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Show if is true or false: if $(f_n)$ converges uniformly to $f$, and $f_n$ is uniformly continuous for all $n$ then $f$ is uniformly continuous

I think is true. My attempt to prove it: if $(f_n)\to f$ uniformly then we can write

$$(\forall\varepsilon>0)(\exists N\in\Bbb N)(\forall x\in\mathcal D):|f_n(x)-f(x)|<\varepsilon,\quad\forall n>N\tag{1}$$

and cause all $f_n$ are uniformly continuous

$$(\forall\varepsilon>0)(\exists\delta>0)(\forall x,y\in\mathcal D):|x-y|<\delta\implies|f_n(x)-f_n(y)|<\varepsilon,\quad\forall n\in\Bbb N\tag{2}$$

and I want to prove that both conditions implies

$$(\forall\varepsilon>0)(\exists\delta>0)(\forall x,y\in\mathcal D):|x-y|<\delta\implies|f(x)-f(y)|<\varepsilon\tag{3}$$

where $\mathcal D$ is the domain of all of them (cause I have the previous knowledge that uniform convergence of continuous functions implies that the limit function is continuous).

Then from $(3)$ I can write

$$|f(x)-f(y)|=|f(x)-f_m(x)+f_m(x)-f(y)|\le |f(x)-f_m(x)|+|f_m(x)-f(y)|$$

Then I will use some $m$ that holds $(1)$ for some $\frac{\varepsilon}{3}$. And from $(2)$ I will use the $\delta$ that holds for the same $\frac{\varepsilon}{3}$. If $|f(y)-f_m(y)|<\frac{\varepsilon}{3}$ then $f(y)<f_m(y)+\frac{\varepsilon}{3}$. And then finally I can write:

$$\begin{align}|f(x)-f(y)|&\le|f(x)-f_m(x)|+|f_m(x)-f(y)|\\&<\frac{\varepsilon}{3}+|f_m(x)-f_m(y)-\frac{\varepsilon}{3}|\\&<\frac{\varepsilon}{3}+|f_m(x)-f_m(y)|+\frac{\varepsilon}{3}\\&<\frac{\varepsilon}{3}+\frac{\varepsilon}{3}+\frac{\varepsilon}{3}=\varepsilon\end{align}$$

then it proves that exists a $\delta$ such that $|f(x)-f(y)|<\varepsilon$ for some $\varepsilon$ in the required conditions. Now, can you check my proof, telling me if it is right or if it lacks something? Thank you in advance.

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    $\begingroup$ you could do $$|f(x)-f(y)|=|f(x)-f_m(x)+f_m(x)-f_m(y)+f_m(y)-f(y)|\leq |f(x)-f_m(x)|+|f_m(x)-f_m(y)|+|f_m(y)-f(y)|$$, you put an estimate inside an absolute value which is not always a good idea. $\endgroup$
    – clark
    Jun 11, 2016 at 23:02
  • $\begingroup$ Oh,ty @clark, what a dumb! Your way is easier and simple. $\endgroup$
    – Masacroso
    Jun 11, 2016 at 23:03
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    $\begingroup$ In equation $(2)$, the $\forall n\in\mathbb{N}$ should be before the $\exists\delta >0$ Because the choice of $\delta$ depends not only on $\epsilon$, but also on which function in the family is being considered (i.e. which $n\in\mathbb{N}$). $\endgroup$ Jun 11, 2016 at 23:09
  • $\begingroup$ Your thinking for this proof seems good, but I'm not satisfied with the details, in particular, how exactly are you selecting delta? Be careful to ensure that it really does work for the given $\varepsilon$. $\endgroup$ Jun 11, 2016 at 23:26
  • $\begingroup$ I made an error in my comment about equation $(2)$ the $\forall n\in\mathbb{N}$ should be after the $\exists\delta >0$ not before, and that fact is what makes uniform continuity different from 'ordinary' continuity. $\endgroup$ Jun 11, 2016 at 23:34

1 Answer 1

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Choose a $\delta$ which works in equation $(2)$ for $\epsilon_0=\frac{\epsilon}{3}>0$. Then, by $(1)$ (uniform convergence) we have an $N$ such that $|f_n(x)-f(x)|<\varepsilon$ and $|f_n(y)-f(y)|<\varepsilon$ holds $\forall m>N$. Now we apply the manipulation in clark's comment to obtain:

$|f(x)-f(y)|\leq |f(x)-f_m(x)|+|f_m(x)-f_m(y)|+|f_m(y)-f(y)|$

From here, we have

$|f(x)-f_m(x)|<\epsilon_0=\dfrac{\epsilon}{3}$ by uniform converge (at $x$)

$|f_m(x)-f_m(y)|<\epsilon_0=\dfrac{\epsilon}{3}$ by uniform continuity (of $f_m$)

$|f_m(y)-f(y)|<\epsilon_0=\dfrac{\epsilon}{3}$ by uniform converge (at $y$)

Thus we now know that

$|f(x)-f_m(x)|+|f_m(x)-f_m(y)|+|f_m(y)-f(y)|<\dfrac{\epsilon}{3}+\dfrac{\epsilon}{3}+\dfrac{\epsilon}{3}=\epsilon$

as required.

Note how carefully I selected my $\delta$, feel free to ask why I did things this way if any of what I did seems unnecessary to you.

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  • $\begingroup$ Uniform convergence is independent of $x$ or $y$. When you choose some $N$ it works for all $x,y$. You dont need use the minimum or maximum. $\endgroup$
    – Masacroso
    Jun 11, 2016 at 23:59
  • $\begingroup$ It is true that $f_n$ converges uniformly to $f$ at both $x$ and $y$, however, that does not mean that for a given $\epsilon$, that the corresponding $N$ is the same. That is why I used the max of the two $N$'s. That is: it depends on the $x\in\mathcal{D}$. $\endgroup$ Jun 12, 2016 at 0:01
  • $\begingroup$ You're right, checked definition on wikipedia: en.wikipedia.org/wiki/Uniform_convergence $\endgroup$ Jun 12, 2016 at 0:04

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