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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.

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  • $\begingroup$ General rule of thumb, $f$ is a function, $f(x)$ is a real number (in this case). $\endgroup$ – Asaf Karagila Apr 28 '15 at 23:21
  • $\begingroup$ @AsafKaragila I believe I have used them correctly, here, what do you think? $\endgroup$ – H5159 Apr 28 '15 at 23:22
  • $\begingroup$ I spot two places where this is not true. One of each kind (one $f(x)$ where it should be $f$; and one $f_n$ where it should be $f_n(x)$). $\endgroup$ – Asaf Karagila Apr 28 '15 at 23:24
  • $\begingroup$ @AsafKaragila Is that what you meant? I'm still a bit confused by the placement. $\endgroup$ – H5159 Apr 28 '15 at 23:28
  • $\begingroup$ That correction is a good start, the second problem is three lines down from the one you just did. $\endgroup$ – Asaf Karagila Apr 28 '15 at 23:29
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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!

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  • $\begingroup$ 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? $\endgroup$ – H5159 Apr 28 '15 at 23:34
  • $\begingroup$ 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. $\endgroup$ – Mark Walth Apr 28 '15 at 23:36
  • $\begingroup$ So if we did want to define $f$, would it would be $f : [0,1] \rightarrow \mathbb{R}$? $\endgroup$ – H5159 Apr 28 '15 at 23:38
  • $\begingroup$ @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 $f(x)=x^n$." $\endgroup$ – Eric Stucky Apr 28 '15 at 23:40
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    $\begingroup$ @Frumpy Your proof is otherwise solid. $\endgroup$ – Mark Walth Apr 28 '15 at 23:46
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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

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

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  • $\begingroup$ 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. $\endgroup$ – H5159 Apr 29 '15 at 0:53
  • $\begingroup$ @Frumpy: It does converge to something in the space, namely the zero function. $\endgroup$ – copper.hat Apr 29 '15 at 1:53
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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.

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