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This is a problem (Exercise 9.2) from Apostol's Mathematical Analysis (second edition) which I couldn't solve.

$\bullet$ Define two sequences $\{f_{n}\}$ and $\{g_{n}\}$ as follows:

$f_{n}(x) = x \Bigl(1 + \frac{1}{n}\Bigr)$ for $x \in \mathbb{R}$, $n \in \mathbb{N}$

$g_{n}(x)= \begin{cases} \frac{1}{n} & x=0 \ \text{or}\ x \in \mathbb{R} - \mathbb{Q} \\ b+\frac{1}{n} & x \in \mathbb{Q},\ \text{say}\ x=\frac{a}{b}, b>0 \end{cases}$

$\mathbb{R}$ is the set of reals, $\mathbb{Q}$ is the set of rational and $\mathbb{R} - \mathbb{Q}$ is the set of irrationals.

I have to show that if $h_{n}(x)=f_{n}(x)g_{n}(x)$, then $h_{n}(x)$ does not converge uniformly on any bounded interval.

How can I show this?

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Have you pursued the lead that $b$ can be arbitrarily large? – timur Jan 6 '11 at 7:24
You seemed to have missed a part of the question. The question is really about the following: Even if $f_n$ and $g_n$ are uniformly convergent, the product $h_n = f_n g_n$ need not be. – Aryabhata Jan 6 '11 at 9:00
up vote 2 down vote accepted

What is the pointwise limit of $f_n(x)g_n(x)$ as $n\rightarrow\infty$? What is the pointwise difference between the limit and the value for a given $n$? Show that, for any $n$, this difference is unbounded on any bounded interval. This follows pretty much immediately from the fact that denominators of rational numbers are unbounded on any bounded interval.

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Pointwise limit of $f_{n}(x)$ is $x$ and pointwise limit of $g_{n}(x)$ is 0 if x is irrational and b is x is rational. – anonymous Jan 6 '11 at 7:39
@Chandru: Isn't the limit at rationals $a$? – Aryabhata Jan 6 '11 at 9:03

Let $I$ be a bounded interval. If $x=0$ or irrational, then $h_n (x) \to 0$. If $x$ is a rational of the form $x=a/b$, $b>0$, then $$ h_n (x) = \frac{a}{b}\bigg(1 + \frac{1}{n}\bigg)\bigg(b + \frac{1}{n}\bigg) = a + \frac{a}{b}\frac{1}{n} + \frac{a}{n} + \frac{a}{b}\frac{1}{{n^2 }}, $$ hence $h_n (x) \to a$. Hence, $h_n$ converges pointwise on $I$; denote the limit by $h$. Since a uniform limit is a pointwise limit, it thus suffices to show that $\sup _{x \in I} |h_n (x) - h(x)| \ge 1$ for infinitely many $n$. Indeed, for any $n \in \mathbb{N}$, there exists a rational number $x_n \in I$ of the form $x_n=a/b$, $b>0$, such that $|a| \geq 2n$. Hence, if we choose $n$ sufficiently large, then $$ |h_n (x_n) - h(x_n)| = \Big|\Big(a + \frac{a}{b}\frac{1}{n} + \frac{a}{n} + \frac{a}{b}\frac{1}{{n^2 }}\Big) - a\Big| > \frac{1}{2}\Big|\frac{a}{n}\Big| \geq 1 $$ (note that $a/b$ is uniformly bounded, since $I$ is a bounded interval). The claim is thus established.

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Assume that $h_n \to h$ uniformly on the domain $D$ of $\{h_n\}$, where

$$h_n(x) = \begin{cases} \frac{x}{n} \biggl(1 + \frac{1}{n} \biggr), & \text{if } x \text{ is irrational} \\ a + \frac{a}{n} + \frac{a}{b} \biggl(1 + \frac{1}{n} \biggr) \biggl(\frac{1}{n} \biggr), & \text{if } x \text{ is rational, say } x = a/b, b > 0 \end{cases}$$

and $$h(x) = \begin{cases} 0, & \text{if }x\text{ is irrational} \\ a, & \text{if }x\text { is rational, say }x = a/b, b > 0. \end{cases}$$

If there is a positive rational in $D$ we can find an interval $A = [s,t]$ contained in $D$ with $s$ rational because $D$ is a bounded interval. Then given $\epsilon = 1$ there exists a positive integer $M$ such that $n \ge M$ implies that

\begin{align} \biggl|h_n\biggl(\frac{a}{b}\biggr) - h\biggl(\frac{a}{b}\biggr) \biggr| &= \ \biggl| \frac{a}{n} + \frac{a}{b} \biggl( \frac{1}{n} + \frac{1}{n^2} \biggr) \biggr| &< 1 \end{align}

whenever $n \ge$ M and $a/b \in A$ where $a$ and $b$ are positive integers. But then

\begin{align} \frac{a}{M} &< \frac{a}{M} + \frac{a}{b} \biggl( \frac{1}{M} + \frac{1}{M^2} \biggr) \ &= \biggl| \frac{a}{M} + \frac{a}{b} \biggl( \frac{1}{M} + \frac{1}{M^2} \biggr) \biggr| \ &< 1. \end{align}

Writing $s = c/d$ where $s$ is the left endpoint of $A$ and $c$ and $d$ are positive integers, we can choose a positive integer $k$ large enough so that both $(ck + 1)/d \in A$ and $ck + 1 > M$.

This contradicts $(ck + 1)/M < 1$. Therefore the assumption that $h_n \to h$ uniformly is false. A similar argument holds if no positive rationals are in $D$

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