0
$\begingroup$

Prove that $X=C([0,1],\mathbb{R})$ with the metric $d(f,g)=\int^1_0|f(x)-g(x)|dx$ is not a complete metric space.

My idea

Consider the sequence of continuous functions $$f_n(x)=\left\{\begin{matrix} 0 & 0\le x \le \frac{1}{2}-\frac{1}{n}& \\ nx+(1-\frac{1}{n})& \frac{1}{2}-\frac{1}{n} \le x \le \frac{1}{2}& \\ 1& \frac{1}{2} \le x \le 1 & \end{matrix}\right.$$

We have $d(f_n,f_m)=\int ^1_0 |f_n-f_m|dx=\frac{1}{n} \rightarrow 0$, and hence $(f_n)$ is a Cauchy sequence.

Now, how do I proceed from here?

$\endgroup$
14
  • $\begingroup$ Do you know what else you would have to prove about $f_n$ in order to show that the space is not complete? $\endgroup$ Apr 6 '18 at 11:35
  • $\begingroup$ $f_n$ is convergent to a function which is not continuous $\endgroup$ Apr 6 '18 at 11:37
  • $\begingroup$ Yes, or at least '$f_n$ does not converge to a continuous function' (so if you show that it does converge to a discontinuous function you're done). $\endgroup$ Apr 6 '18 at 11:43
  • 1
    $\begingroup$ But... I don't think that $f_n$ as you defined it is a continuous function for each $n$ (let alone the fact that the definition might be a little ambiguous for $n=1$, but that's not a big deal; the continuity of $f_n$ is). $\endgroup$ Apr 6 '18 at 11:44
  • $\begingroup$ @AlejandroNasifSalum [If you show that it does converge to a discontinuous function] ---> converge not pointwise but for $d(.,.)$ otherwise $x^n$ tends to zero in this space. $\endgroup$ Apr 6 '18 at 12:01
2
$\begingroup$

Your sequence $f_n$ does not work since those are not continuous functions. If you don't know how to fix it, a good option is to define $$f_n(x)=\left\{\begin{matrix} nx & 0\le x \le \frac{1}{n}\\ 1 & \frac 1n <x\le 1.\\ \end{matrix}\right.$$

Now, instead of proving that it converges point-wise to a discontinuous function (which says nothing) you have to prove that there is no $f\in C\big([0,1]\big)$ such that $d(f_n,f)\to 0$ (for the distance of the metric space).

One option is to suppose, on the contrary, that there is a continuous $f\colon [0,1]\to\mathbb R$ such that $$\int_0^1 \big|f_n(x)-f(x)\big|\,dx \to 0.$$ Suppose that $f(0)=y_0$ and take $\delta>0$ (and maybe $\delta<1$) such that $$0\le x<\delta\implies |f(x)-y_0|<\tfrac12,$$ (that is, 'take $\varepsilon=\tfrac12$') which is possible since $f$ is continuous at $x=0$.

Now show that there is an $N$ such that for $n\ge N$ $$\int_0^1 |f_n(x)-f(x)| \, dx$$ is greater than a certain positive lower bound (I guess you can try to integrate in $[0,\delta]$ and take $N$ such that $\tfrac1N<\delta$, so $f_n(0)=0$ and $f_n(\delta)=1$ for $n\ge N$).

Then is not true that $f_n \to f$ in the space, which is absurd.

$\endgroup$
2
  • $\begingroup$ I assume that "converges punctually" means "converges point-wise". $\endgroup$ Apr 6 '18 at 23:31
  • $\begingroup$ Yes, thanks, btw. $\endgroup$ Apr 8 '18 at 4:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.