Consider the normed vector space $(C^1[a,b],||\cdot||_{\infty})$, we have to prove that it is not Banach.
From theory a space is Banach if it is complete, ie if every Cauchy sequence in the space converges to an element of the same space.
So what we have to do is:
- find a Cauchy sequence in the space
- show that the squence does not converge to an element of the space
Consider $[a,b]=[-1,1]$ and the sequence defined by $f_k(x)=\sqrt{x^2+1/k},\ k\in\mathbb{N}$. Since the argument of the root is always non-negative, the sequence is always differentiable and both $f_k$ and $f'_k$ are continuous. So $\{f_k\}\in C^1[-1,1]$.
Intuitively I see that $\lim_{k\rightarrow+\infty} f_k(x)=\sqrt{x^2}=|x|\notin C^1[-1,1]$, but I first have to show that the sequence is Cauchy.
From theory a sequence $\{x_n\}$ is Cauchy iff $\forall\epsilon>0\ \exists N\in\mathbb{N}:\forall m,n>N$ we have $||x_n-x_m||<\epsilon$.
In our case we have $||\sqrt{x^2+1/n}-\sqrt{x^2+1/m}||\le||\sqrt{x^2+1/n}||+||\sqrt{x^2+1/m}||$ (*).
The sum (*) should be less than $\epsilon$, so each term should be less than $\epsilon/2$.
Moreover, since we are working with the uniform norm and $||f||_{\infty}=max\{|f(x)|:x\in [a,b]\}$, I think that $||f_k(x)||_{\infty}=\sqrt{2}$ because $max\{x^2:x\in [-1,1]\}=1$ and $max\{1/k:k\in\mathbb{N}\}=1$. So the sum (*) should be less or equal to $2\sqrt{2}$.
How can I use these information to prove that $\{f_k\}$ is a Cauchy sequence?
Applying the triangle inequality like this is too coarse. We can proceed like this:
$$\|\sqrt{x^2+{1/n}}-\sqrt{x^2+{1/m}}\|=\left\|\frac{1/n-1/m}{\sqrt{x^2+{1/n}}+\sqrt{x^2+{1/m}}}\right\|= \\ = |1/n-1/m|\cdot\left\|\frac{1}{\sqrt{x^2+{1/n}}+\sqrt{x^2+{1/m}}}\right\|$$
Now, $|1/n-1/m|=\frac{|n-m|}{nm}$, and
$$\left\|\frac{1}{\sqrt{x^2+{1/n}}+\sqrt{x^2+{1/m}}}\right\| = \frac{1}{1/\sqrt n+1/\sqrt m} = \frac{\sqrt{nm}}{\sqrt n + \sqrt m}$$
(the denominator is easily seen to be at minimum for $x=0$).
So, assuming $N<n<m$, the product is
$$\frac{|n-m|}{\sqrt{nm}(\sqrt n + \sqrt m)} \leq \frac{m}{\sqrt{nm}(\sqrt n + \sqrt m)}=\frac{1}{\sqrt{n}(\sqrt{n/m}+1)}\leq \\ \leq \frac{1}{\sqrt{N}}$$
(since $\sqrt{n/m}+1 \geq 1$).
Solving $\frac{1}{\sqrt N}<\varepsilon$ gives $N > \frac{1}{\varepsilon^2}$. Thus, for
$$n,m > \frac{1}{\varepsilon^2}$$
we get
$$\|\sqrt{x^2+{1/n}}-\sqrt{x^2+{1/m}}\| \leq \varepsilon$$