is this sequence Cauchy in the Banach space of continuous functions with infinity norm? In the Banach space $(C[0,1], ||\cdot||_{\infty})$ where $||\cdot||_{\infty} = max_{[0,1]}|f(x)|$ let the sequence of functions $\{f_n(x)\}$ be given by $f_n(x) = \frac{n\sqrt{x}}{1+nx}$.
State whether the sequence is Cauchy in this space.
Here's my work: 
If the sequence is Cauchy, then given any $\epsilon > 0$, there exists $N \in \mathbb{N}$ such that for all $m, n \geq N$ we have $||f_n(x) - f_m(x)|| \leq \epsilon$.  
Since $f_n(x) \rightarrow \frac{1}{\sqrt{x}} = f(x)$ on $[0,1]$ we can find $N$ such that for $m, n \geq N$, $||f_n(x) - f(x)|| \leq \frac{\epsilon}{2}$ and $||f_m(x) - f(x)|| \leq \frac{\epsilon}{2}$.  Thus we will have: 
$||f_n(x) - f_m(x)|| = ||f_n(x) - f(x) + f(x) - f_m(x)|| \leq ||f_n(x) - f(x)|| + ||f_m(x) - f(x)|| \leq \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon$. 
So the sequence is Cauchy assuming convergence, but I'm not sure if I can assume this.   

Suppose $(f_n)$ is Cauchy in $C([0,1]).$ Then there exists $N\in \mathbb N$ such that $\|f_n-f_N\| < 1$ for $n>N.$ This $N$ is now fixed. It follows that
$$\tag 1 \|f_n\| \le \|f_n-f_N\| + \|f_N\| < 1 + \|f_N\|$$
for $n>N.$ But notice $f_n(1/n)=\sqrt n/2.$ Thus $\|f_n\|\to \infty,$ violating $(1),$ contradiction.
