Space of Continuously Differentiable on Closed Interval Real-Valued Functions with C^1 Norm is Banach Space

== Theorem ==
Let $I := \closedint a b$ be a closed real interval.
Let $\map C I$ be the space of real-valued functions continuous on $I$.
Let $\map {C^1} I$ be the space of real-valued functions, continuously differentiable on $I$.
Let $\norm {\, \cdot \,}_{1, \infty}$ be the $\CC^1$ norm.
$\struct {\map {C^1} I, \norm {\, \cdot \,}_{1, \infty} }$ be the normed space of real-valued functions, continuously differentiable on $I$.
Then $\struct {\map {C^1} I, \norm {\, \cdot \,}_{1, \infty} }$ is a Banach space.
== Proof ==
Let $\sequence {x_n}_{n \mathop \in \N}$ be a Cauchy sequence in $\struct {\map {C^1} I, \norm {\, \cdot \,}_{1, \infty} }$:
:$\forall \epsilon \in \R_{>0}: \exists N \in \N: \forall m, n \in \N: m, n \ge N: \norm {x_n - x_m}_{1, \infty} < \epsilon$
=== $\sequence {x_n}_{n \mathop \in \N}$ converges in $\struct {\map C I, \norm {\, \cdot \,}_\infty}$ to $x$ ===
$\forall \epsilon \in \R_{>0}: \exists N \in \N: \forall m, n \in \N: m, n \ge N$ we have that:
Hence, $\sequence {x_n}_{n \mathop \in \N}$ is a Cauchy sequence in $\struct {\map C I, \norm {\, \cdot \,}_\infty}$.
We have that $\struct {\map C I, \norm {\, \cdot \,}_\infty}$ 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 $x \in \map C I$.
=== $\sequence {x'_n}_{n \mathop \in \N}$ converges in $\struct {\map C I, \norm {\, \cdot \,}_\infty}$ to $y$ ===
$\forall \epsilon \in \R_{>0}: \exists N \in \N: \forall m, n \in \N: m, n \ge N$ we have that:
Hence, $\sequence {x'_n}_{n \mathop \in \N}$ is a Cauchy sequence in $\struct {\map C I, \norm {\, \cdot \,}_\infty}$.
We have that $\struct {\map C I, \norm {\, \cdot \,}_\infty}$ 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 $y \in \map C I$.
=== $x$ 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:
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:
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, $x' = y \in \map C I$.
Hence, $x \in \map {C^1} I$.
=== $\sequence {x_n}_{n \mathop \in \N}$ converges in $\struct {\map {C^1} I, \norm {\, \cdot \,}_{1, \infty} }$ 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:
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:
== Sources ==
* : $\S 1.4$: Normed and Banach spaces. Sequences in a normed space; Banach spaces
Category:Banach Spaces