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I am trying to construct a sequence of functions in $C[0,1]$ with the $L^{1/2}$ metric $$d(f, g) = \int_0^1 \sqrt{|f(x) - g(x)|} \,dx$$ that is Cauchy but not convergent.

There are a whole bunch of questions on the site with examples for $L^1$ (though they're all basically the same), and I've been trying to adapt them for this case. Unfortunately, no matter how I tweak my sequence it makes calculating the actual distance $d(f_n,f_m)$ (and showing that this goes to $0$, etc.) very difficult, because of the square root.

Does anyone have a concrete and reasonably computable example to show that $C[0,1]$ is not complete with the $L^{1/2}$ metric? Your help is appreciated.

RobPratt
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fish
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1 Answers1

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For $n=1,2,\dots,$ define $f_n(x)= 1/x, 1/n\le x \le 1,$ $f_n(x)= n, 0\le x \le 1/n.$ This is a sequence in $C[0,1]$ that converges to $1/x$ in $L^{1/2}.$

zhw.
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  • But $d(f_n,f_m)$ still requires me to integrate $\sqrt{1/x - m}$, which I can't do... – fish Apr 09 '20 at 18:39
  • You don't need to do that. $f_n$ converges in $L^{1/2}$ to $1/x,$ therefore $f_n$ is Cauchy in $L^{1/2}.$ – zhw. Apr 09 '20 at 18:44
  • But to show that it converges to $1/x$, don't I need to show that $\int_0^1 \sqrt{1/x - f_n(x)} ,dx = \int_0^{1/n} \sqrt{1/x - n} ,dx$ goes to $0$? – fish Apr 09 '20 at 18:49
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    Yes you do. That integral is less than

    $$\int_0^{1/n} x^{-1/2}, dx = 2x^{1/2}\big |_0^{1/n} =2/\sqrt n \to 0.$$

     – zhw. Apr 09 '20 at 18:53