# $C'[0,1]$ is a Banach space with the norm $||f||=\max_{t\in[0,1]} \{|f(t)|,|f'(t)|\}$

Let $$C'[0,1]$$ be the space of real functions defined in $$[0,1]$$ continuously differentiable in $$(0,1)$$ which derivative can be extended continuously to $$[0,1]$$. Show this is a Banach space with the norm $$||f||=\max_{t\in[0,1]} \{|f(t)|,|f'(t)|\}$$.

My approach is really simple, since $$||f||=\max_{t\in[0,1]} \{|f(t)|,|f'(t)|\},$$ we can take $$(f_n) \subseteq C'[0,1]$$ a Cauchy sequence. Then for $$N>0$$ and $$n,m>N$$, $$||f_n-f_m||=\max_{t\in[0,1]} \{ |f_n(t)-f_m(t)|, |f_n'(t)-f_m'(t)| \}<\varepsilon.$$ In particular, $$|f_n(t)-f_m(t)|<\varepsilon$$, so $$(f_n(t))$$ is a Cauchy sequence for each $$t\in[0,1]$$. Then I basically repeat the argument used to prove that $$C[0,1]$$ is Banach space with the $$\sup$$ norm. Constructing $$F(t)=\lim_{n\to\infty}f_n(t)$$ for each $$t\in[0,1]$$ as the candidate function which $$(f_n)$$ converges.

Sadly, I'm struggling to conclude that $$f$$ is continuously differentiable in $$(0,1)$$. I suppose that I have to use the derivatives defined in the norm. However, a similar argument (considering $$(f_n')$$ a Cauchy sequence) as the one I used would yield to a $$G$$ function which $$f_n'(t)\to G$$. However, I cannot conclude $$G=F'$$.

Additionally, what does "which derivative can be extended continuously to [0,1]" exactly means? That I can define the limit of the derivative without breaking continuity?

• The proof in this link works here too: math.stackexchange.com/questions/920070/… Sep 10, 2023 at 5:17
• The space $C^1[0,1]$ is isomorphic with $C[0,1]\times \mathbb{C}$ through the mapping $f\mapsto (f',f(0)).$ Sep 10, 2023 at 12:03

$$\left|\frac{F(t+h)-F(t)}{h} - G(t) \right| =$$ $$\left|\frac{F(t+h)-F(t)}{h} - \frac{f_n(t+h)-f_n(t)}{h} + \frac{f_n(t+h)-f_n(t)}{h} - f_n'(t) + f_n'(t) - G(t) \right| \leq$$ $$\left|\frac{F(t+h)-F(t)}{h} - \frac{f_n(t+h)-f_n(t)}{h}\right| + \left|\frac{f_n(t+h)-f_n(t)}{h} - f_n'(t)\right| + \left|f_n'(t) - G(t)\right|$$ $$\left|\frac{F(t+h)-F(t)}{h} - \frac{f_n(t+h)-f_n(t)}{h}\right| + \left|f_n'(\psi) - f_n'(t)\right| + \left|f_n'(t) - G(t)\right|$$ for some $$\psi \in [t,t+h]$$.

First choose $$n$$ such that for all $$n \geq m$$ we have, $$|f_n'(t)-G(t)| < \epsilon$$

Lets say $$h$$ in the above is chosen such that by uniform continuity of $$f'_{m}(t)$$ we have $$|f_m'(\psi) - f_m'(t)| < \epsilon, \forall \psi \in [t,t+h]$$.

Hence for all $$n \geq m$$, $$|f_n'(\psi) - f_n'(t)| \leq |f_n'(\psi) - f_m'(\psi) + f_m'(\psi) - f_m'(t) + f_m'(t)- f_n'(t)| \leq |f_n'(\psi) - f_m'(\psi)| + |f_m'(\psi) - f_m'(t)| + |f_m'(t)- f_n'(t)| < 5 \epsilon$$

Now with above chosen $$h$$, further choose $$n$$ such that for all $$n \geq M_2 \geq m$$, we have,

$$\left|\frac{F(t+h)-F(t)}{h} - \frac{f_n(t+h)-f_n(t)}{h}\right| < \epsilon$$

Hence we have, $$\left|\frac{F(t+h)-F(t)}{h} - G(t) \right| \leq$$ $$\left|\frac{F(t+h)-F(t)}{h} - \frac{f_n(t+h)-f_n(t)}{h}\right| + \left|f_n'(\psi) - f_n'(t)\right| + \left|f_n'(t) - G(t)\right| \leq 7 \epsilon$$

$$m,M_2$$ does not depend on $$t$$ by uniform continuity and uniform convergence.