Exercise 21.9

Question

Let $f_n: {\mathbb{R}} 	o {\mathbb{R}}$ be the function
  \begin{gather*}
    f_n(x) = \frac{1}{n^3 [x-(1/n)]^2 + 1} \,.
  \end{gather*}
  See Figure~21.1.
  Let $f: {\mathbb{R}} 	o {\mathbb{R}}$ be the zero function.
  \begin{enumerate}[label=(\alph*)]
  \item Show that $f_n(x) 	o f(x)$ for each $x \in {\mathbb{R}}$.
  \item Show that $f_n$ does not converge uniformly to $f$.
    (This shows that the converse of Theorem~21.6 does not hold; the limit function $f$ may be continuous even though the convergence is not uniform.)
  \end{enumerate}

Dan Whitman · Accepted Answer

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\def\b{\beta}
\def\d{\delta}
\def\e{\epsilon}
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(a)
  \begin{proof}
    This is easy to show by evaluating the limit using techniques from elementary calculus.
    Fix $x \in {\mathbb{R}}$ and first suppose that $x 
eq 0$.
    Clearly $1/n 	o 0$ as $n 	o \infty$ so that $[x-(1/n)]^2 	o x^2$.
    Since $x^2 > 0$ it follows that $n^3 [x - (1/n)]^2 	o n^3 x^2 	o \infty$ as $n 	o \infty$.
    Hence the overall function
    \begin{gather*}
      f_n(x) = \frac{1}{n^3 [x-(1/n)]^2 + 1} 	o \frac{1}{\infty + 1} 	o 0 = f(x)
    \end{gather*}
    as $n 	o \infty$.
    Of course this is a little informal, but it can be justified rigorously using nothing more than Exercise~21.5.
    If $x = 0$ then we clearly have
    \begin{gather*}
      f_n(x) = f_n(0) = \frac{1}{n^3 [-(1/n)]^2 + 1} = \frac{1}{n^3/n^2 + 1} = \frac{1}{n+1} 	o 0 = f(x)
    \end{gather*}
    as $n 	o \infty$.
  \end{proof}

(b)
  \begin{proof}
    Let $\e = 1/2$ and consider any $N \in {{\mathbb{Z}}_+}$ and let $n = N+1$ so of course $n > N$.
    Also set $x = 1/n$.
    Then we have
    \begin{gather*}
      f_n(x) = \frac{1}{n^3 [x-(1/n)]^2 + 1} = \frac{1}{n^3 [(1/n)-(1/n)]^2 + 1} = \frac{1}{n^3 \cdot 0 + 1} = \frac{1}{1} = 1
    \end{gather*}
    whereas of course $f(x) = 0$.
    We therefore have
    \begin{gather*}
      d(f_n(x), f(x)) = d(1,0) = \left|1-0ight| = 1 \geq 1/2 = \e \,.
    \end{gather*}
    This shows the negation of the definition of uniform convergence so that $(f_n)$ does not converge uniformly to $f$ as desired.
    \end{proof}
  Note that this also shows that $(f_n)$ does not uniformly converge to any function at all since, if it did, it can only converge uniformly to $f$ since this is the only function to which it converges pointwise.
  This follows from \href{./exercise_21-6}{Lemma~21.6.1} and \href{./exercise_21-6}{Lemma~21.6.2} as in Exercise~21.6.

Exercise 21.9

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Exercise 21.9

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