2
$\begingroup$

$C^1_0[0,1]$ 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!

$\endgroup$
  • $\begingroup$ does $d_u$ also means uniform norm of the derivative? or just the supremum of $f$? $\endgroup$ – supinf Jul 12 '17 at 12:53
  • $\begingroup$ @supinf It is the uniform norm on the metric space $C^1[0,1]$. $\endgroup$ – PropositionX Jul 12 '17 at 12:56
  • $\begingroup$ does this mean that the uniform norm is $\sup |f|+\sup |f'|$ in this case? or just $\sup |f|$? $\endgroup$ – supinf Jul 12 '17 at 13:02
  • $\begingroup$ @supinf Sorry about the misleading expression! $d_u$ denotes the uniform metric, not the uniform norm(which is induced by the uniform metric). it is defined as: for $f,g\in C^1[0,1], d_u(f,g)=sup_{x\in[0,1]}|f(x)-g(x)|$. $\endgroup$ – PropositionX Jul 12 '17 at 13:08
  • 1
    $\begingroup$ ok now i understand. In this case the metric space $(C^1[0,1],d_u)$ is not complete, because the limit $f$ of $f_n$ does not need to be differentiable. However, $(C^1[0,1],d_1)$ is a complete space, where $d_1(f,g,)=d_u(f,g)+d_u(f',g')$. $\endgroup$ – supinf Jul 12 '17 at 13:13
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ I have tried for a while and I have updated what I have got so far in the edited question. But there remains some unsolved problems, could you please have a look? $\endgroup$ – PropositionX Jul 13 '17 at 0:27
  • 1
    $\begingroup$ @PropositionX Your proposed solution seems confusing. I do not know what you are doing. I've given you a roadmap to a much easier solution. If you don't undertand my hint, please ask questions. $\endgroup$ – zhw. Jul 13 '17 at 2:30
  • $\begingroup$ I have attempted to understand your hint. I deleted my former argument because I found it is not correct. Could you please check if I have got the hint correct? May I please ask how to deal with the integral? $\endgroup$ – PropositionX Jul 13 '17 at 5:32
  • $\begingroup$ @PropositionX I've edited my answer to a more complete solution. $\endgroup$ – zhw. Jul 13 '17 at 16:12
3
$\begingroup$

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 $$ \int_0^1 |h(x)|\mathrm dx \leq \sup | h(x) | $$

$\endgroup$
  • $\begingroup$ Could you please have a look to the edit? I used the fundamental theorem of Calculus, but I still have some problem. May I please ask if my track is correct? But maybe it is different from the method in your hint... $\endgroup$ – PropositionX Jul 13 '17 at 0:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.