How do I prove the multivariable limit $\max\{ \frac{1}{\sqrt n} -\frac{\sqrt m}{ n},\frac{\sqrt n}{m} -\frac{1}{ \sqrt m}\}\to 0$?

Question

While proving that $C([0,1])$ whith the norm $\|f\|=\sup_{t\in [0,1]}(t|f(t)|)$ is not complete the tutor used the following sequence:

$x_n(t)=\cases{\frac{1}{\sqrt t}, t\in [\frac{1}{n},1]\\ \sqrt n , t \in [0,\frac{1}{ n}]} $ And he proved it is Cauchy:

$\forall n>m$:

$\|x_n-x_m\|=\sup_{t\in [0,1]}(t|x_n(t)-x_m(t)|)$

$=\max\{ \sup_{t\in [0,1/n]}(t(\sqrt n - \sqrt m)),\sup_{t\in [1/n,1/m]}(t(\frac{1}{\sqrt t }- \sqrt m)),0\}$

$\le \max\{ \frac{1}{n}(\sqrt n - \sqrt m),\frac{1}{m}(\sqrt n- \sqrt m)\} = \max\{ \frac{1}{\sqrt n} -\frac{\sqrt m}{ n},\frac{\sqrt n}{m} -\frac{1}{ \sqrt m}\}\to 0$

How do I prove that limit is zero step by step? I don't know how to deal with the max and how to take the multivariable limit.

My questions are:

1 Since the max of continuous functions is a continuous function , is it true that

$\lim_{(n,m)\to(\infty,\infty)}\max(f(n,m),g(n,m))=\max(\lim_{(n,m)\to(\infty,\infty)} f(n,m), \lim_{(n,m)\to(\infty,\infty)} g(n,m))$ ?

2 How do I find the multivariable limit ? I feel like taking first the limit as$ n \to \infty$ (holding m constant) and the limit as $ m \to \infty$ in the first element of the max and the other way around for the second element wouldn't be correct since it is well-known that the multivariable limit is different than the iterated limit.

Answer 1

By construction of answer $\frac{1}{n} < \frac{1}{m}$, i.e., $n>m$, so $\sqrt{n}-\sqrt{m} \leq 2 \sqrt{n}$. Thus we get from your second last step $$=max\{2\sqrt{n}/n, 2\sqrt{n}/m \}.$$ By Archimedian property, there exists $k \in \mathbb{N}$ such that $km>n ~ => ~ \frac{1}{m}<\frac{k}{n}$ thus above is less or equal to $$ \leq max\{\frac{2}{\sqrt{n}},\frac{2k}{\sqrt{n}} \}$$ which goes to zero

Answer 2

Replacing $\sup_{t\in [1/n,1/m]}(t(\frac{1}{\sqrt t }- \sqrt m))$ by the weaker upper bound $\frac{1}{m}(\sqrt n- \sqrt m)$ kills the proof because, for example, with $n=2m^2$ the expression grows beyond every bound.

A more sensible approach observes that $\max_{t\in [0,1/m]}(t(\frac{1}{\sqrt t }- \sqrt m))=\frac1{2m}(1-\frac1{2 \sqrt m})$ at $t=\frac1{4m^2}$.