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Let $f_n:A\subseteq\mathbb{R}\to\mathbb{R}$

$f_n$ is uniformly Cauchy $\implies$ $\exists f:A\to\mathbb{R}$ : $f_n\xrightarrow{u}f$ in A

proof.

$\forall\varepsilon>0$ $\exists\nu$ : $\forall n,m>\nu$ $\sup_{x\in A}|f_n(x)-f_m(x)|<\varepsilon$

I don't know why we start like that, I know by definition that a uniformly Cauchy is the second step of this proof

$\implies \forall\varepsilon>0$ $\exists\nu\in\mathbb{N}$ : $\forall n,m>\nu$ $\forall x\in A$ $|f_n(x)-f_m(x)|<\varepsilon$

Where is sup now?

So if $m\to+\infty$

$\implies \forall\varepsilon>0$ $\exists\nu\in\mathbb{N}$ : $\forall n>\nu$ $\forall x\in A$ $|f_n(x)-f(x)|\leq\varepsilon$

Ok it's clear why m disappears but I don't understand why $<\varepsilon$ becomes $\leq\varepsilon$

$\implies \forall\varepsilon>0$ $\exists\nu\in\mathbb{N}$ : $\forall n>\nu$ $\sup_{x\in A}|f_n(x)-f(x)|\leq\varepsilon$

This is not clear, where does sup come from?

$\implies\lim_{n\to\infty}sup_{x\in A}|f_n(x)-f(x)|=0$

This is directly from the definition of limit, and this means that

$\implies f_n\xrightarrow{u}f$ in $A$

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  • $\begingroup$ Please consider accepting my answer if it has helped :) $\endgroup$ – user667 Nov 19 '18 at 14:11
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Recall the definition of supremum (it is the least upper bound). Thus, if we can choose $v$ such that $\sup_{x\in A}|f_n(x)-f_m(x)|<\epsilon$ whenever $n,m>v$ then we are guaranteed that $\forall x\in A, |f_n(x)-f_m(x)|<\epsilon$ whenever $n,m>v$. Going the other way around, strict inequality could become non strict since the supremum is an upper bound itself. Thus if $|f_n(x)-f(x)|<\epsilon$ for all $x\in A$, we have that $\epsilon$ is an upper bound over all $x$. In particular, it could be the least upper bound (supremum). Thus we have to write $\sup_{x \in A}|f_n(x)-f|\leq\epsilon$ since equality could hold.

I agree that this proof is somewhat unclear. To be completely rigorous, we use the triangle inequality. Choose arbitrary $x \in A$. We have, $$|f_n(x)-f(x)|\leq|f_n(x)-f_m(x)|+|f_m(x)-f(x)|$$where $f(x)$ is the pointwise limit of the sequence $\{f_n(x)\}$. We know that there is a pointwise limit because $\mathbb{R}$ is complete. By definition of pointwise convergence, there exists some $N_1$ such that if $m>N_1$ we have $|f_m(x)-f(x)|<\frac{\epsilon}{2}$. By assumption, we can choose $N$ such that if $n,m>N$ we have $|f_n(x)-f_m(x)|<\frac{\epsilon}{2}$. Now set $m>\max\{N_1,N\}$ (this step is a rigorous way of saying $m \rightarrow \infty$). Putting it all together $$|f_n(x)-f(x)|\leq|f_n(x)-f_m(x)|+|f_m(x)-f(x)|<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon$$Because $x$ was an arbitrary point of $A$, we have found an $N$ (independent of $x\in A$) such that, $$\forall x\in A, n>N, |f_n(x)-f(x)|<\epsilon$$

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