Is $(C^1[0,1],\|\cdot\|)$ with $\|f\|:=|f(0)|+\sup_{0\le{t\le{1}}}{|f'(t)|}$ a Banach space? Let $(C^1[0,1],\|\space{}.\|)$ be a normed space where $C^1[0,1]$ is the set of functions with continuous derivatives and let $\|\space{}.\|$ be the norm on this set defined by: 
$$\|f\|:=|f(0)|+\sup_{0\le{t\le{1}}}{|f'(t)|}.$$
Is this space Banach?
My attempt:
I don't think it is. Here is my counter example:
Let $(f_n)_{n=1}^{\infty}$ be a sequence of functions defined by $f_n(t)=\sqrt{(t-\frac{1}{2})^{2}+\frac{1}{n}}$.
This clearly belongs to $C^1[0,1]$ but its limit does not, namely $f(t)=|t-\frac{1}{2}|$. My issue however is this hasn't shown the sequence converges to $f$ with respect to the norm $\|\space{}.\|$. I don't know how to use this example, since the norm doesn't make sense with $f$, since f is not differentiable on t = 1/2. But does this counter example work?
 A: Hint : it's a Banach Space.
Why ? Because $(C^1 [0,1], \| \cdot \|_{C_1} )$ where $\|f\|_{C_1} = \|f\|_{\infty} + \|f^{'} \|_{\infty}$ is a Banach Space (more classical).
And norms of both spaces are equivalent :
It's obvious that :
$$\|f\| \leq \|f\|_{C_1} $$
But since : $|f(x)| = |\int_0^{x}f'(t) dt + f(0)| \leq \|f^{'}\|_{\infty} + |f(0)|$ we also have :
 $$\|f\|_{C_1} \leq 2\|f\| $$
A: The space is complete. Your sequence is not a  Cauchy sequence.
Let $(f_n)$ be a Cauchy sequence. Then $\lim f_n(0)$ exists and $f_n'$ converges uniformly to some continuous function $g$. Now $f_n(x)=f_n(0)+\int_0^{x} f_n'(t)dt$. From this we see that $(f_n)$ is uniformly Cauchy and hence $f_n $ tends to a continuous function $f$ uniformly. Uniform convergence of $f_n$ to $f$ and $f_n'$ to $g$ implies that $f$ is differentiable and $f'=g$. It should now be easy to see that $f_n \to f$ in the given norm.  
A: Consider the Cauchy sequence $\{f_n\}$, i.e. $\forall\varepsilon>0$ $\exists N\in\mathbb{N}:$ $\forall n,m>N$ $\|f_n-f_m\|<\varepsilon$. Since $|f_n(0)-f_m(0)|+\sup\limits_{t\in[0,1]}|f_n'(t)-f_m'(t)|<\varepsilon$, then $|f_n(0)-f_m(0)|<\varepsilon$ and $\forall t\in[0,1]$ $|f_n'(t)-f_m'(t)|<\varepsilon$. Thus, $\{f_n(0)\}$ is Cauchy and $\{f_n'(t)\}$ is uniformly Cauchy. Therefore $\exists\lim\limits_{n\to\infty}f_n(0)$ and $\{f_n'(t)\}$ is uniformly convergent. By the well-known theorem, this means that $\{f_n(t)\}$ converges uniformly to $f(t)$ and $\{f_n'(t)\}$ converges uniformly to $\{f'(t)\}$. By definition of uniformly convergence we have $\forall\varepsilon>0$ $\exists N\in\mathbb{N}:$ $\forall n>N$ $\forall t\in[0,1]$ $|f_n(t)-f(t)|<\varepsilon$ and $|f_n'(t)-f'(t)|<\varepsilon$. Thus, $\|f_n-f\|\leq2\varepsilon$. So $\{f_n\}$ is converges in $C^1[0,1]$ and $C^1[0,1]$ is Banach.
