I have to prove this:
Show that the space $\mathcal{C}^{(1)}([a,b])$ with respect to the norm $||\cdot||^{(1)}$ is a Banach Space
where $||\cdot||^{(1)} = |f(a)|+||f'||_{\infty}$. In the proof I use this theorem (D):
Let $I$ be a real limited interval, $c\in I$. Let $f_n:I\rightarrow\mathbb{R}$. Suppose for each $n$ that $f_n\in\mathcal{C}^{(1)}(I)$, $\lim_{n\rightarrow\infty}f_n$ exists for some $x_0\in I$ and $\{f'_n\}_{n\in\mathbb{N}}$ converges uniformly on $I$, then $\{f_n\}_{n\in\mathbb{N}}$ converges uniformly on $I$ and $$\frac{d}{dx}\left(\lim_{n\rightarrow\infty}f_n(x)\right) = \lim_{n\rightarrow\infty}\left(\frac{d}{dx}f_n(x)\right)$$
My proof:
Let $\{f_n\}_{n\in\mathbb{N}}\subset\mathcal{C}^{(1)}([a,b])$ be a Cauchy sequence in $\mathcal{C}^{(1)}([a,b])$, then $$ \forall\epsilon > 0\;\exists n_0=n_0(\epsilon) : |f_n(a)-f_m(a)|+||f'_n(x)-f'_m(x)||_{\infty}<\epsilon \quad\forall n,m>n_0 $$ in particular $$ |f_n(a)-f_m(a)|<\epsilon \text{ and } ||f'_n(x)-f'_m(x)||_{\infty}<\epsilon $$ so $\{f_n\}_{n\in\mathbb{N}}$ is a Cauchy sequence in $(\mathbb{R},|\cdot|)$ and $\{f'_n\}_{n\in\mathbb{N}}$ is a Cauchy sequence in $\mathcal{C}^{(0)}([a,b]),||\cdot||_{\infty})$.
Since $(\mathbb{R},|\cdot|)$ and $\mathcal{C}^{(0)}([a,b]),||\cdot||_{\infty})$ are complete spaces, \begin{align*} f_n(a)&\xrightarrow{n\rightarrow+\infty}f(a)\in\mathbb{R}\\ f'_n&\xrightarrow{n\rightarrow+\infty}g\in\mathcal{C}^{(0)}([a,b]). \end{align*}
By theorem (D) $\{f_n\}_{n\in\mathbb{N}}$ converges uniformly to $f:[a,b]\rightarrow\mathbb{R}$, $f\in\mathcal{C}^{(1)}([a,b])$ and $f'(x)=g(x)\;\forall x\in[a,b]$.
I have to prove that $f$ is the limit with respect to the norm $||\cdot||^{(1)}$, i.e. that $$ ||f_n(x)-f(x)||^{(1)}= |f_n(a)-f(a)|+||f'_n(x)-f'(x)||_{\infty}\rightarrow 0 \text{ for } n\rightarrow+\infty $$
I have already shown that $f_n(a)\xrightarrow{n\rightarrow+\infty}f(a)\in\mathbb{R}$. Regarding $$ \sup_{x\in[a,b]}|f'_n(x)-f'(x)| = \sup_{x\in[a,b]}|f'_n(x)-g(x)|\xrightarrow{n\rightarrow+\infty}0, $$ therefore $||f_n(x)-f(x)||^{(1)}\xrightarrow{n\rightarrow+\infty}0$.
Is this proof correct? Thanks!