For $n \in \mathbb{N}$, the function $f_n:[0, \infty) \rightarrow \mathbb{R}$ is defined by $$ f_n(y)=\frac{n^2 y}{1+n y+n^4 y^2} $$

I am trying to prove that the sequence $\{f_{n}\}_{n\in \mathbb{N}}$ does not converge uniformly to the identically-zero function (the function $f$ given by $f(y)=0$, $\forall$y). Is the following correct?

Set $\epsilon=\frac{1}{3}$. Consider arbitrary $N\in \mathbb{N}$. Choose an arbitrary $n>N$. Set $y=\frac{1}{n^2}$. Thus

$$\left|\frac{n^2 y}{1+n y+n^4 y^2}\right|\geq\left|\frac{n^2 y}{1+n^2 y+n^4 y^2}\right|= \left|\frac{1}{1+1+1 }\right|=\epsilon$$

  • $\begingroup$ Yes. Your reasoning is correct. $\endgroup$
    – Riemann
    Oct 21, 2022 at 4:57
  • 5
    $\begingroup$ $\lim_{n\to\infty}f_n(1/n^2)=1/2.$ $\endgroup$ Oct 21, 2022 at 5:00


You must log in to answer this question.