Let $f_n \in C([0,\, +\infty))$ be defined by $f_n(t) = \sin (\sqrt{t + 4n^2\pi})$ for $n\in\mathbb{N}$, $t \geq 0$.
Prove that $f_n$ converges pointwise to $f \in C([0, +\infty))$ and determine $f$.
Study the uniform convergence on bounded intervals and on $[0, +\infty)$.
Prove that the set $\{f_n : n \geq 1\}$ is equi-continuous. Is it true that this set is also compact in the space $C_b([0, +\infty))$ of bounded continuous functions, endowed with the $||\cdot||_\infty$ norm?
The first question was treated in this old post: About the convergence of $f_n:=\sin{\sqrt{t+4n^2\pi^2}}$. Can someone give me some hints for the last two points? Unfortunately I do not know what to do.
Thanks in advance.
$|f_n(t)-f_n(s)| \leq |\sqrt {t+4n^{2}\pi } -\sqrt {s+4n^{2}\pi }|$ by MVT.But $|\sqrt {t+4n^{2}\pi } -\sqrt {s+4n^{2}\pi }|=\frac {|t-s|} {\sqrt {t+4n^{2}\pi } +\sqrt {s+4n^{2}\pi } } \leq \frac {|t-s|} {\sqrt {t} +\sqrt {s}} <\frac {|t-s|} {2}$ if $t,s \geq 1$. The fact that $\f_n\}$ is uniformly convergent and (hence equicontinuous ) on bounded sets (in particular $[0,1]$) was already mentioned in the previous post, so we can now conclude that $\{f_n\}$ is equicontinuous. To show that $f_n$ does not tend to 0 uniformly on $[0,\infty )$ take $t_n=((2n+1)^{2})\pi /2)^{2}-4n^{2} \pi$ and note that $|f_n(t_n)|=1$. This also proves that $\{f_n\}$ is not compact in $C_b([0,\infty ))$.