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up vote 3 down vote favorite

I have to prove that $(C^1[0,1],d_1)$ is complete metric space, where $d_1(f,g)=\max|f(x)-g(x)|+\max|f'(x)-g'(x)|,x\in[0,1]$

Firstly, I take an arbitrary Cauchy sequence of functions from $C^1[0,1]$, $\langle f_n(x):x\in[0,1]\rangle$.
Since it is Cauchy's, it holds$$(\forall\epsilon>0)(\exists n_0\in N)(\forall n,m\geq n_0)(d_1(f_n,f_m)<\epsilon)$$ i.e. $$(\forall\epsilon>0)(\exists n_0\in N)(\forall n,m\geq n_0)(d_1(f,g)=\max|f(x)-g(x)|+\max|f'(x)-g'(x)|,x\in[0,1])$$

From this I have

  1. $(\forall\epsilon>0)(\exists n_0\in N)(\forall n,m\geq n_0)\max|f_n(x)-f_m(x)|<\epsilon$
  2. $(\forall\epsilon>0)(\exists n_0\in N)(\forall n,m\geq n_0)\max|f_n'(x)-f_m'(x)|<\epsilon$

I will skip what I was doing now and I will just write that I found that there exists $f(x)$ and $g(x)$ such that $\lim\limits_{n\rightarrow\infty}f_n(x)=f(x)$ and $\lim\limits_{n\rightarrow\infty}f_n'(x)=g(x)$.

I have written in my notebook that I have to show now that $f\in C[0,1]$, $g\in C[0,1]$ and $f'=g$.

It's clear to me why do I have to show that $g\in C[0,1]$ and $f'=g$, but don't I have to show for $f$ that $f\in C^1[0,1]$?

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If you did the first two, then by virtue of finding a continuous derivative for $f$ on $[0,1]$ you've proven $f$ is continuously differentiable there. – Jonathan Y. Aug 25 '13 at 12:25
    
Note that in order to show $d_1$ convergence to $f$ we'd need more than having $f,g$ the pointwise limit of $f_n,f_n^\prime$, respectively. We require uniform convergence on $[0,1]$. – Jonathan Y. Aug 25 '13 at 12:27
    
Yes, I have proven that $f$ is uniformly convergent at $[0,1]$. So, that's correct? I thought that I maybe rewrote it wrong from the blackboard. – gov Aug 25 '13 at 12:30
    
That the $f_n$'s uniformly converge to $f$, and their derivatives uniformly converge to $f^\prime$, yes. – Jonathan Y. Aug 25 '13 at 12:37
    
Sorry, I have proven that $f$ is uniformly continuous. – gov Aug 25 '13 at 12:41

2 Answers 2

up vote 2 down vote accepted

$C^1[0,1]$ is the set of all function on $[0,1]$ having continuous first order partial derivative.

Consider a Cauchy sequence $\{f_n(x)\}_n$ on $C^1[0,1]$. So for any $\epsilon > 0$ $\exists$ $m_1 \in \mathbb{N}$, s.t.$\forall$ natural numbers $p , q > m_1$ $d_1(f_p,f_q) < \epsilon$.

$d_1(f_p(x),f_q(x)) = \max_{x \in [0,1]}|f_p(x) - f_q(x)| + \max_{x \in [0,1]}|f'_p(x) - f'_q(x)| < \epsilon$.

So we are getting the followings.

$$\max_{x \in [0,1]}|f_p(x) - f_q(x)| < \epsilon$$

and

$$\max_{x \in [0,1]}|f'_p(x) - f'_q(x)| < \epsilon$$

We know the set on all continuous function on a closed bounded set of $\mathbb{R}$ with the metric $d(f(x),g(x)) = \max|f(x) - g(x)|$ is complete. Thus both the sequences $\{f_n(x)\}$ and $\{f'_n(x)\}$ are convergent and let converges to $f(x)$ and $g(x)$. Also the convergence is uniform.

