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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.

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  • $\begingroup$ If it converges uniformly, then the limit is continuous on $[0,1]$. $\endgroup$ Commented May 11, 2017 at 15:32
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    $\begingroup$ You have pointwise convergence to $f(x)=1/\sqrt{x}$ on $[0,1]$, you're sure? Your $f$ is not even defined at $0$. $\endgroup$
    – Clement C.
    Commented May 11, 2017 at 15:34

2 Answers 2

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We have $f_n$ converges to $f$ pointwise where $f:[0,1]\rightarrow \mathbb{R}$ defined by $f(x)=\frac{1}{\sqrt{x}}$ if $\neq0$ and $f(0)=0$.So $f$ $\notin$ $C[0,1]$ which is a Banach space. If the sequence was Cauchy we would have $f_n$ converges to $g$ where $g\in C[0,1]$. Hence a contradiction.

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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.

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