Prove $C([0,1]),||.||)$ is not complete where $||x||= \text{sup}_{t \in [0,1]}|tx(t)|$ is a norm.
The solution given in my lectures:
We take a Cauchy sequence defined as follows:
$$ f_n(t)= \begin{cases} \frac{1}{\sqrt t} && ,\frac{1}{n} \leq t \leq 1 \\ \sqrt n &&, 0 \leq t \leq \frac{1}{n} \end{cases} $$
To prove it is Cauchy we take n>m
$$||f_n(t)-f_m(t)||=\text{sup}_{t \in [0,1]}|t(f_n(t)-f_m(t))|=\text{max}\{\text{sup}_{t \in [0,\frac{1}{n}]}|t(\sqrt n -\sqrt m)| ,\text{sup}_{t \in [\frac{1}{n},\frac{1}{m}]}|t(\frac{1}{\sqrt t}-\sqrt m)|,0\} $$
Since in $[\frac{1}{n},\frac{1}{m}], \frac{1}{\sqrt t}-\sqrt m \leq \sqrt n-\sqrt m$
$$||f_n(t)-f_m(t)||= \text{max}\{\frac{1}{n}(\sqrt n -\sqrt m) ,\frac{1}{m}(\sqrt n -\sqrt m) \} ….(\alpha)$$
$$||f_n(t)-f_m(t)|| \to 0 , \text{ when } n,m\to \infty….(\beta)$$
If $\exists f \in C([0,1]) \text{ such that } ||f_n-f|| \to 0$, then
$ \text{sup}_{t \in [0,1]}|t(f_n-f)|\to 0 $,
$\forall t_0 \in [0,1], |f_n(t_0)-f(t_0)| \to 0$,
$\forall t_0 \in (0,1] ,f(t_0)=\lim_{n\to \infty} f_n(t_0)=\frac{1}{\sqrt t_0} ….(\gamma)$
, but no continuous function in $[0,1]$ exists , that coincides with $\frac{1}{\sqrt t}$ in $(0,1] $
Can someone clarify these points:
-In $(\alpha)$ how is that they found that second sup? It doesn't seem right that $ \text{sup}_{t \in [\frac{1}{n},\frac{1}{m}]}|t(\frac{1}{\sqrt t}-\sqrt m)||= \frac{1}{m}(\sqrt n -\sqrt m)$, that would be like requiring $t$ to be $1/m$ and $1/n$ at the same time
-In $(\beta)$ why are they taking a double limit when $m, n \to \infty$ is this legal?, I 've only seen that in Cauchy sequences proofs we have to find the expresion that is less that an $\varepsilon$ for some $n$ and $m$,whya aren't they doing that?
-In $(\gamma)$ they are just rewriting the previous line using $lim$, so why are they leaving out the $0$?
Best Answer
($\alpha$) It doesn't matter. All you need is an inequality, and $\frac{1}{m}(\sqrt n -\sqrt m)$ is an upper bound.
($\beta$) It's the same. Being Cauchy is $\lim_{m,n\to\infty}\|f_n-f_m\|=0$, which is the same as the common phrasing of "Cauchy" that you use.
($\gamma)$ Because it doesn't fit the expression $1/\sqrt{t_0}$, and it doesn't hurt the argument.