Convergence in space of all square integrable functions on [-1,1]
Problem:
1. Given the sequence  = (|s|+\frac{1}{n})^{-1/3})
and function =|s|^{-1/3})
. show that 
in )
2. How can we use this to deduce that ![C[-1,1]](http://latex.codecogs.com/png.latex?C[-1,1])
(with the max norm) is not complete.
Pathetic Attempt:
1. -x(s)||_{L_2} = lim_{n\rightarrow \infty} \int^1_{-1} |x_n(s)-x(s)|^2 = 0)
because 
.
2. The limit )
isn't in C[-1,1] so its not complete...
Very bad I know.
Any suggestions are welcome.
Re: Convergence in space of all square integrable functions on [-1,1]
A somewhat better attempt:
 - x(s)|^2 ds)
I split the integral up in two for simplicity.
^{-1/3}ds = 6(1/n+1)^{1/3} - 6(1/n)^{1/3})
Let n go to infinity and we have proved convergence.
Still working on the other part though.
