Uniform continuity of a sequence of function

Question

Discuss the uniform convergence of the sequence of function $f_n(x)=n sin (\sqrt{4π^2n^2+x^2})$, on $\mathbb{R}$ and on $[0,a].$

Answer 1

let's consider $\mathbb{R}$ first. As $n$ increases, we know that $f_n(x)$ becomes unbounded, we don't even need to consider what is going on inside the argument for sine. This means the $f_n$ don't even converge pointwise, since they diverge.

Now let's consider $[0,a]$. Here is some work, but not an entire discussion. Note that for $f_n(0)$, we have $f_n(0)=nsin (\sqrt{4\pi^2 n^2})$. However, $\forall n$ we have that the $f_n$ also do not converge if $a$ is "large enough". Let $u=\sqrt{4\pi^2}$ and if $nsin(nu)$ is monotonically increasing on $[0,a]$, then there might be pointwise convergence, but it will almost certainly won't be uniform. In addition, for $[0,1]$, we have that $f_n(\frac{\pi}{4n})=nsin(nu)$ and the limit doesn't exist if we have $n=2k$.

To summarize, the convergence might be pointwise for $[0,a]$ but it depends heavily on $a$, and convergence will likely not be uniform.