Example to prove that $ C^1[0,1] $ is not a Banach space for the uniform norm? The space $ C^1[0,1] $- the space of all continuously differentiable functions on $ [0,1]$ is not a Banach space with respect to the sup norm,$ \|.\|_{\infty} $ since the uniform limit of a continuously differentiable function need not be differentiable.
How can I illustate this statement using a counter example? Can I use $ f_{n}=\frac1 n \sin nx $ as a counter example?
Also, is $ C^1[0,1] $ is same as the space given by $ X=\{f\in C^1[0,1]:f(0)=0 \}$. Can I use the same example to show that this is not a Banach space?
More help is appreciated!
Thanks!
 A: The example you gave converges uniformly to the zero function, which is continuously differentiable. 
Every continuous function on $[0,1]$ is a uniform limit of polynomial functions (by the Weierstrass approximation theorem), and polynomial functions are continuously differentiable.
For an explicit example, you could also consider the sequence $f_n(x)=\left|x-\frac12\right|^{(n+1)/n}$.

I might not fully understand the last question.  Those are not the same, because elements of $C^1[0,1]$ do not generally have to vanish at $0$.  But you can use similar examples.  E.g., you could still think about the Weierstrass approximation theorem for dramatic counterexamples, or you could modify the example above by taking, say, $g_n(x)=f_n(x)-f_n(0)$.
A: The ${\cal C}^1[0,1]$ functions are a dense subset of $\cal{C}[0,1]$ in the sup norm.  However the inclusion is proper so the $\cal{C}^1$ functions are not a complete subspace of ${\cal C}[0,1]$.  
The continuous functions on $[0,1]$ that vanish at 0 are a Banach space, they are the kernel of the continuous map $f\mapsto f(0)$.   They form a closed subspace.  But those in ${\cal C}^1[0,1]$ are not.  
