Exercise 21.6

Question

Define $f_n : [0,1] 	o {\mathbb{R}}$ by the equation $f_n(x) = x^n$.
  Show that the sequence $(f_n(x))$ converges for each $x \in [0,1]$, but that the sequence $(f_n)$ does not converge uniformly.

Dan Whitman · Accepted Answer

\def\a{\alpha} \def\b{\beta} \def\d{\delta} \def\e{\epsilon} \def { ho} First, we build up a little more theory \begin{definition} \label{def:metricc:pointwise} Let $f_n: X o Y$ be a sequence of functions from a set $X$ to topological space $Y$. We say that the sequence of functions converges \emph{pointwise} to a function $f: X o Y$ if the sequence $(f_n(x))$ converges to $f(x)$ for every $x \in X$. \end{definition} \begin{lemma} \label{lem:metricc:pwunique} Let $f_n : X o Y$ be a sequence of functions from a set $X$ to topological space $Y$. If $Y$ is a Hausdorff space and $(f_n)$ converges pointwise to $f$, then $f$ is unique. \end{lemma} \begin{proof} Suppose that $(f_n)$ converges pointwise to two distinct functions $f$ and $g$. Since they are distinct, there is an $x_0 \in X$ where $f(x_0) eq g(x_0)$. But then the sequence $(f_n(x_0))$ converges to both of the distinct points $f(x_0)$ and $g(x_0)$ since $(f_n)$ converges pointwise to both $f$ and $g$. As $Y$ is Hausdorff, this violates Theorem~17.10 so that $(f_n)$ can only converge pointwise to a unique function. \end{proof} \begin{lemma} \label{lem:metricc:unifpw} Let $f_n : X o Y$ be a sequence of functions from a set $X$ to metric space $Y$ with metric $d$. If $(f_n)$ converges uniformly to a function $f$, then it also converges pointwise to $f$. \end{lemma} \begin{proof} Suppose that $(f_n)$ converges uniformly to $f$ and consider any $x_0 \in X$ and $\e > 0$. Then there is an $N \in {{\mathbb{Z}}_+}$ where $d(f_n(x), f(x)) < \e$ for all $n > N$ and $x \in X$. Then, since $x_0 \in X$, we have $d(f_n(x_0), f(x_0)) < \e$ for all $n \geq N+1$ since then $n > N$. This shows that the sequence $(f_n(x_0))$ converges to $f(x_0)$ since $\e$ was arbitrary. Since $x_0$ was arbitrary, this shows that $(f_n)$ converges pointwise to $f$ as desired. \end{proof} extbf{Main Problem.} \begin{proof} First we show that the sequence $(f_n)$ converges pointwise to the function $g: [0,1] o {\mathbb{R}}$ defined by \begin{gather*} g(x) = \begin{cases} 0 & x < 1 \ 1 & x = 1 \end{cases} \end{gather*} for $x \in [0,1]$. This of course shows that $(f_n(x))$ converges for each $x \in [0,1]$ by Definition~ ef{def:metricc:pointwise}. So consider any such $x \in [0,1]$ so that we have the following exhaustive cases: Case: $x = 0$. Then $f_n(x) = x^n = 0^n = 0$ for any $n \in {{\mathbb{Z}}_+}$. Thus clearly the sequence $(f_n(x)) = (0,0,\ldots)$ converges to $0$. Case: $x = 1$. Somewhat similarly we have $f_n(x) = x^n = 1^n = 1$ for every $n \in {{\mathbb{Z}}_+}$ so that clearly the sequence $(f_n(x)) = (1,1,\ldots)$ converges to $1$. Case: $0 < x < 1$. Here we show that, for any $x \in (0,1)$, that the sequence $(f_n(x))$ is strictly decreasing and that it is bounded below by $0$. We prove this using induction on $n$. So first for $n = 1$ we have \begin{align*} 0 < x &< 1 \ 0 < x^2 &< x & ext{(since $x > 0$)} \ 0 < f_2(x) &< f_1(x) \ 0 < f_{n+1}(x) &< f_n(x) \,. \end{align*} For the inductive step suppose that $f_n(x) > 0$ and $f_{n+1}(x) < f_n(x)$. Then we of course have $0 < f_n(x) = x^n$ so that clearly $0 = x \cdot 0 < x \cdot x^n = x^{n+1} = f_{n+1}(x)$ as well since $x > 0$. We also clearly have \begin{align*} f_{n+1} &< f_n(x) \ x^{n+1} &< x^n \ x \cdot x^{n+1} &< x \cdot x^n & ext{(since $x > 0$)} \ x^{n+2} &< x^{n+1} \ f_{n+2}(x) &< f_{n+1}(x) \,, \end{align*} which completes the inductive step. Hence the sequence is strictly decreasing (and therefore of course non-increasing) and bounded below by zero, from which it follows that it converges by a theorem analogous in an obvious way to that shown in Exercise~21.11 part~(a). So suppose that the sequence converges to the value $a$. Then of course also the sequence $(f_{n+1}(x))$ must also converge to $a$. Thus we have \begin{gather*} a = \lim_{n o \infty} f_{n+1}(x) = \lim_{n o \infty} x^{n+1} = \lim_{n o \infty} x \cdot x^n = \lim_{n o \infty} x f_n(x) = x \lim_{n o \infty} f_n(x) = xa \,, \end{gather*} where we note that clearly the function $h(y) = xy$ for $y \in {\mathbb{R}}$ is clearly continuous so that \begin{gather*} \lim_{n o \infty} x f_n(x) = \lim_{n o \infty} h(f_n(x)) = h\left(\lim_{n o \infty} f_n(x) ight) = x \lim_{n o \infty} f_n(x) \end{gather*} by Theorem~21.3. Therefore we have $a = xa$ so that it would be that $1 = x$ if $a$ were any nonzero value. However, we know that $x < 1$ so it must be that $a = 0$. This completes the proof that $(f_n)$ converges pointwise to $g$. It then becomes fairly easy to show that $(f_n(x))$ does not converge uniformly. Consider any function $f: [0,1] o {\mathbb{R}}$. If $f eq g$ then $(f_n)$ cannot converge pointwise to $f$ since it was shown to converge pointwise to $g$ above and $g$ is unique by Lemma~ ef{lem:metricc:pwunique} since ${\mathbb{R}}$ is Hausdorff. Hence $(f_n)$ cannot converge uniformly to $f$ by the converse of Lemma~ ef{lem:metricc:unifpw} since it does even not converge pointwise. On the other hand if $f = g$ then clearly $f = g$ has a discontinuity at $1$ so that it is not continuous. However each $f_n(x) = x^n$ is continuous on $[0,1]$ by elementary calculus. This shows that $(f_n)$ cannot converge uniformly to $f=g$ since this would violate Theorem~21.6. Therefore $(f_n)$ does converge uniformly to any function since $f$ was arbitrary. \end{proof}

Exercise 21.6

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Exercise 21.6

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