $f_n(0) = 1$ for all $n$.

• $f_n$ does converge to the zero function with the metric $d$, because $\int_0^1 |f_n - 0| \to 0$.

Moral: Different notions of convergence are not always compatible. Uniform convergence (on a bounded interval) implies convergence in the metric $d$. Pointwise convergence does not. Convergence in $d$ doesn't imply anything more than that a subsequence converges pointwise almost everywhere.

Answer 2

Your sequence does not converge uniformly to the null function. Why did you write that? And the simplest proof of the fact that it doesn't converge uniformly to the null function is precisely the fact that $\lim_{n\in\mathbb N}f_n(0)\neq0$.