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

  • $\begingroup$ 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 $C^1[0,1]$. $\endgroup$
    – Y.X.
    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 $f,g\in C^1[0,1], d_u(f,g)=sup_{x\in[0,1]}|f(x)-g(x)|$. $\endgroup$
    – Y.X.
    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


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

  • $\begingroup$ 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... $\endgroup$
    – Y.X.
    Jul 13 '17 at 0:26

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.

  • $\begingroup$ 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? $\endgroup$
    – Y.X.
    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? $\endgroup$
    – Y.X.
    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

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