Convergence in space of all square integrable functions on [-1,1]

**Mollier** · February 11th 2013, 10:10 PM

Problem:
1. Given the sequence $x_n(s) = (|s|+\frac{1}{n})^{-1/3}$ and function $x(s)=|s|^{-1/3}$ . show that $x_n\rightarrow x$
in $L_2(-1,1)$

2. How can we use this to deduce that $C[-1,1]$ (with the max norm) is not complete.

Pathetic Attempt:

1. $lim_{n\rightarrow \infty}||x_n(s)-x(s)||_{L_2} = lim_{n\rightarrow \infty} \int^1_{-1} |x_n(s)-x(s)|^2 = 0$
because $lim_{n\rightarrow \infty} 1/n = 0$ .

2. The limit $x(s)$ isn't in C[-1,1] so its not complete...

Very bad I know.
Any suggestions are welcome.

**Mollier** · February 12th 2013, 11:51 AM

A somewhat better attempt:

$\int^1_{-1} |x_n(s) - x(s)|^2 ds$

I split the integral up in two for simplicity.

$\int^1_{-1} (|s|+1/n)^{-1/3}ds = 6(1/n+1)^{1/3} - 6(1/n)^{1/3}$

$\int^1_{-1} |s|^{-1/3}ds = 6$

Let n go to infinity and we have proved convergence.

Still working on the other part though.

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