Exercise 6.2.3

Question

For each $n \in \mathbf{N}$ and $x \in[0, \infty)$, let
  $$
  g_{n}(x)=\frac{x}{1+x^{n}} \quad 	ext { and } \quad h_{n}(x)= \begin{cases}1 & 	ext { if } x \geq 1 / n \ n x & 	ext { if } 0 \leq x<1 / n\end{cases}
  $$
  Answer the following questions for the sequences $\left(g_{n}ight)$ and $\left(h_{n}ight)$;
  \begin{enumerate}[label=(\alph*)] 
  \item Find the pointwise limit on $[0, \infty)$.
  \item Explain how we know that the convergence cannot be uniform on $[0, \infty)$.
  \item Choose a smaller set over which the convergence is uniform and supply an argument to show that this is indeed the case.
   \end{enumerate}

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] \item $$ \lim g_n(x) = \begin{cases} x & ext{if } x \in [0, 1) \ \frac{1}{2} & ext{if } x = 1 \ 0 & ext{if } x \in (1, \infty) \ \end{cases} \quad ext{and}\quad \lim h_n(x) = \begin{cases} 1 & ext{if } x > 0 \ 0 & ext{if } x = 0 \end{cases} $$ \item They can't converge uniformly since it would contradict Theorem 6.2.6 as both $g_n$ and $h_n$ are continuous but the limit functions are not. \item Over $[1,\infty)$ we have $h_n(x) = h(x) = 1$ for all $n$, thus $|h_n(x) - h(x)| = 0$ for all $x \in [1,\infty)$ so $h_n$ converges uniformly. Now for $g_n$. Let $t \in [0,1)$, I claim $g_n(x) o x$ uniformly over $[0, t)$ since $$ \left|\frac{x}{1 + x^n} - x ight| = \left|\frac{x - x(1+x^n)}{1 + x^n} ight| = \left|\frac{x^{n+1}}{1 + x^n} ight| < \left|t^{n+1} ight| < \epsilon \quad\forall x $$ After setting $n > \log_t \epsilon$. \end{enumerate}

Exercise 6.2.3

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Exercise 6.2.3

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