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I have the following problem: Let $\mathcal{C}^1([a,b],\mathbb{K})$ be the linear space of continuously differentiable functions provided with the norm $ \| f \|_1 = \| f \|_{\infty} + \| f' \|_{\infty} $. Show that $\mathcal{C}^1([a,b],\mathbb{K})$ is a Banach space. Note that $\mathcal{C}([a,b],\mathbb{K})$ is a Banach space with the sup norm.

My solution: Consider the Cauchy sequence $(f_n) \in \mathcal{C}^1$ such that $f_n \rightarrow f$. Now we have to show that $f \in \mathcal{C}^1$. We have that $\forall \epsilon > 0 $ $\exists N \in \mathbb{N}$ such that $ \|f_n - f\|_1<\epsilon $ $\forall n \geq N$, which implies $\| f_n -f \|_1 = \| f_n -f \|_{\infty} + \| f_n' -f' \|_{\infty} < \epsilon_1 + \epsilon_2 = \epsilon$. Then $f_n'\rightarrow f'$. Since $\mathcal{C}$ is a Banach Space with the sup norm and $f_n' \in \mathcal{C}$, then $f'\in \mathcal{C}$ and so $f \in \mathcal{C}^1$.

My doubts stem from whether I can just ignore $\| f_n -f \|_{\infty}$, since I don't use it in my proof. However, I don't so since I think I don't need it. Is my proof sufficient, or is $\| f_n -f \|_{\infty}$ necessary? Any feedback is appreciated!

enfield
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    Start with just $(f_n)$. No $f$ is given to you. – Kavi Rama Murthy Feb 16 '25 at 12:47
  • @geetha290krm would I first have to show that $(f_n)$ converges to some element $f$? I thought Cauchy sequences always converged to element, though perhaps not in the space. – enfield Feb 16 '25 at 14:12
  • In some way yes (there is the notion of completion of a normed space), however your $f$ might not even be a function (in this case it will a fortiori be a function but you can't use that). The best course of action would instead be to find the right candidate among the functions and show that your sequence converges to said candidate. – Bruno B Feb 16 '25 at 14:22
  • That is also where the $|f|_\infty$ part of $|\cdot|_1$ comes in: without it, all constant functions would be of zero norm, and so you wouldn't have a normed space, but this would also mean that the "limit" you'd consider would have to be an equivalence class of functions (modulo constant functions here), since you would have to make sense of $(x \mapsto n + \frac{1}{n} x)_n$ converging to $0$ for example, when pointwise it's going to $+\infty$. – Bruno B Feb 16 '25 at 14:31
  • @BrunoB Isn't there a theorem in Topology where if $\mathbb{K}$ is a complete metric space, then there exists a function $f$ for which $f_n$ converges uniformly to? – enfield Feb 16 '25 at 16:19
  • The point is that you need some kind of result similar to what you've described or manually find the right candidate to be the limit of your sequence, otherwise your proof will be flawed from the get-go. Not every norm is a "uniform" norm on a space of continuous functions, so you should bear that in mind. Another main issue is that you tried to look at $f'$ when a continuous function doesn't have to be differentiable even at any one point, so you need some argument or result there too. – Bruno B Feb 16 '25 at 17:56

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Set $A:=\mathcal{C}^1([a,b],\mathbb{K})$. Let $(f_n)\subset A$ be a $\|\cdot\|_1$-Cauchy sequence. Since $\|f_n' - f_m'\|_\infty =\|(f_n-f_m)'\|_\infty \le \|f_n-f_m\|_1$, we have that the sequence $(f_n')_{n=1}^\infty\subset \mathcal{C}([a,b])$ is $\|\cdot\|_\infty$-Cauchy, so it converges to some $g\in\mathcal{C}([a,b])$. Also, since $|f_n(a)-f_m(a)|\le \|f_n-f_m\|_1$, the sequence $(f_n(a))_{n=1}^\infty\subset\mathbb{K}$ is Cauchy, and thus converges to some $\lambda\in\mathbb{K}$.

set $f(t):= \int_a^tg(s)\mathrm{d}s + \lambda$. Then, $f'(t) = g(t)$, so $f$ has a continuous derivative, and thus $f \in A$. Moreover, for any $t\in[a,b]$ we have

$$ |f_n(t) - f(t)| = \bigg|( \int_a^tf_n'(s)\mathrm{d}s + f_n(a)) - (\int_a^tg(s)\mathrm{d}s +\lambda) \bigg| = $$ $$\bigg| \bigg(\int_a^t(f_n'(s)-g(s))\mathrm{d}s\bigg)+ (f_n(a)-\lambda)\bigg|\le $$ $$ \int_a^b|f_n'(s)-g(s)|\mathrm{d}s +|f_n(a)-\lambda| \le |b-a|\cdot \|f_n'-g\|_\infty +|f_n(a)-\lambda| $$ and thus $\|f_n-f\|_\infty \le |b-a|\cdot \|f_n'-g\|_\infty+|f_n(a)-\lambda|$, therefore $\|f_n-f\|_1\le (1+|b-a|)\cdot \|f_n'-g\|_\infty +|f_n(a)-\lambda|\to0$, which shows that $f_n\to f\in A$, so $A$ is complete.

Just dropped in
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  • Wouldn't it be $f(t) := \int_a^t g(s) \mathrm{d}s + \lim_{n \to \infty} f_n(a)$ and $f_n(t) = \int_a^t f'_n(s) \mathrm{d}s + f_n(a)$? Otherwise the rest is fine since the method is there. – Bruno B Feb 16 '25 at 17:51
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    @BrunoB You're right, I guess I got confused while writing the answer - I edited the post – Just dropped in Feb 16 '25 at 20:37