Showing that $C^1([0,1])$ space with the norm $||f||^2=\int_{0}^{1}|f|^2+\int_0^{1}|f'|^2$ is not a Banach Space I am given the $C^1([0,1])$ space containing $f$-s such that $f,f'\in C[0,1]$, with the norm $||f||^2=\int_{0}^{1}|f|^2+\int_0^{1}|f'|^2$ and asked to show that it is not a Banach Space.
What I am not so sure about is how; do I need to find $f_n$ such that $\int_0^{1}|f_n'|^2$ is infinite, or should I rather find $f'_n$ such that $\lim_{n\to \infty}f'n\notin C[0,1]$, or is any of the two will be sufficient? I am having trouble to find ANY Cauchy sequence of functions, especially the conditioned ones. Should I start looking for derivatives sequence and integrate them?
 A: If you have a linear space $X$ and two norms  $||\cdot ||_1 ,||\cdot ||_2 $ on $X$ such that are comparable i.e. there exist $C_1>0 $ such that $$\forall_{x\in X} ||x||_1 \leq C_1 ||x||_2 $$
or
there exist $C_2>0 $ such that $$\forall_{x\in X} ||x||_2 \leq C_2 ||x||_1 $$
but not equivalent then only on of these two norms can be complete.
To see this assume conversly that $(X, ||\cdot ||_1 ) $ and $(X, ||\cdot||_2 )$ are Banach spaces. Then the one of the itentity maps $$I_1 : (X, ||\cdot ||_1 )\to (X, ||\cdot||_2 )$$ or $$I_2 : (X, ||\cdot ||_2 )\to (X, ||\cdot||_1 )$$
is continous and hence by Banach inverse operator theorem it is a linear homeomorphism and therefore the norms are equivalent which gives contradiction.
It is well known that the norm $||\cdot ||^1_{\infty}: C^1[a,b]\to\mathbb{R},$ $$||f||^1_{\infty}=\sup_{x\in [a,b]} |f(x)| +\sup_{x\in [a,b]} |f'(x)| $$ is complete. So it is enough to show that $||\cdot ||$ is not eqivalent to $||\cdot ||_{\infty }^1 .$
To prove this take $f_n (x) = \frac{x^n}{n} ,$ then $f_n ' (x) =x^{n-1} $ for $x\in [0,1].$ Therefore $$||f_n ||^2=\frac{1}{2n^3 } +\frac{1}{2n} \to 0$$
but $$||f_n ||^1_{\infty } =1+\frac{1}{n} \geq 1$$
so these norms cannot be equivelent , but always $$||f||\leq \sqrt{2} ||f||^1_{\infty} $$ so the norm $||\cdot ||$ coannot be complete.
