I have to show that the real linear space $C^{1}[a,b]$ of all continuously differential functions defined on $[0, 1]$ equipped with the norm given by $\lVert x\rVert _{\infty} = \sup_{t\in [0, 1]} \lvert x(t) \rvert$ is an incomplete normed space.

I have taken sequence $\{x_{n}\} = \sqrt{t^2 +\frac{1}{n}}$ in $C^{1}[a,b]$.

Please tell me whether my procedure to show that $\{x_{n}\}$ is Cauchy in $C^{1}[a,b]$ is correct or not. Here is my attempt.

Let $n \geq m$

$\lVert x_n - x_m\rVert = \lVert \sqrt{t^2 +\frac{1}{n}} - \sqrt{t^2 +\frac{1}{m}}\rVert = \frac { \frac{1}{n} - \frac{1}{m}}{\sqrt{t^2 +\frac{1}{n}} + \sqrt{t^2 +\frac{1}{m}}}\leq \frac{1}{n} + \frac{1}{m} \leq \frac{1}{2m}\leq \frac{1}{m} \leq \epsilon $

Further I have shown that $\{x_{n}\}$ converges pointwise to $x(t) = \frac{1}{\sqrt n}$. But $x(t)$ is not differentiable at $t = 0$ and hence the proof.

Also could anybody provide me any other example?

  • $\begingroup$ See also math.stackexchange.com/q/271884 $\endgroup$ – Jonas Meyer May 27 '13 at 7:10
  • $\begingroup$ In addition to the reason Zach L gives, you could note that your sequence converges uniformly to $|t|$ by Dini's theorem. However, I guess you want your interval to be $[-1,1]$ or something else containing $0$ in its interior, so that $|t|$ is not $C^1$. $\endgroup$ – Jonas Meyer May 27 '13 at 7:13
  • $\begingroup$ @JonasMeyer Thanks to you. That is very nice idea. Instead of showing cauchy we may prove that $x_n$ is uniform convergent if possible. I am trying to prove by using Dinni's theorem. $\endgroup$ – mathscrazy May 27 '13 at 7:31

It does look like there are some typos. For instance, if $n \geq m$, then $\frac{1}{m} \geq \frac{1}{n}$ and so your fraction is negative. This is bad, since it's supposed to be equal to the absolute value of something! The next inequality also seems fishy, since if it were true we would have $$\frac{1}{m} - \frac{1}{n} \leq \bigg(\frac{1}{n} + \frac{1}{m}\bigg)\bigg(\sqrt{\frac{1}{n}} + \sqrt{\frac{1}{m}}\bigg).$$ If we let $n \rightarrow \infty$, this will end up being $$\frac{1}{m} \leq \frac{1}{m\sqrt{m}}.$$ So $\sqrt{m} \leq 1$, which doesn't hold up.

But your idea works fine. Try this: you know that $t^2 + \frac{1}{n}$ converges uniformly to $t^2$. You also know that $\sqrt{x}$ is uniformly continuous on intervals $[0,A]$. Perhaps use these to show that $\sqrt{t^2 + \frac{1}{n}}$ converges uniformly to $\sqrt{t^2} = |t|$? Then you do not even need to bother showing it's Cauchy, since all convergent sequences are automatically Cauchy!

Here's my example: Let $g_n$ be the sequence of functions defined by $$g_n(x) = \begin{cases} 0 &\mbox{ if } -1 \leq x \leq \frac{-1}{n} \\ \frac{n}{2}x + \frac{1}{2} &\mbox{ if } \frac{-1}{n} \leq x \leq \frac{1}{n} \\ 1 &\mbox{ if } \frac{1}{n} \leq x \leq 1 \end{cases}.$$

If I've done it right, this should be a piecewise function defined by staying at zero until you hit $\frac{-1}{n}$, going linearly up to 1 until you hit $\frac{1}{n}$, and then staying at 1. These functions are continuous, and so $$f_n(x) = \int_{-1}^x g_n(t) dt$$ are all differentiable.

However, the limit of the $g_n$'s is the step function $g = \chi_{[0,1]}$, whose integral $$f(x) = \int_{-1}^x g(t) dt$$ is not differentiable. But $$\|f_n - f\| \rightarrow 0.$$ So $f_n$ converges in the $C^0$ norm and is therefore Cauchy. The idea is to make the bad behavior "one level down" and integrate it.

  • $\begingroup$ Thank you very much. You are absolutely right. I need to go for the uniform convergence of the sequence. This idea is very helpful to me. I am reading your answer. Thanks again. $\endgroup$ – mathscrazy May 27 '13 at 7:29

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