$f_n\rightarrow f$ uniformly in an interval Let $f_n:\mathbb{R}\rightarrow \mathbb{R}$ a sequence of functions and $f:\mathbb{R}\rightarrow \mathbb{R}$.
Which of the following statements are correct?
(a) If $f_n\rightarrow f$ uniformly in each interval of the form $(a, +\infty)$, then $f_n\rightarrow f$ uniformly in the whole $\mathbb{R}$.
(b) If for each $a>0$ it holds that $f_n\rightarrow f$ uniformly in $(a, +\infty)$, then $f_n\rightarrow f$ uniformly in $[0,+\infty)$.
(c) If $f_n\rightarrow f$ uniformly in each closed and bounded interval, then $f_n\rightarrow f$ uniformly in the whole $\mathbb{R}$.
(d) If $f_n\rightarrow f$ uniformly in the whole $\mathbb{R}$ and $f_n$ are continuous, then $f$ can be continuous.
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At uniform convergence the properties of $f_n$ hold also for $f$. So since $f_n$ are continuous, then $f$ must be also continuous. So (d) doesn't hold because $f$ is continuous, right?
At (a) and (c) we have to check if the union of the respective intervals is equal to $\mathbb{R}$, right?
At (b) we have to check if $\bigcup_a (a,  +\infty)=[0,\infty)$, right?
 A: It is not difficult to find counterexamples that show that (a), (b), and (c) are false.
For (c) consider the sequence $f_n(x) = \frac{x}{n}$ which converges pointwise to $f(x) = 0$.  On any closed and bounded interval $[a,b]$  we have
$$\left|f_n(x) - f(x) \right| = \left| \frac{x}{n}\right| \leqslant \frac{\max(a,b)}{n}$$
For any $\epsilon > 0$ take $N(\epsilon)  \in \mathbb{N}$ such that $N(\epsilon) > \max(a,b)/\epsilon$. For all $n \geqslant N(\epsilon)$ and for all $x \in [a,b]$ we have
$$|f_n(x) - f(x)|\leqslant \frac{\max(a,b)}{n} \leqslant \frac{\max(a,b)}{N(\epsilon)} \leqslant \epsilon,$$
and, thus, $f_n \to f$ uniformly on $[a,b]$.  However, the convergence is not  uniform on $\mathbb{R}$, since
$$\sup_{x \in \mathbb{R}}\left| f_n(x) - f(0) \right| = \sup_{x \in \mathbb{R}}\left| \frac{x}{n} \right| = +\infty \underset{n \to \infty}{\not\to}0$$
A counterexample for (b) is $f_n(x) = \begin{cases}(1-x)^n,&0 \leqslant x \leqslant 1\\ 0, & x > 1 \end{cases}$. Try yourself to understand this and also find a counterexample for (a).
