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: