Suppose $f_n: E \longrightarrow \mathbb{R}$ is uniformly continuous on $E (E \subset \mathbb{R})$ for every $n \in \mathbb{N}$ and $f_n \rightarrow f$ uniformly on $E$.
$(a)$ Prove that $f$ is uniformly continous on $E.$
$(b)$ Does the result in (a) remain true if $f_n \rightarrow f$ point-wise only? Prove or give a counterexample.
Now let $f_n:[0, \infty) \longrightarrow \mathbb{R}$ by $f_n(x)=\frac{sin(nx)}{1+nx}.$
$(c)$ Show that $f_n$ converges point-wise on $[0, \infty)$ and find the point-wise limit $f$.
$(d)$ Show that $f_n \rightarrow f$ uniformly on $[a, \infty)$ for every a>0, but that $f_n$ does not converge uniformly to $f$ on $[0,\infty).$
My answer is as follows:
If $f_n→f$ uniformly then we can write
$(∀ε>0)(∃N∈\mathbb{N})(∀x∈D):$
$|f_n(x)−f(x)|<ε,∀n>N$ .......$(1)$
and since $f_n$ is uniformly continuous for all n,
$(∀ε>0)(∃δ>0)(∀x,y∈D)$:
$|x−y|<δ⟹|f_n(x)−f_n(y)|<ε,∀n∈\mathbb{N}$ ...........$(2)$
and both conditions imply
$(∀ε>0)(∃δ>0)(∀x,y∈D):|x−y|<δ⟹|f(x)−f(y)|<ε$.........$(3)$
where $D$ is the domain of all of them (uniform convergence of continuous functions implies that the limit function is continuous).
Then from $(3)$
$|f(x)−f(y)|=|f(x)−f_m(x)+f_m(x)−f(y)|≤|f(x)−f_m(x)|+|f_m(x)−f(y)|$
Then I will use some $m$ that holds $(1)$ for some $ε_1$. And from $(2)$ I will use the $δ$ that holds for the same $ε_1$.
If $|f(y)−f_m(y)|<ε_1$ then $f(y)<f_m(y)+ε_1$
And then finally I can write:
$|f(x)−f(y)|≤|f(x)−f_m(x)|+|f_m(x)−f(y)|$
$<ε_1+|fm(x)−f_m(y)−ε_1|$
$<ε_1+|fm(x)−fm(y)|+ε_1$
$<ε_1+ε_1+ε_1=ε$
Hence, $f$ is uniformly continuous.
Please check whether part (a) is correct and please give a counterexample for part (b) since I think the result above need not remain true if $f_n$ is pointwise convergent
Answer for b): Take $E=[0,1], f_n(x)=x^{n}$ and $f(x)$ for $x<1$, $f(1)=1$.
Answer for c): The point-wise limit is $0$.
Answer for d): For $x \geq a$ we have $0\leq |\frac {sin (nx)} {1+nx}| \leq \frac 1 {1+na}$. Hence we get uniform convergence on $[a,\infty)$. Also $\sup |\frac {sin (nx)} {1+nx}| \geq \frac {\sin 1 } {1+1}$ so the convergence is not uniform.