Prove that $C^1([a,b];X)$ is a Banach space

Question

Let $X$ be a Banach space with norm $\|\cdot\|$. I want to show that $C^1([a,b];X)$ (the space of continuously differentiable functions on $[a,b]$ with values in $X$) provided with the norm $$\|u\|_{C^1([a,b];X)}:=\max_{t\in[a,b]}(\|u(t)\|+\|u'(t)\|)$$ is a Banach space.

I think I need to show that any given Cauchy sequence converges in $C^1([a,b];X)$. But how do I begin? I havent got too comfortable with Banach spaces yet.

Answer 1

Hint. Consider a Cauchy sequence $f_n$ in $C^1([a,b];X)$. Show that both $f_n$ and their derivatives $f_n'$ converge in $C^0([a,b]; X)$ with the uniform norm. Suppose $f_n\to f$ and $f_n'\to g$ in $C^0([a,b];X)$. Now prove that $f'=g$, which is a well-known conclusion in real-analysis, hence $f_n\to f$ in $C^1([a,b];X)$.

Answer 2

So let $(f_n)$ be an arbitrary Cauchy sequence in $C^1([a,b]:X)$. Since $X$ is complete $(f_n)$ converges in $C^0([a,b]);X)$ with the uniform norm to $f$. Similarly, $(f'_n)$ converges in $C^0([a,b]);X)$ with the uniform to a function $g$. We observe that $$\int g =\int \lim f'_n=\lim \int f'_n=\lim f_n =f$$

Hence, $g=f'$ and we can conclude that $f'\in C^0([a,b];X)$.

Is my proof right? If so, where do I make us of the norm that I mentioned in my question?