Verification of a peculiar Banach space. Suppose $H$ is the space of absolute continuous functions $f$, and both $f$ and $f'$ are in $L^2(\mathbb{R})$. And the norm is defined as:
$$\|f\|=\sqrt{\int_{\mathbb{R}}|f|^2\mathrm{d} x+\int_{\mathbb{R}}|f'|^2\mathrm{d} x}$$
I'm trying to prove it's a Banach space with the standard method, which is, find the limit of an arbitrary Cauchy sequence$\{f_n\}$.
Here is my attempt.
First, both $f_n$ and $f_n'$ should converges to a function in $L^2(\mathbb{R})$(denoted by $f$ and $\tilde{f}$ respectively), and $\int_{a}^b \tilde{f}\mathrm{d}x=f(a)-f(b)$ should hold almost everywhere. But I'm stuck here.
At first I thought it could be proved that $\forall f$ that is absolutely continuous and integrable satisfies $f(-\infty)=0$, yet I constructed a counterexample soon; similiarly, I constructed a Cauchy sequence that is nowhere pointwise convergent.
So now I have no clue and any kind of hint or help would be very useful.
 A: I figured it out lately and my friend also gave a solution.
My solution:
$\forall a,b$, $f_n,f_n'$ converge in $L^2[a,b]$, so in $L^1[a,b]$ to $f,\tilde{f}$. So $\lim\limits_n\int_{[a,x]}f_n\mathrm{d} m=\lim\limits_n(f_n(x)-f_n(a))=\int_{[a,x]}\tilde{f}\mathrm{d}m$. Therefore $\{g_n(x)=f_n(x)-f_n(a)\}$ converge pointwise\uniformly\in the sense of $L^1$ on $[a,b]$, $\{h_n(x)=f_n(x)-g_n(x)\equiv f_n(a)\}$ converge in $L^1[a,b]$. Therefore $\{f_n(x)\}$ converges pointwise in $[a,b]$. So $\lim\limits_n(f_n(b)-f_n(a))=f(b)-f(a)=\int_{[a,b]}\tilde{f}\mathrm{d}m$, $f\in H$ and $f'=\tilde{f}$. $\|f-f_n\|\leq (\|f-f_n\|_2+\|f'-f'_n\|_2)\rightarrow 0$
My friend's solution:
$\forall h\in H$, $x\in \mathbb{R}$, $\exists x_0< x\ s.t.|h(x_0)|\leq \frac{|h(x)|}{2}$(otherwise it won't be integrable). Therefore $\|h\|\geq \sqrt{\int_{x_0}^x|h|^2\mathrm{d}m+\int_{x_0}^x|h'|^2\mathrm{d}m}\geq\sqrt{(\int_{x_0}^x|h|^2\mathrm{d}m\int_{x_0}^x|h'|^2\mathrm{d}m)^\frac{1}{2}}\geq\sqrt{\int_{x_0}^x|h·h'|\mathrm{d}m}\geq\sqrt{|h^2(x)-h^2(x_0)|}\geq \sqrt{\frac{3}{4}}|h(x)|$.
Replace $h$ with $f_n-f$, we reach the conclusion that $\{f_n\}$ converges uniformly on $\mathbb{R}$. The rest would be natural.
It's the uniformly convergent part that shocks me, I think it's because the convergence of derivative does not 'allow' the original function to bounce over anywhere. And we can conclude that convergent in the sense of this norm is stronger than convergent in the sense of $L^\infty$ norm.
