Show that $C^1(I; \mathbb{R})$ is a Banach space

Question

Let $I$ be a finite closed interval of $\mathbb{R}$.

Consider the normed vector space $\left ( C^1(I; \mathbb{R}), ||.||_{C^1} \right ) $ where $||f||_{C^1} = ||f||_\infty + ||f'||_\infty$.

My solution:

Consider a Cauchy sequence $\left ( f_n \in C^1(I; \mathbb{R}) \right )_{n \in \mathbb{N}} $. For every $\epsilon > 0$, there exists an $N(\epsilon)$ such that for all $n, m > N(\epsilon)$, we have $$||f_n - f_m||_\infty + ||f_n' - f_m'||_\infty < \epsilon$$

This certainly implies $||f_n - f_m||_\infty < \epsilon$ so for all $x \in I$, $|f_n(x) - f_m(x)| < \epsilon$ for all $n, m > N(\epsilon)$. Hence, for every $x \in I$, $f_n(x)$ is Cauchy so $f(x) = \lim \limits_{n \to \infty} f_n(x)$ exists. It can also be shown that $$||f_n - f ||_\infty < \epsilon$$ for all $n > N(\epsilon)$. So now we have $f$ approaching $f_n$ with respect to $||.||_\infty$ norm or simply, $f_n \xrightarrow{\text{unif}} f$.

$$||f_n' - f_m'||_\infty = \sup_{x \in I} \Big | f_n'(x) - f_m'(x) \Big |$$

$$ = \sup_{x \in I} \Big | f_n'(x) - \lim_{h \to 0} \frac{f_m(x + h) - f_m(x)}{h} \Big | $$ Letting $m \to \infty$, we get:

$$ = \sup_{x \in I} \Big | f_n'(x) - \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} \Big |$$

$$ = \sup_{x \in I} \Big | f_n'(x) - f'(x) \Big | $$ $$ = ||f_n' - f'||_\infty$$

The last term is smaller than $\epsilon$ for all $n > N(\epsilon)$. So we have $f_n' \xrightarrow{\text{unif}}$ $f'$.

EDIT: After I show that any Cauchy sequence in $C^1$ converges to $f$ w.r.t $||.||_{C^1}$ norm, how do I show $f \in C^1$ as well?

Solution

We showed that a cauchy sequence $\left ( f_n \in C^1(I; \mathbb{R}) \right )_{n \in \mathbb{N}} $ converges uniformly to $f$. This means the

cauchy sequence $f_n$ viewed as $\left ( f_n \in C^0(I; \mathbb{R}) \right )_{n \in \mathbb{N}} $ $\xrightarrow{\text{unif}}$ $f$. By completeness of $C^0(I; \mathbb{R})$, $f \in C^0(I; \mathbb{R})$. Why is $C^0(I; \mathbb{R})$ complete? Because it is the set of continuous functions on a bounded, closed interval $I$; any continuous function on a compact set is bounded; the space of bounded continuous functions is complete. Similarly, we also showed a cauchy sequence $\left ( f_n' \in C^0(I; \mathbb{R}) \right )_{n \in \mathbb{N}} $ $\xrightarrow{\text{unif}}$ $f'$. By a similar argument, $f' \in C^0(I; \mathbb{R})$.

Hence, $f \in C^1(I; \mathbb{R})$.

Answer 1

You are almost there, you only need to use the following theorem: Given a sequence of function $f_n$ and $f'_n$ on an interval, we have that if

$$f'_n\underset{\text{unif}}{\rightarrow} g\\ f_n(x_0)\ \ \text{converges}$$

Then $f_n$ converges uniformly to a function $f$ and $f'=g$.

For a proof, see here.

Note that for our problem we may use a weaker result, namely:

Let $f_n$ be a sequence of $C^1(I)$ functions such that $f_n'$ converges uniformly and $f_n$ converges. Then $f'=\lim_{n\to \infty}f'_n$

Proof:

$$f_n(x)=f_n(x_0)+\int_{x_0}^x f_n'(t)dt\\ f(x)=f(x_0)+\int_{x_0}^x g(t)dt\\ f'(x)=g(x)$$

You have already proved that $f'_n$ is a Cauchy sequence in $(C^0(I),||\cdot||_{\infty})$, and by completeness of this space we have that $f'_n\to g$, and more: the convergence is in the $\infty$ metric, and it's thus uniform. Applying the theorem I stated yelds the result.