# Why the sequence is not convergent?

1. Jun 20, 2009

### symbol0

In a book I am reading, they mention the following as an example of a Cauchy sequence which is not convergent:
Consider the set of all bounded continuous real functions defined on the closed unit interval, and let the metric of the set be
d(f,g)=$$\int_0^1 \! |f(x)-g(x)| \, dx$$.
Let $$(f_n)$$ be a sequence in this space defined as:
$$f_n(x)=1$$ if $$0 \leq x \leq 1/2$$
$$f_n(x)=(-2)^n(x-1/2)+1$$ if $$1/2 \leq x \leq 1/2+(1/2)^n$$
$$f_n(x)=0$$ if $$1/2+(1/2)^n\leq x \leq 1$$

I can see that this is a Cauchy sequence, but I can't see how this sequence does not converge. I would say that it converges to:
$$f_n(x)=1$$ if $$0 \leq x \leq 1/2$$
$$f_n(x)=0$$ if $$1/2 < x \leq 1$$

I would appreciate if anybody can help me to see why this sequence does not converge.

2. Jun 20, 2009

### Hurkyl

Staff Emeritus
I'm going to assume you meant to say something like the following.
I would say that it converges to the function f defined by
$$f(x)=1$$ if $$0 \leq x \leq 1/2$$
$$f(x)=0$$ if $$1/2 < x \leq 1$$[/quote]​

A good way to see these things is to try and prove that it does, and see where things fail. You need to:
(1) Show that f is well-defined
(2) Show that $\lim_{n \rightarrow +\infty} d(f_n, f) = 0$
Try it out, and let us know how it goes!

3. Jun 21, 2009

### symbol0

First of all , I think this sequence is not really in the mentioned space because when n is even, the functions are not continuous.
But even if we take only the functions with odd n, I think the sequence converges to the function I wrote, but this function is not continuous, so it is not in the space.
Thus the sequence does not converge in the space.

Am I right?

4. Jun 21, 2009

### Hurkyl

Staff Emeritus
That last part is the key thing here -- f is not in the space.

In fact, technically speaking, d was only defined for pairs of elements in the space, so it would be wrong to say that the sequence converges to f! You only get that convergence when looking at the larger space

5. Jun 22, 2009

### symbol0

Thanks Hurkyl

6. Feb 9, 2011

### buckie447

Sorry to bump an old thread, but I'm having a problem seeing how this sequence is cauchy. I used the metric on an arbitrary f_n(x) and f_m(x), but I ended up with constants and other terms that I couldn't make arbitrarily small by taking m and n large enough. Could someone give me a brief overview of how to prove it?

7. Feb 11, 2011

### HallsofIvy

$f_n$ and $f_m$, with m> n, differ only on the interval from $1/2+ 1/2^m$ to $1/2+ 1/2^n$ a length of
$$\frac{2^m- 2^n}{2^{m+n}}$$
which goes to 0 as m and n go to infinity.

The integrand doesn't have to go to 0, it only has to be bounded. As the length of the interval goes to 0, the integral will go to 0.