Exercise 6.2.1

Question

Let
  $$
  f_{n}(x)=\frac{n x}{1+n x^{2}} .
  $$
  \begin{enumerate}[label=(\alph*)] 
  \item Find the pointwise limit of $\left(f_{n}\right)$ for all $x \in(0, \infty)$.
  \item Is the convergence uniform on $(0, \infty)$?
  \item Is the convergence uniform on $(0,1)$?
  \item Is the convergence uniform on $(1, \infty)$?
   \end{enumerate}

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] 
  \item $\lim f_n(x) = \lim \frac{nx}{1+nx^2} = \lim \frac{x}{1/n + x^2} = 1/x$
  \item Examine the difference $|f_n(x) - f(x)|$
    $$
    \left|\frac{nx}{1+nx^2} - \frac{1}{x}\right|
    = \left|\frac{nx^2 - (1+nx^2)}{x(1+nx^2)}\right|
    = \frac{1}{x(1+nx^2)}
    $$
    Consider $x_n = 1/n$, then
    $$
    \left|f_n(x_n) - f(x_n)\right| = \frac{1}{(1/n)(1+n(1/n^2)} = \frac{1}{n/2} = \frac{n}{2}
    $$
    Which shows that no matter how big $n$ is, we can find $x = 1/n$ such that $|f_n(x) - f(x)| \ge 1/2$ meaning $\epsilon$ cannot be made smaller then $1/2$. So $f$ isn't uniformly continuous.
  \item No, same logic as (b)
  \item Yes, because $x \ge 1$ implies
    $$
    |f_n(x) - f(x)| = \frac{1}{x(1 + nx^2)} \le \frac{1}{n}
    $$
    Meaning for all $\epsilon > 0$, setting $N > 1/\epsilon$ implies every $n \ge N$ has $|f_n(x) - f(x)| \le 1/N < \epsilon$ for every $x \in (1,\infty)$.
   \end{enumerate}

Exercise 6.2.1

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Exercise 6.2.1

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