Let $\sequence {x_n}_{n \mathop \in \N}$ be a Cauchy sequence in $\struct {\map {\CC^1} I, \norm {\, \cdot \,}_{1, \infty} }$: :$\forall \epsilon \in \R_{>0}: \exists N \in \N: \forall m, n \in \N: m, n \ge N: \norm {x_n - x_m}_{1, \infty} < \epsilon$ === $\sequence {x_n}_{n \mathop \in \N}$ converges in $\struct {\map \CC I, \norm {\, \cdot \,}_\infty}$ to $x$ === $\forall \epsilon \in \R_{>0}: \exists N \in \N: \forall m, n \in \N: m, n \ge N$ we have that: {{begin-eqn}} {{eqn | l = \norm {x_n - x_m}_\infty       | o = \le       | r = \norm {x_n - x_m}_\infty + \norm {x'_n - x'_m}_\infty }} {{eqn | r = \norm {x_n - x_m}_{1, \infty}       | c = {{defof|C^k Norm}} }} {{eqn | o = <       | r = \epsilon       | c = {{defof|Cauchy Sequence}} }} {{end-eqn}} Hence, $\sequence {x_n}_{n \mathop \in \N}$ is a Cauchy sequence in $\struct {\map \CC I, \norm {\, \cdot \,}_\infty}$. We have that $\struct {\map \CC I, \norm {\, \cdot \,}_\infty}$ is a Banach space. Therefore, $\sequence {x_n}_{n \mathop \in \N}$ converges. Denote this limit as $x$: :$\ds \lim_{n \mathop \to \infty} x_n = x$ where $x \in \map \CC I$. {{qed|lemma}} === $\sequence {x'_n}_{n \mathop \in \N}$ converges in $\struct {\map \CC I, \norm {\, \cdot \,}_\infty}$ to $y$ === $\forall \epsilon \in \R_{>0}: \exists N \in \N: \forall m, n \in \N: m, n \ge N$ we have that: {{begin-eqn}} {{eqn | l = \norm {x'_n - x'_m}_\infty       | o = \le       | r = \norm {x_n - x_m}_\infty + \norm {x'_n - x'_m}_\infty }} {{eqn | r = \norm {x_n - x_m}_{1, \infty}       | c = {{defof|C^k Norm}} }} {{end-eqn}} Hence, $\sequence {x'_n}_{n \mathop \in \N}$ is a Cauchy sequence in $\struct {\map \CC I, \norm {\, \cdot \,}_\infty}$. We have that $\struct {\map \CC I, \norm {\, \cdot \,}_\infty}$ is a Banach space. Therefore, $\sequence {x'_n}_{n \mathop \in \N}$ converges. Denote this limit as $y$: :$\ds \lim_{n \mathop \to \infty} x'_n = y$ where $y \in \map \CC I$. {{qed|lemma}} === $x$ is differentiable, and $y' = x$ === Let $t \in I$. By fundamental theorem of calculus $\paren \star$: :$\ds \map {x_n} t - \map {x_n} a = \int_a^t \map {x'_n} \tau \rd \tau$ Consider the following difference: :$\ds \size {\map {x_n} t - \map {x_n} a - \int_a^t \map y \tau \rd \tau}$ Then: {{begin-eqn}} {{eqn | l = \size {\map {x_n} t - \map {x_n} a - \int_a^t \map y \tau \rd \tau}       | r = \size {\int_a^t \paren {\map {x'_n} \tau - \map y \tau} \rd \tau}       | c = Substitution of $\paren \star$ }} {{eqn | o = \le        | r = \int_a^t \size {\map {x'_n} \tau - \map y \tau} \rd \tau }} {{eqn | o = \le        | r = \int_a^t \sup_{T \mathop \in I} \size {\map {x'_n} T - \map y T} \rd \tau }} {{eqn | o = \le       | r = \norm {x'_n - y}_\infty \paren {t - a}       | c = {{defof|Supremum Norm}} }} {{end-eqn}} Absolute value is a continuous function. Therefore, we can take the limit of the composite function. Passing the limit, as $n$ goes to infinity, gives: {{begin-eqn}} {{eqn | l = \size {\map x t - \map x a - \int_a^t \map y \tau \rd \tau}       | o = \le       | r = \norm {y - y}_\infty \paren {t - a}       | c = $\ds \sequence {x'_n}_{n \mathop \in \N}$ converges to $y$ }} {{eqn | r = 0 }} {{end-eqn}} In other words: :$\ds \map x t = \map x a + \int_a^t \map y \tau \rd \tau$ By fundamental theorem of calculus, $x$ is a primitive of $y$. Thus, $x' = y \in \map \CC I$. Hence, $x \in \map {\CC^1} I$. {{qed|lemma}} === $\sequence {x_n}_{n \mathop \in \N}$ converges in $\struct {\map {\CC^1} I, \norm {\, \cdot \,}_{1, \infty} }$ to $x$ === Let $\epsilon \in \R_{> 0}$. Let $N \in \N : \forall m, n \in \N : m, n > N \implies \norm { x_n - x_m}_{1, \infty} < \epsilon$. Then for all $t \in I$ we have that: {{begin-eqn}} {{eqn | l = \size {\map {x_n} t - \map {x_m} t} + \size {\map {x'_n} t - \map {x'_m} t}       | o = \le       | r = \sup_{t \mathop \in I} \size {\map {x_n} t - \map {x_m} t}+ \sup_{t \mathop \in I} \size {\map {x'_n} t - \map {x'_m} t} }} {{eqn | r = \norm {x_n - x_m}_\infty + \norm {x'_n - x'_m}_\infty       | c = {{defof|Supremum Norm}} }} {{eqn | r = \norm {x_n - x_m}_{1, \infty}       | c = {{defof|C^k Norm}} }} {{eqn | o = <       | r = \epsilon }} {{end-eqn}} Absolute value is a continuous function. Therefore, we can take the limit of the composite function. Letting $m$ go to infinity, it follows that: :$\forall n > N : \size {\map {x_n} t - \map x t} + \size {\map {x'_n} t - \map {x'} t} < \epsilon$ Since $t \in I$ was arbitrary: {{begin-eqn}} {{eqn | l = \size {\map {x_n} t - \map x t} + \size {\map {x'_n} t - \map {x'} t}        | o = \le       | r = \sup_{t \mathop \in I} \size {\map {x_n} t - \map x t} + \sup_{t \mathop \in I} \size {\map {x'_n} t - \map {x'} t} }} {{eqn | r = \norm {x_n - x}_{1, \infty}       | c = {{defof|Supremum Norm}} }} {{eqn | o = <       | r = \epsilon }} {{end-eqn}} {{qed}}