Convergence of functions in a metric space

Question

Let $C([0,1])$ be the space of all continuous functions from $[0,1]$ to $\mathbb{R}$ under the metric $$ \lVert f \rVert_1 = \int_0^1 \lvert f(x) \rvert \, dx. $$

Now consider $f_n(x) = e^{-nx}$. I have proven that $\{f_n\}$ is a Cauchy sequence. And pointwise I know that $\lim_{n \to \infty}f_n(0) = 1 \ne \lim_{x \to 0} \lim_{n \to \infty} f_n(x) = 0$ so that $\lim_{n \to \infty} f_n$ is not continuous.

I am inclined to conclude then that I have proved $C([0,1])$ is not complete. However, by the metric $\lVert f_n - 0 \rVert_1 \to 0$, so that $f_n$ does converge to $0$.

Does this mean that a function can converge to another function by a metric, but not converge pointwise to that function? Can someone explain what exactly the significance of this is?

If we instead just consider the space of all functions on $[0,1]$, then the metric says that $f_n$ converges to any function that is $0$ except on a set of measure $0$. Does this means that equality of functions is dependent on the metric? Or does this mean that the metric violates $\lVert f - g\rVert_1=0 \implies f = g$?

Answer 1

The main issue here is that the concept of a limit is dependent on the metric. On one hand, we have the pointwise limit, where

$$ \lim_{n \rightarrow \infty} f_n = g \Leftrightarrow \sup_{x \in [0,1]} |f_n(x) - g(x)| \rightarrow 0$$ On the other hand, we have the $L^1$ limit, which is given by convergence in the $1$-norm:

$$ \lim_{n \rightarrow \infty} f_n = h \Leftrightarrow \|f_n - h\|_1 \rightarrow 0$$

These limits are not the same, which is why you have found that $g \neq h$.

Answer 2

The problem here is that the pointwise convergence does not imply convergence for the metric. This is why such phenomenon can occur.

Here the limit is $0$ for the reason you said.

But the space is not complete. For a counterexample, try with the following sequence: $$ f_n(x) = \begin{cases} -1&\text{if}& x < \frac 12 - \frac 1n\\ nx - n\frac 12 &\text{if}& \frac 12 - \frac 1n \le x < \frac 12 + \frac 1n \\ 1&\text{if}& x \ge \frac 12 + \frac 1n \end{cases} $$