$C^1_0[0,1]$ 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!
Best Answer
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 $$ \int_0^1 |h(x)|\mathrm dx \leq \sup | h(x) | $$