I want to determine whether $C^1[0,1]$ is a Banach space with respect to the norm $||f|| = |f(0)| + \sup_{t \in [0,1]} |f'(t)|$.
I have proven that this is indeed a norm. Then I took a Cauchy sequence $f_n$ in $(C^1[0,1], ||\cdot||)$ so that given $\varepsilon > 0$ there exists an $N$ such that $m,n>N \Rightarrow ||f_m-f_n||<\varepsilon$, i.e. $|f_m(0)-f_n(0)| + \sup_{t \in [0,1]} |f_m'(t)-f_n'(t)| < \varepsilon$, so $|f_m(0)-f_n(0)| < \varepsilon$ and $\sup_{t \in [0,1]} |f_m'(t)-f_n'(t)| < \varepsilon$.
Hence $(f_n(0))_{n \in \mathbb{N}}$ is a Cauchy sequence of real numbers, so we can define some function $f$ on $[0,1]$ with $f(0) = \lim_{n \rightarrow \infty} f_n(0)$.
Moreover, $(f_n'(t))_{n \in \mathbb{N}}$ is a Cauchy sequence in the Banach space $(C[0,1], ||\cdot||_\infty)$ and so converges uniformly to some $g$.
Then I considered trying to define $f$ on $(0,1]$ such that $g=f'$, but then I'm not sure if this is even possible. Any tips would be great, or if it isn't a Banach space then is there a counterexample?
Best Answer
You're very close. Assume $g_n\to g$ uniformly. Then,
$$ \left|\int_0^x g_n(t)\textrm{d}t-\int_0^xg (t) \textrm{d}t\right|\leq \int_0^x |g_n(t)-g(t)| \textrm{d}t\leq ||g_n-g ||_{\infty}, $$ implying that $\int_0^x g_n(t)\textrm{d}t\to\int_0^x g(t)\textrm{d}t$ uniformly. Hence, in your case, $$ f_n(x)-f_n(0)= \int_0^x f_n'(t)\textrm{d}t\to \int_0^x g(t)\textrm{d}t$$ uniformly. Using that you've shown that $f_n(0)$ converges, you now get that $f_n$ converges uniformly to $\int_0^x g(t)\textrm{d}t+\lim_{n\to\infty} f_n(0)$.