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Let $C^1[a,b]$ be the space of continuous differentiable functions on $[a,b]$ equipped with the following norm $$\|x\|=|x(a)| + \sup_{t\in [a,b]}|x'(t)|.$$ Prove that $(C^1[a, b],\|\cdot\|)$ is a Banach space.

Here is my attempt:

Let $(x_n)$ be a Cauchy sequence in $C^1[a,b]$, i.e., $$\lim \limits_{m,n\to \infty}\|x_m-x_n\|=0.$$ This means $$|x_m(a) -x_n(a)| + \sup_{t\in [a,b]}|x_m'(t) -x_n'(t)| \to 0,$$ as $m,n\to\infty$. Therefore, $\{x_n(a)\}$ is a Cauchy sequence in $ \mathbb R$ and $\{x'_n\}$ is a Cauchy sequence in $C[a,b]$ with $\sup$ norm. Thus $\{x_n\}$ converges at $a$, and $\{x'_n\}$ uniformly converges to some $y$ in $C[a,b]$.

I don't know how to show that $y\in C^1[a,b]$ and $x_m \to y$ in $C^1[a,b]$.

Richkent
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    If $x_n'$ converges to $x$ in $C[a,b]$, then you don't have $x_m \to x$ in $C^1[a,b]$. $x$ will be the derivative of $\lim x_n$. – Daniel Fischer Mar 23 '14 at 10:52
  • I'd better say that $\left{ x_n \right}$ converges to some $y$ in $C[a,b]$ so we do not get confused with the original $x$. Now, if $y \in C[a,b]$, what can be said about its Riemann integrability? Does a primitive exist? Where does it lie? In the other hand, I do not see why $x_n(a) \to a$, you mean the left bound of the interval? – busman Mar 23 '14 at 10:58
  • I mean ${x_n(a)}$ is a convergent sequence of numbers, not $x_n(a) \to a$. – Richkent Mar 23 '14 at 11:03
  • But, to me $a$ is the one in $[a,b]$, shouldn't it be that $x_n(a) \to c \in [a,b]$ so you must show that $c =x(a)$? I am just letting my thoughts out, it could be I am wrong. – busman Mar 23 '14 at 11:05
  • Yes, you're right! – Richkent Mar 23 '14 at 11:06

2 Answers2

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In your notation, let $c=\text{lim}_{n\rightarrow \infty}x_n(a)$: Define $$ Y(x)=\int_a^x y(t)dt+c. $$ Then by the fundamental theorem of calculus $Y\in C^1(a,b)$, $Y'(x)=y(x)$ and furthermore $Y(a)=c$. Now show that $Y$ is the limit of $x_n$

MichalisN
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Well, regarding one of your questions, you have already shown that $\left\{ x^{\prime}_n \right\}$ is a Cauchy sequence within $(C[a,b],\| \cdot \|_{\infty})$, so since this is a Banach space in the first place, you have that $y \in C[a,b]$ (but not necessarily $C^1[a,b]$). The thing is that if that is the case, then there exists $\tilde{x} \in C^1[a,b]$ such that $y=\tilde{x}^{\prime}$ since all continuous functions are Riemann-integrable. Can you show that $\tilde{x} = x + \lambda$, where $\lambda \in \mathbb{R}$?

EDIT: Michalis' answer is a thousand times better than mine.

busman
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