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Consider $X$ to be the set of all continuous real-valued functions on $C[0,1]$ with metric

$$d(x,y) = \int_{0}^{1} |x(t) - y(t)| dt$$

Show that $x_n(t) = \left\{\begin{matrix} n& 0 \leq t \leq n^{-2} \\ t^{-1/2}& n^{-2} \leq t \leq 1 \end{matrix}\right.$ is Cauchy, but does not converge.

So here is what I did, I basically assumed that $m > n$, (for some $m$) $$d(x_n, x_m) = \int_{0}^{m^{-2}}|x_n-x_m|dt + \int_{m^{-2}}^{n^{-2}}|x_n-x_m|dt + \int_{n^{-2}}^{1}|x_n-x_m|dt = \int_{0}^{m^{-2}}|n-m|dt + 0 + 0 = (m - n)m^{-2}$$

Now according to the book, the answer should've been $m^{-1} - n^{-1}$ for $n > m$.

As for convergence, I got $d(x_n,x) = \int_{0}^{n^{-2}} |x_n - x| dt + \int_{n^{-2}}^{1} |t^{-1/2} - x| dt \to 0$ implies that $x(t) =1/\sqrt{t}$ on $[0,1]$. But $x(0)$ is undefined, so not continuous.

I am not sure if I did this right, I have more confidence in the convergence part over the Cauchy part.

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I think the lower limit of the third integral (in the RHS of $d(x_m,x_n)$) will be $1/n^2$. –  Sayantan Jan 27 '13 at 2:01
1  
For the second integral, you have that going to $0$, but I think it should be $$\int_{m^{-2}}^{n^{-2}}|n-t^{-1/2}|dt$$ since one of the functions will have a square root and not the other. –  Clayton Jan 27 '13 at 2:02
    
Why? I have $m >n$, so according to my definition of $x(t)$, it is $t^{-1/2}$. hence both I get 0 –  sidht Jan 27 '13 at 2:09
    
@sizz: Since $m>n$, we have $m^{-2}<n^{-2}$, hence $x_n(t)=n$ while $x_m(t)=t^{-1/2}$. –  Clayton Jan 27 '13 at 2:14
    
I am guessing my conclusion for the convergence part was right since no one said anything about it? –  sidht Jan 27 '13 at 3:23

1 Answer 1

up vote 1 down vote accepted

If $m>n$, you have $d(x_m,x_n) = \int_{0}^\frac{1}{m^2} |m-n| dt + \int_{\frac{1}{m^2}}^\frac{1}{n^2} |\frac{1}{\sqrt{t}}-n| dt + \int_{\frac{1}{n^2}}^1 |\frac{1}{\sqrt{t}}-\frac{1}{\sqrt{t}}| dt = \frac{m-n}{mn}$. It follows that the sequence is Cauchy.

You have to do a little more work to show that the sequence $x_n$ does not converge to a continuous function. Your idea is right, but you need to show that $d(x,x_n) \to 0$ implies that $x_n(t) \to \frac{1}{\sqrt{t}}$ on $(0,1]$. Another way is to show directly that $x_n$ cannot converge to any continuous function $f$. As you observed, it hinges on the behavior at $t=0$.

Let $f$ be a continuous function. Then for some $k>0$, and $\epsilon>0$, we have $f(t) \leq k$ for $t \in [0,\epsilon]$. Without loss of generality, we may choose $\epsilon < \frac{1}{k^2}$. Let $n > \frac{2}{\sqrt{\epsilon}}$, then \begin{eqnarray} d(x_n,f) &\geq& \int_{[0,\epsilon]} (x_n(t)-k) dt \\ &=& \int_{[0,\frac{1}{n^2}]}(n-k)dt +\int_{[\frac{1}{n^2},\epsilon]}(\frac{1}{\sqrt{t}}-k)dt \\ &=& \frac{1}{n^2}(n-k)+2\sqrt{\epsilon}-k\epsilon+\frac{k}{n^2}-\frac{2}{n} \\ &=& 2\sqrt{\epsilon}-k\epsilon-\frac{1}{n} \\ &=& \sqrt{\epsilon}(2-k \sqrt{\epsilon})-\frac{1}{n} \\ &\geq & \sqrt{\epsilon} -\frac{1}{n} \\ & = & \frac{\sqrt{\epsilon}}{2} > 0 \end{eqnarray} It follows that $x_n$ cannot converge to any continuous function.

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So what did I do wrong? I am not sure what you did there with the epsilon argument. What happened to $1$? –  sidht Jan 27 '13 at 3:58
    
You didn't integrate correctly for the first part. For the second part, you have the right idea, but you need to formalize your answer a little more. I gave an alternative approach above. –  copper.hat Jan 27 '13 at 4:01
    
No, no. For the Cauchy part, I fixed it. I see what you did. I just don't understand your answer to the convergence part. Is your answer generalizing for a larger domain of $t$? –  sidht Jan 27 '13 at 4:06
    
A slightly more rigorous approach for your answer would be to suppose $x_n \to x$, where $x$ is continuous. Since $x$ is continuous, it is bounded on $[0,1]$. Then show (rigorously) that $x(t)= \frac{1}{\sqrt{t}}$ on $(0,1]$, which contradicts boundedness. –  copper.hat Jan 27 '13 at 4:06
    
I think your answer for $d(x_n,x_m)$ is still incorrect? No, I have not generalized, if anything, I have made it more specific. I just show that for any continuous $f$, there exists a constant $\eta>0$ such that $d(x_n,f) \geq \eta$. I did this by focusing on the behavior at $t=0$, $f$ is bounded locally (by $k$ above), but the $x_n$ are unbounded near $0$ in some uniform way. –  copper.hat Jan 27 '13 at 4:11

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