$f$ and $g$ are in $C^1([0,1])$.
I can show that the space of functions, along with this metric, is indeed a metric space. But showing that it's complete is proving to be a bit more complicated.
Say I'm given a cauchy sequence in the space, say $\{f_n\}_{1\leq n}$. Since the sequence is cauchy, I can define $f(x)=\lim_{n\to\infty}f_n(x).$ But how can I show that $f$ is both continuous and differentiable?
On
Let $X=(C^1[0,1],d)$ and $Y$ be $C[0,1]$ endowed with a metric $d’$ such that $d’(f,g)=\sup_{x\in [0,1]} |f(x)-g(x)|$. It is well-known (see, for instance, Theorem 4.3.13 from [Eng]) and easy to show (see, for instance, this thread) that $(Y,d’)$ is complete. Clearly, that a map $\partial: X\to Y$, $f\mapsto f’$ is Lipshitz with the constant $1$.
Let $\{F_n\}$ be a Cauchy sequence in $X$. Since the map $\partial$ is Lipshitz, $\{\partial F_n\}$ is a Cauchy sequence in $Y$. Since the space $Y$ is complete, the sequence $\{\partial F_n\}$ has a limit $f$. Since a sequence $\{F_n(0)\}$ is Cauchy, there exists a limit $F(0)=\lim_{n\to\infty} F(0)$. For each $x\in [0,1]$ put $F(x)=F(0)+\int_0^x f(t)dt$. By Newton-Leibnitz formula, $F’(x)=f(x)$ for each $x\in [0,1]$.
We claim that $\{F_n\}$ converges to $F$. Indeed, let $\varepsilon>0$ be any number. Since $\{F_n(0)\}$ converges to $F(0)$, there exists natural $N$ such that $|F_n(0)-F(0)|\le\varepsilon/2$ for each $n>N$. Since $\{\partial F_n\}$ converges to $f$, there exists natural $N’\ge N$ such that $d’(\partial F_n, f)\le\varepsilon/2$ for each $n>N’$. Fix any such $n$ and any $x\in [0,1]$. By Newton-Leibnitz formula, $F_n(x)=F_n(0)+\int_0^x \partial F_n(t)dt$. Thus $$|F_n(x)-F(x)|=$$ $$\left|F_n(0)+\int_0^x \partial F_n(t)dt - F(0)-\int_0^x f(t)dt \right|\le$$ $$|F_n(0)-F(0)|+\left|\int_0^x \partial F_n(t)dt -\int_0^x f(t)dt \right|\le$$ $$\varepsilon/2+\left|\int_0^x (\partial F_n(t)-f(t))dt \right|\le$$ $$\varepsilon/2+\left|\int_0^x |\partial F_n(t)-f(t)|dt \right|\le$$ $$\varepsilon/2+\int_0^x \varepsilon/2\le$$ $$ \varepsilon/2+\varepsilon/2=\varepsilon.$$
Thus $d’(F_n,F)\le \varepsilon$ and so $d(F_n,F)=d’(F_n,F)+ d’(\partial F_n, f)\le 2\varepsilon$. Therefore $\{F_n\}$ converges to $F$ and so the space $X$ is complete.
References
[Eng] Ryszard Engelking, General Topology, 2nd ed., Heldermann, Berlin, 1989.
On
First note that $|f’(x)-g’(x)|\le d(f, g)$ for any $x\in [0, 1]$ and $f, g\in C^1[0, 1].$ This shows that if $f_n$ is $d$-Cauchy, then $f_n’$ is Cauchy in the $\sup$-norm. Similarly, it is clear that if $f_n$ is Cauchy in $d$-norm, then $f_n$ is Cauchy in the $\sup$-norm.
Now we know that $C[0, 1]$ with the usual sup norm is complete, therefore, we conclude that $f_n’\to h$ And $f_n\to H$ in sup norm for some $h, H\in C[0, 1].$
Apriori we only know that $H$ Is continuous. But we claim that $H$ is differentiable and moreover $H’=h.$ To this end, we simply note that $$f_n(x)-f_n(y)=\int_{x}^y f_n’(t)dt.$$ By using the uniform convergence of $f_n’$ to $h,$ and the uniform convergence of $f_n$ to $H,$ we obtain that $$H(x)-H(y)=\int_{x}^{y}h(t)dt.$$
Now observe that for a continuous function $h,$ we have $\frac{1}{x-y}\int_{x}^yh(t)dt\to h(x)$ as $|x-y|\to 0.$ In particular, it follows that $\lim\limits_{|x-y|\to 0}\frac{H(x)-H(y)}{x-y}=h(x).$ This proves that $H$ Is $C^1$ and $H’=h.$
The last piece is to now show that $f_n\to H$ in the $d$-metric. But this follows immediately by observing that for $f, g\in C^1,$ We have $$d(f, g)\le ||f-g||_{\infty}+||f’-g’||_{\infty}.$$
HINT:
It is not hard to show that $(f_n)$ is Cauchy in the given metric is equivalent to: $(f_n)$ is uniformly convergent to a function $f$, and $f'_n$ is uniformly convergent to a function $g$. Now, both $f$ and $g$ are continuous ( uniform limits sequences of continuous functions). The only thing left is to show that $$f'=g$$ For this, use the Leibniz-Newton equality $$f_n(x)-f_n(0)= \int_0^x f'_n(t) dt$$ for all $x\in [0,1]$ and pass take the limit as $n\to \infty$. We get $$f(x)-f(0) =\int_0^x g(t)dt$$ for all $x\in [0,1]$, and this implies $$f'(x) = g(x)$$ for all $x\in [0,1]$.
$\bf{Added:}$ Why the convergence of $(f_n)$ and $(f_n')$ are uniform: $|f_n(x)-f_m(x)|+ |f_n'(x)-f_m'(x)|\le \epsilon$ for all $m,n \ge n(\epsilon)$ and $x\in [0,1]$ means that for every $x$ $(f_n(x))$ and $(f_n'(x))$ are Cauchy so convergent to $f(x)$, respectively $g(x)$. Passing to limit $m \to \infty$ in the above inequality we get $$|f_n(x)-f(x)|+ |f_n'(x)-g(x)|\le \epsilon$$ for all $n\ge n(\epsilon)$ and $x \in [0,1]$. That implies that $(f_n)$ converges uniformly to $f$ and $f_n'$ converges uniformly to $g$.