Let $C^1([-1,1])$ denote the space of continuously differentiable functions on $[-1,1]$. Define \begin{align*} \|f\|_{C^1} = \left(\displaystyle\int_{-1}^1 (|f|^2+|f'|^2)dx\right)^{\frac{1}{2}}, \quad \forall f\in C^{1}([-1,1]). \end{align*} Show that $\|\cdot \|_{C^1}$ is a norm on $C^1([-1,1])$. If $(C^1([-1,1]),\|\cdot\|_{C^1})$ complete?
I am able to show that $\| \cdot \|_{C^1}$ is a norm and I choose the sequence of functions as $f_n(x) = \sqrt{x^2+\dfrac{1}{n^2}}$ for $x \in [-1,1]$. I am also able to show this is a Cauchy Sequence. Now I want to prove that this sequence does not converge by contraction. Assume that it converges to some function $F$.
And I think that $F(x) = |x|$ so $F \not \in C^1[-1,1]$. So this Cauchy sequence is not a convergent sequence. But I'm not really sure to show that $F(x) = |x|$ properly (my professor said that saying something like $\lim\limits_{n\to \infty} f_n(x) = |x|,\quad \forall x \in [-1,1]$ is inappropriate).