We also know if $\{f_n(x)\}$ converges at least one point in its domain of definition, $\{f'_n(x)\}$ exists and converges uniformly to a function $g(x)$, then the sequence of functions $\{f_n(x)\}$ converges uniformly to a function $f(x)$ s.t $f'(x) = g(x)$.

So for any fixed $\epsilon > 0$ and $\forall x \in [0,1]$ $|f_n(x) - f(x)| < \frac{\epsilon}{2}$ when $n > m_2$ and $|f'_n(x) - f'(x)| < \frac{\epsilon}{2}$ when $n > m_3$.

So $\exists k > \max{m_1, m_2}$ s.t.

$$\max_{x \in [0,1]}|f_n(x) - f(x)| < \frac{\epsilon}{2}$$

and

$$\max_{x \in [0,1]}|f'_n(x) - f'(x)| < \frac{\epsilon}{2}$$

Combining them we shall get $d_1(f_n(x),f(x)) < \epsilon$ whenever $n > k$

Thus any Cauchy sequence in $(C^1[0,1], d_1)$ is convergent and hence the space is complete.

The results I have used here will be available in any standerd text book on Real Analysis.

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up vote 0 down vote

We'll follow the stages you've outlined.

  1. Take some $d_1$-Cauchy sequence $(f_n)\subseteq C^1([0,1])$. So we know that for all $x_0\in[0,1]$ the sequecnes $\left(f_n(x_0)\right)_{n\geq 0}$, $\left(f_n^\prime(x_0)\right)_{n\geq 0}$ are Cauchy sequences in $\mathbb{R}$, hence converging. We will denote their limits $f(x_0),g(x_0)$, respectively.

  2. For all $\epsilon>0$, we take $N>0$ such that for all $n,m>N$ it holds that $\max_{t\in[0,1]}|f_n(t)-f_m(t)|\leq\epsilon$. Then for all $t\in[0,1]$, $n>N$ we have $$|f_n(t)-f(t)| \mathop{\leq}_{\forall k} |f_n(t)-f_{n+k}(t)| + |f_{n+k}(t)-f(t)| \leq \epsilon + |f_{n+k}(t)-f(t)|\mathop{\longrightarrow}_{k\to\infty} \epsilon$$ Put differently $\sup_{t\in[0,1]}|f_n(t)-f(t)|\to 0$.

  3. Similarly, $\sup_{t\in[0,1]}|f_n^\prime(t)-g(t)|\to 0$.

  4. For all $n\geq 0$, $x\in[0,1]$ it holds that $f_n(x) = \int_0^x f_n^\prime(t)dt$, hence using parts 2-3 we take for all $x\in[0,1]$, $\epsilon>0$ an index $N>0$ such that $n>N$ implies $\sup_{t\in[0,1]}|f_n(t)-f(t)|,\sup_{t\in[0,1]}|f_n^\prime(t)-g(t)|\leq\frac{\epsilon}{2}$. We then have $$|f(x) - \int_0^x g(t)dt| = \left|f(x) - f_n(x) + \int_0^x f_n^\prime(t)dt - \int_0^x g(t)dt\right| \leq\\ \leq |f(x) - f_n(x)| + \int_0^x\left|f_n^\prime(t)-g(t)\right|dt \leq \frac{\epsilon}{2} + (x-0)\frac{\epsilon}{2}\leq\epsilon.$$ Such is the case for all $\epsilon>0$, hence $f(x)=\int_0^x g(t)dt$ (for all $x\in[0,1]$). This implies that $f\in C^1([0,1])$ with $f^\prime = g$.

  5. Finally, using parts 2-3 again, we have $d_1(f_n,f)\mathop{\longrightarrow}_{n\to\infty}0$, as desired.

So, again, it's not any uniform continuity as the fact that $f_n\to f$ uniformly and $f_n^\prime\to g=f^\prime$ uniformly, and that this also implies $f\in C^1([0,1])$, so that everything works as it should.

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