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I want to prove the following:

Let be the sequence of functions in $C([0,1])$ given by $$\displaystyle f_n(x) = \begin{cases}\sqrt n & 0\le x<\dfrac{1}{n}\\\dfrac{1}{\sqrt x} & \dfrac{1}{n}\le x\le 1.\end{cases}$$ Then $f_n$ is a Cauchy sequence in $(C([0,1]),||\cdot||_1)$ that does not converge. Thus, $(C([0,1]),||\cdot||_1)$ is not complete.

So to prove that it is Cauchy I did the following:

$$\int_{0}^{1}|f_{n}(x)-f_{m}(x)|dx \leq \int_{0}^{1}|f_n(x)|+\int_{0}^{1}|f_m(x)|=\int_{0}^{1/n}|\sqrt(n)|+\int_{1/n}^{1}|1/ \sqrt(x)|+\int_{0}^{1/m}|\sqrt(m)|+\int_{1/m}^{1}|1/ \sqrt(x)|=1/\sqrt(n)+1/\sqrt(m)+(2-2/\sqrt(n))+(2-1/\sqrt(m))=4-(1/\sqrt(n)+1/\sqrt(m))$$

So the thing is, I think there is something wrong here since I can't bound $1/\sqrt(n)+1/\sqrt(m)$ as both $n,m$ goes to zero, and the norm could be negative since $1/\sqrt(n)+1/\sqrt(m)$ could be larger that 4.

Now another thing, How can I prove that is not convergent?.

And finally, I was trying to figure out if $(C([0,1]),||\cdot||_2)$ is complete? but the thing is that proving it should be worse that rearrange the above function to give a counter example, but Am I right?.

Can someone help with this questions please?

user162343
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    Actually $n$ is tending to infinity, not zero. But never mind that. You start by saying $||f_n-f_m||\le||f_n||+||f_m||$, and that's never going to work. Instead, start by considering where the two functions are the same! You can say $\int_0^1|f_n-f_m| = \int_0^?|f_n-f_m|\le\int_0^?|f_n|+\int_0^?|f_m|$ and that will tend to zero... – David C. Ullrich Aug 20 '15 at 02:16
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    Hint: If $n<m$, $f_n-f_m$ is only non-zero on the interval $[0,1/n]$, so $$ |f_n - f_m|1 = \int_0^{1/m} |\sqrt{n} -\sqrt{m}| + \int{1/m}^{1/n} |\sqrt{n}-x| $$ – Prahlad Vaidyanathan Aug 20 '15 at 02:29

2 Answers2

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For simplicity, let $n \wedge m := \min \{n,m\}$ and let $n \vee m := \max \{ n,m \}$ for all integers $n,m \geq 1$.

We may do it softly. Note that $$ \int_{0}^{1} |f_{n} - f_{m}| = \int_{0}^{\frac{1}{n \vee m}} | \sqrt{n} - \sqrt{m}| dx + \int_{\frac{1}{n \vee m}}^{\frac{1}{n \wedge m}} | \frac{1}{\sqrt{x}} - \sqrt{n \wedge m} | dx \leq \frac{\sqrt{n}}{n \vee m} + \frac{\sqrt{m}}{n \vee m} + 2\frac{1}{n \wedge m} + 2\frac{1}{n \vee m} + \frac{\sqrt{n \wedge m}}{n \wedge m} + \frac{\sqrt{n \wedge m}}{n \vee m} \to 0 $$ as $m,n$ grows.

