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Show that the set of continuous functions $C[a,b]$ under the metric $\rho(f,g)=\displaystyle{\sqrt{\int_a^b|f(t)-g(t)|^2dt}}$ is not a complete metric space for $f,g\in C[a,b]$ and $t\in[a,b]$. To prove that this is incomplete, we must find a Cauchy sequence which converges to a function that is not continuous. Let $\{f_n\}$ be a Cauchy sequence. Then, we define the following: \begin{align*} f_n(x) = x^n\mbox{on}[0,1]. \end{align*} Now if we take the limit as $n$ approaches infinity, we have:

$$f(x)= \lim_{n\rightarrow \infty}f_n(x) = \begin{cases} 0,& \text{if } x\in[0,1) \\ 1, & x=1 \end{cases} $$ We know that $\{f_n\}$ is a Cauchy sequence which converges, because for any $\epsilon>0, \exists \thinspace N\in\mathbb{N}$ such that $|f_n-f_m| < \epsilon$ for all $n,m\geq N$. However, the point-wise limit seen above is not continuous, since $\{f_n\}\rightarrow f$. Thus, we have that not every Cauchy sequence converges into the set of continuous functions $C[a,b]$. We have shown that this metric space is not complete.

My problem here is I am mixing up $f_n$ with $f_n(x)$ and $f(x)$, some clarification on whether I did them right would be helpful and if there are any mistakes in the proof advice would be appreciated. Also if there are any problems with the proof please comment.

H5159
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3 Answers3

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Looks mostly fine to me. The thing to keep in mind is that $f(x)$ or $f_n (x)$ refers to the value of the function at the point $x$, which are real numbers. By contrast, $f_n$ or $f$ by themselves represent the function as an object unto itself.

However, the point-wise limit seen above is not continuous, since ${f_n}→f(x)$.

This should technically read $f_n\rightarrow f$ since the sequence approaches $f$ itself, not just a value of $f$. Similarly, your display equation should technically read $$f(x)=\lim_{n\rightarrow\infty}f_n(x)=\begin{cases} 1 &\mbox{if } x\in[0,1)\\ 0 &\mbox{if } x=1\end{cases}$$ since you are giving a pointwise definition of $f$. Hope this helps!

mwalth
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  • I understand why $f_n \rightarrow f$ but why is it $f(x)$ and not $f$ for the display? Also, is the first line defining $f_n(x)$ fine? – H5159 Apr 28 '15 at 23:34
  • In the display you need $f(x)$ because you are giving a formula for what value $f$ takes at a particular value of $x$. Your initial definition of the $f_n (x)$ is fine. – mwalth Apr 28 '15 at 23:36
  • So if we did want to define $f$, would it would be $f : [0,1] \rightarrow \mathbb{R}$? – H5159 Apr 28 '15 at 23:38
  • @Frumpy: That's not a definition of $f$; it's just giving some information about it. I've never seen a non-awkward way to define a function rigorously, but the most palatable options are usually along the lines "The function $f:[0,1]\to\Bbb R$ is defined by $f(x)=x^n$." – Eric Stucky Apr 28 '15 at 23:40
  • @EricStucky Thank you. Aside from my notation, does the proof itself look correct? I believe it does (proving its not a complete metric space). – H5159 Apr 28 '15 at 23:42
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    @Frumpy Your proof is otherwise solid. – mwalth Apr 28 '15 at 23:46
  • The proof is not solid, unfortunately. – copper.hat Apr 29 '15 at 01:45
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No, there is a error in the proof. In fact the functions $f_n(x) = x^n$ converge to the zero function in the metric $\rho$ on $C([0,1]):$

$$ \sqrt {\int_0^1 |x^n-0|^2\,dx} = 1/\sqrt {2n+1} \to 0.$$

You need to find a sequence $f_n$ in $C([0,1]),$ Cauchy in the $\rho$ metric, such that for every $f\in C([0,1]),$ $\rho(f_n,f)$ does not $\to 0.$ Suggestion: consider something like

$$f_n(x) = \begin{cases} 0,\, x \in [0, 1/2-1/n] \\ f_n(x) = 1,\, x\in [1/2,1] \\ f_n(x) = \text {the obvious straightline}, \,x\in[1/2-1/n,1/2] \end{cases}$$

zhw.
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  • My teacher accepted this proof but asked to fix a few things. I believe because the piece wise limit is discontinuous, then that function isn't a part of our set $C[0,1]$; which means it converged to something outside our space. – H5159 Apr 29 '15 at 00:53
  • @Frumpy: It does converge to something in the space, namely the zero function. – copper.hat Apr 29 '15 at 01:53
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The above $f_n$ converge to zero in the given metric, so they do not provide a counterexample.

Consider $f_n$ to be the function whose graph is the straight line connection of the points $(0,0), ({1 \over 2}-{1 \over n} ,0), ({1 \over 2}+{1 \over n} ,1), (1,1)$. It is straightforward to show that $f_n$ is Cauchy with the $L^2[0,1]$ metric.

Now suppose $f_n \to f$, with $f$ continuous. It is straightforward to show that for $x< {1 \over 2}$ we must have $f(x) = 0$ and for $x> {1 \over 2}$ we must have $f(x) = 1$. This contradicts $f$ being continuous, hence no such $f$ exists.

copper.hat
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