Is this metric space complete?

Question

Let $a, b \in \mathbb{R}$ such that $a < b$, and let $X$ be the metric space of all the (real- or complex-valued) functions defined and continuous on the closed interval $[a,b]$ with the metric $d$ defined as follows: $$d(x,y) \colon= \ \int_a^b \ \vert x(t) - y(t) \vert \ \mathrm{d} t \ \mbox{ for all } \ x, y \in X.$$

Then is $X$ complete?

Let $(x_n)_{n\in \mathbb{N}}$ be a Cauchy sequence in $(X,d)$. Then, given $\epsilon > 0$, there exists a natural number $N$ such that $$d(x_m, x_n) < \epsilon \ \mbox{ for all } \ m, n \in \mathbb{N} \ \mbox{ such that} \ m > N \ \mbox{ and } \ n > N. $$ That is, $$\int_a^b \ \vert x_m(t) - x_n(t) \vert \ \mathrm{d} t < \epsilon \ \mbox{ for all } \ m, n \in \mathbb{N} \ \mbox{ such that} \ m > N \ \mbox{ and } \ n > N. $$ What next?

Answer 1

No, consider $[a,b]=[0,1]$ and

$$y_n(t)= \begin{cases} 0 & 0\le t\le {1\over 2}-{1\over n+1} \\ {n+1\over 2}(x-{1\over 2}+{1\over n+1})& {1\over 2}-{1\over n+1}\le t\le {1\over 2}+{1\over n+1}\\ 1 & {1\over 2}+{1\over n+1}\le t\le 1\end{cases}$$

Then each $y_n$ is continuous is clear. The limit in the completion (which may be our original set or not, that's what we're investigating) has a point-wise representative

$$y(t) =\begin{cases} 0 & 0\le t < {1\over 2} \\ {1\over 2} & t={1\over 2} \\ 1 & {1\over 2}<t\le 1\end{cases}.$$

If there is a continuous representative, $y_0(t)$ then we note that $y_0(t)$ has the property that ${1\over \epsilon}\int_{1/2}^{1/2+\epsilon}y(t)\,dt\to y(1/2)=1/2$ by the FTC. However, it is clear that that integral is identically $1$ for all $\epsilon$, hence no such $y_0$ exists.

The big picture: continuous functions have a nice averaging property in integration. By finding a function which is the pointwise limit we are able to figure out what averages (i.e. integration) properties a candidate limit would have. But since we know we can mess with the average around the discontinuity, we see a way forward to construct a counter-example.

Answer 2

The ultimate answer to the question is "no". One must be careful in proving this, because it is possible to furnish a function which converges in your metric to a discontinuous function and to a continuous function. The problem here is that when you enlarge the domain to include certain discontinuous functions, the resulting metric becomes a pseudometric, so its limits are not unique. For instance this occurs with the example of $f_n(x)=x^n$ on $[0,1]$. This appears to converge to the pointwise limit, which has a jump at $1$, but it also converges in this metric to the zero function, which is continuous. Thus this sequence does not help us to prove that this space is not complete.

We should instead consider approximating a function with a jump in the interior of the domain. For instance consider the sequence

$$f_n(x)=\begin{cases} 0 & x \in [0,1/2-1/n] \\ n(x-1/2+1/n)/2 & x \in [1/2-1/n,1/2+1/n] \\ 1 & x \in [1/2+1/n,1] \end{cases}$$

for $n=2,3,\dots$. These rise linearly from $0$ to $1$ very rapidly close to $1/2$. (It is much easier to graph these than it is to write down a formula.) In the pointwise limit there is a jump at $1/2$. You should argue that if $f_n \to f$ in your metric then $f$ cannot be continuous at $1/2$.

Answer 3

The metric is not complete.

Say $[a.b]=[0,2]$. Say $y_n(t)=t^n$ for $0\le t\le 1$ and $y_n(t)=1$ for $1\le t\le 2$. Then $(y_n)$ is a Cauchy sequence. But it is not convergent in our metric.

Showing that it's not convergent is trickier than you might think. Define $y(t)=\lim y_n(t)$. Then $y(t)=0$ for $0\le t < 1$ while $y(t)=1$ for $1\le t\le 2$. So $y$ is not continuous. But that's not a proof. It's really not a proof; if that were a proof then a certain wrong example would be right. See below for the wrong example.

First show this: If $f$ is continuous then $$\epsilon=\int_0^2|f(t)-y(t)|\,dt>0.$$ Now choose $N$ so $$\int_0^2|y_n-y|<\epsilon/2\quad(n>N).$$ It follows that if $n>N$ then $$d(f,y_n)=\int_0^2|f-y_n|\ge\int_0^2|f-y|-\int_0^2|y-y_n|>\epsilon-\epsilon/2=\epsilon/2.$$So $d(f,y_n)$ does not tend to zero.

Wrong Example: Say $[a,b]=[0,1]$, and let $y_n(t)=t^n$. As before let $y(t)=\lim y_n(t)$. Then $y$ is not continuous. But that's irrelevant, because the pointwise limit is irrelevant; what matters is the limit in the metric $d$. If we let $x(t)=0$ for all $t$ then in fact $d(y_n,x)\to0$. So this sequence is convergent in our metric space, even though the pointwise limit is discontinuous.