If $(f_n)\to f$ uniformly and $f_n$ is uniformly continuous for all $n$ then $f$ is uniformly continuous

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.

Solution 1:

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.