Yes
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Let $\varepsilon > 0$, and choose $N >\frac{1}{\varepsilon^2}$. Then for $n > m > N$, we have \begin{align*} \int_0^1 |f_n(x) - f_m(x)| \, \mathrm{d}x &= \int_0^1 f_n(x) - f_m(x) \, \mathrm{d}x \\ &= \int_0^1 f_n(x) \, \mathrm{d}x - \int_0^1 f_m(x) \, \mathrm{d}x \\ &= \left(\int_0^{1/n} \sqrt{n} \, \mathrm{d}x + \int_{1/n}^1 \frac{1}{\sqrt{x}} \, \mathrm{d}x\right) - \left(\int_0^{1/n} \sqrt{n} \, \mathrm{d}x + \int_{1/n}^1 \frac{1}{\sqrt{x}} \, \mathrm{d}x \right) \\ &= \left(\frac{\sqrt{n}}{n} + 2 - \frac{2}{\sqrt{n}} \right) - \left (\frac{\sqrt{m}}{m} + 2 - \frac{2}{\sqrt{m}} \right ) \\ &= \frac{1}{\sqrt n} - \frac{2}{\sqrt{n}} - \frac{1}{\sqrt m} + \frac{2}{\sqrt{m}} \\ &= \frac{1}{\sqrt m} - \frac{1}{\sqrt n} \\ &< \frac{1}{\sqrt m} < \varepsilon, \end{align*} showing the sequence is Cauchy w.r.t. $||\cdot||_1.$ Now suppose $f_n$ has limit $f \in C([0,1])$. Then $f(x) = \frac{1}{\sqrt x}$ on $(0,1)$, which cannot be extended continuously onto $[0,1]$. Thus the space is incomplete.

user217285
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  • Answering the latter part of your question, $C([0,1])$ is incomplete w.r.t. $|| \cdot ||_p$ for $1 \leq p < \infty$ (the inequality excludes the sup norm). Some solutions are given here: http://math.stackexchange.com/questions/156904/showing-that-the-space-c0-1-with-the-l-1-norm-is-incomplete – user217285 Aug 20 '15 at 03:00
  • Thanks a lot of this, then we can not extend continously f because we are allowed to reach zero which is infinite right? and why $f(x)=1/ \sqrt(x)$ – user162343 Aug 20 '15 at 03:12
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    More formally, there is a sequence $x_n \to 0$ such that $f(x_n)$ is unbounded. The reasoning $f(x) = \frac{1}{\sqrt x}$ should be clear from plots of $f_n$ for large $n$. If you consider $f_n$ over the interval $[a,1]$ any $0 < a < 1$, the limit is exactly $f(x) = \frac{1}{\sqrt x}$. Letting $a \to 0$, we necessarily have $f(x) = \frac{1}{\sqrt x}$, which is a contradiction. – user217285 Aug 20 '15 at 03:27
  • Ok, thanks a lot, le me check it out, If I get stuck, can I let you know, here is late, so it will be tomorrow :) – user162343 Aug 20 '15 at 03:41
  • Why do we can forget the absolute value in the integral? – user162343 Aug 20 '15 at 16:11
  • We let $n > m$. – user217285 Aug 20 '15 at 17:08
  • Right, thanks a lot, now, If the functions are complex valued, Does the proof change? – user162343 Aug 20 '15 at 18:02
  • How can you argue that the limit as $n$ goes to $\infty$ is $\frac{1}{\sqrt(x)}$ – user162343 Aug 20 '15 at 18:48
  • And well it can't converge :) right that is what we want to prove, my definition of completeness is different right? – user162343 Aug 20 '15 at 23:14
  • The proof is specifically for the $f_n$ described in the question but shows a counterexample to how the space of real, continuous functions on a compact interval is incomplete in the $||\cdot||_1$ norm. If we ask the question "is the space of continuous functions from $[0,1]$ into $\mathbb C$ complete wrt $||\cdot||_1$", the answer is no, and we can just use this example since the imaginary part of the function is 0. And no, the limit as $n \to \infty$ does not exist! On the interval $[a,1]$, $a > 0$, the limit is $1/\sqrt x$, but we cannot extend any further (see previous comment). – user217285 Aug 20 '15 at 23:24
  • Thanks Nitin but the thing is that I want to prove that $C([0,1])$ is not complete, therefore the sequences given above should not converge, this is it does not have to be getting close to anything so how can you prove in detail that this funtion is not close to some number :) thaat is my question – user162343 Aug 20 '15 at 23:43
  • The sequence does not converge in the space $C([0,1])$. This was made very clear in the previous comments. Also, the function is not going to be close to 'some number' -- if the sequence were to converge, it would be 'close' to some function in the sense that the integral of the difference between the functions is very small for large $n$. In this case, we assume there is a limit and derive a contradiction, the contradiction being that the closest possible limit is not continuous on the interval $[0,1]$. – user217285 Aug 21 '15 at 00:14
  • Ok give me some minutes right an I will be back :) – user162343 Aug 21 '15 at 00:23