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

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

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.

• When you write $f_n' \to_\text{unif} g$, this is uniform convergence. And $f_n \to f$ is just pointwise convergence? Can you provide a link to this theorem on this website or any other website?
– rims
Dec 7, 2019 at 15:47
• @Black Yes, it is only pointwise convergence (nothing else is required). I'll write a proof of the theorem in a little bit
– user515010
Dec 7, 2019 at 15:50
• But if my work above is correct, I think I showed already that $\lim \limits_{n \to \infty} ||f_n - f||_\infty = 0$ as well as $\lim \limits_{n \to \infty} ||f_n' - f'||_\infty = 0$ for any Cauchy sequence $f_n$
– rims
Dec 7, 2019 at 15:52
• @Black Your work is correct: it is simply somewhat stronger than needed
– user515010
Dec 7, 2019 at 15:55
• Can we show that this space is separable?
– José
Oct 29 at 18:34