Could you please verify whether my attempt is fine or contains logical gaps/errors? Any suggestion is greatly appreciated!
Prove that the $C^1([a,b],\mathbb R)$ together with the $C^1$-norm $\| \cdot \|$ is a Banach space.
My attempt:
Lemma: Let $C^1([a,b],\mathbb R)$ be endowed with the supremum norm $\| \cdot \|_\infty$. Let $X$ be an open (or convex) perfect subset of $\mathbb R$ and $(f_n)$ be a sequence of functions in $C^1([a,b],\mathbb R)$. Suppose that there are $f,g \in \mathbb R^{[a,b]}$ such that
i) $(f_n)$ converges pointwise to $f$, and
ii) $(f'_n)$ converges locally uniformly to $g$.
Then $f \in C^1([a,b],\mathbb R) $, $f'=g$, and $(f_n)$ converges locally uniformly to $f$.
Assume that $(f_n)$ is a Cauchy sequence in $\langle C^1([a,b],\mathbb R),\|\cdot\| \rangle$. Because $\|f_n\| = \| f_n \|_\infty + \| f'_n \|_\infty$, $(f_n)$ and $(f'_n)$ are Cauchy sequences in $\langle C([a,b],\mathbb R),\|\cdot\|_\infty \rangle$, which is a Banach space. Then there are $f,g$ in $C([a,b],\mathbb R)$ such that $f_n \to f$ and $f'_n \to g$ w.r.t $\|\cdot\|_\infty$. Clearly, $(f_n),(f'_n)$ converges uniformly to $f,g$ w.r.t $\|\cdot\|_\infty$ respectively. By our Lemma, $f \in C^1([a,b],\mathbb R)$ and $f'=g$.
Next we prove that $(f_n)$ converges to $f$ w.r.t $\|\cdot\|$. Given $\epsilon > 0$. Because $(f_n)$ converges to $f$ w.r.t $\|\cdot\|_\infty$, there is $N_1$ such that $\|f_n-f\|_\infty < \epsilon/2$ for all $n > N_1$. Similarly, there is $N_2$ such that $\|f'_n-f'\|_\infty < \epsilon/2$ for all $n > N_1$. Take $N =\max\{N_1,N_2\}$. Then $\|f_n-f\| =\| f_n -f\|_\infty + \| f'_n -f'\|_\infty < \epsilon/2 + \epsilon/2 =\epsilon$ for all $n > N$. This completes the proof.