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?
Best Answer
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 : $$\|f\|_{C_1} \leq 2\|f\| $$