Prove that $C^1_0[0,1]$ with a certain metric is a complete metric space. $C^1_0[0,1]$ is the set of functions $f:[0,1]\rightarrow \Bbb R$ such that $f,f'$ are continuous on $[0,1]$ and $f(0)=0.$
The metric $d$ is given by:
$d(f,g)=\int^1_0 |f(x)-g(x)|\,\mathrm dx+\sup_{x\in[0,1]}|f'(x)-g'(x)|$
Now I aim to prove that:
$C^1_0[0,1]$ with the matric $d$ is a complete metric space.
My attempt:
I need to pick a Cauchy sequence in $C^1_0[0,1]$ and prove that it converges.
Call the Cauchy sequence $(f_n)$, then we have:
$(\forall \epsilon>0)(\exists N\in \Bbb N)(m,n\ge N\Rightarrow d(f_m,f_n)=\int^1_0 |f_m(x)-f_n(x)|dx+sup_{x\in [0,1]}|f'_m(x)-f'_n(x)|<\epsilon)$
To conclude that the space is complete, I aim to show that it converges, that is:
$(\exists f\in C^1_0[0,1])(\forall \epsilon>0)(\exists N\in\Bbb N)(n\ge N\Rightarrow d(f,f_n)=\int^1_0 |f(x)-f_n(x)|dx+sup_{x\in [0,1]}|f'(x)-f'_n(x)|<\epsilon)$
So I think I need to conclude both $\int^1_0 |f(x)-f_n(x)|dx$ and $sup_{x\in [0,1]}|f'(x)-f'_n(x)|$ can be arbitarily small. 
For the latter:
Consider the sequence $(f'_n)\in C[0,1]$ (we have this sequence because $(f_n)\in C^1_0[0,1]$). From that fact that both of the $2$ parts in $\int^1_0 |f_m(x)-f_n(x)|dx+sup_{x\in [0,1]}|f'_m(x)-f'_n(x)|<\epsilon)$ are non-negative, we have $sup_{x\in [0,1]}|f'_m(x)-f'_n(x)|<\epsilon$, so $(f'_n)$ is a Cauchy sequence in $C[0,1]$.
I know that $(C[0,1],d_u)$ where $d_u$ is the uniform metric is complete, so $(f'_n)$ converges to some $f'\in C[0,1]$, so $d_u(f',f'_n)=sup_{x\in [0,1]}|f'(x)-f'_n(x)|$ can be arbitarily small. 
But I have stuck for a while for how to deal with the integral part... Maybe it is because of the fact that I am not such familiar to the fundamental theorem of calculus... Could some one help with that part? Thanks!
 A: Hint:
Consider the metric $d_1$ defined as
$$
d_1(f,g) = \sup |f(x)-g(x) | + \sup |f'(x)-g'(x)|.
$$
What you need to show is that $d_1$ and $d$ are equivalent,
that means there are constants $c_1,c_2$ such that
$d(f,g) \leq c_1 d_1(f,g)$ and $d_1(f,g) \leq c_2 d(f,g)$.
Then you can use that $(C_0^1[0,1],d_1)$ is a complete metric space
(you can prove that very similar to what you did so far)
For the inequality $d_1(f,g) \leq c_2 d(f,g)$ you can use an inequality of the following type:
$$
  \sup | h(x) | \leq \sup | h'(x) |
$$
for functions $h$ with $f(0)=0$.
This inequality can be proven using
$$
 h(x)-h(0) = \int_0^x h'(x) \mathrm dx.
$$
For the other inequality $d(f,g) \leq c_1 d_1(f,g)$ you can use
$$
 \int_0^1 |h(x)|\mathrm dx \leq \sup | h(x) |
$$
A: Suppose $f_n$ is Cauchy in $(C_0^1([0,1]),d).$ Then $f_n'$ is Cauchy in $(C[0,1],d_u).$ Now the latter metric space is complete. Thus there exists $g\in C[0,1]$ such that $f_n' \to g$ in $(C[0,1],d_u).$ Define $G(x) = \int_0^x g(t)\, dt.$ Then $G(0)=0$ and $G'=g$ by the FTC. Thus $G\in (C_0^1([0,1]),d).$
Claim: $f_n \to G$ in $(C_0^1([0,1]),d).$ (This proves $(C_0^1([0,1]),d)$ is complete.)
Proof: We already know $d_u(f_n',G') = d_u(f_n',g) \to 0.$ To handle the integral condition, note
$$\int_0^1|f_n(x) - G(x)|\, dx = \int_0^1|\int_0^x(f_n'(t) - g(t))\, dt\,|\, dx $$ $$ \le \int_0^1\int_0^x|f_n'(t) - g(t)|\, dt\, dx \le \int_0^1\int_0^1|f_n'(t) - g(t)|\, dt\, dx .$$
We used the FTC and the fact that $f_n(0)=0= G(0)$ to obtain the first line. Since $|f_n'(t) - g(t)| \le d_u(f_n',g)$ for every $t,$ the last iterated integral is bounded above by $d_u(f_n',g),$ which we already know $\to 0.$ This proves the claim.
