Is the sequence of functions, $f_n(x) = x^n$ Cauchy in $C([0,1])$? Is it Cauchy in $L^2([0,1])$? Is the sequence of functions, $f_n(x) = x^n$ Cauchy in $C([0,1])$? Is it Cauchy in $L^2([0,1])$?
I know that it is not Cauchy in the space of continuous functions. I know that the L-space is complete and suspect that the sequence will be Cauchy in that space then. 
My idea to prove it is by contradiction. I have not been provided with a particular norm to work with in the continuous function space. 
The norm in the L-space is 
$\left\lVert f\right\rVert = (\int_a^b x(t)^2 dt)^{1/2}$
I'm wondering what the distinguishing factor between the sequence being in either space will be. 
It must be that the  sequence does not converge to a continuous function in the first case. I'm not sure what this function will be. 
 A: If it were Cauchy in $C[0,1]$ then it would converge. This is convergence in uniform sense and the limit function would be continuous. The limit function would have to agree with the pointwise limit function. 
However, it converges pointwise to a discontinuous function which is 0 except at 1 where it is 1. 
For $L^2$ the question is already answered. 
A: Hint: For $L^{2}([0,1])$ start by computing the integral $\int_{0}^{1}(f_{n}(x)-f_{m}(x))^2 dx$.
\begin{align*}
\int_{0}^{1} (f_n - f_m)^2 dx &= \int_{0}^{1} (x^n-x^m)^2 dx \\
           &= \int_{0}^{1} x^{2n}-2x^{n+m}+x^{2m} dx \\
           &= \frac{1}{2n+1}-\frac{2}{n+m+1} + \frac{1}{2m+1}
\end{align*}
If $n$ and $m$ are both big... this integral is small.
Hint: For $C([0,1])$ with $\Vert\cdot\Vert_{\infty}$ consider points that are very close to $x=1$.  If you fix an $n$, you can get $x^{n}$ as close to 1 as you like with $x<1$.  Fix this $x_{0}$ so that $x_{0}<1$ and $x_{0}^{n}$ is very close to $1$.  Then $x_{0}^{m}$ can be very close to zero by taking $m$ much larger than $n$ (since $x_0<0$ $x_{0}^{m}\to 0$ as $m\to \infty$).
