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\begin{document}
\begin{center}
{\bf \Large Math 112 Homework 8 Solutions}\medskip
\end{center}

{\bf Problem 1.}

To show $f\not\in\mathcal{R}$ for any $[a,b]$, we show that for all
partitions $P$, $U(P,f)=b-a$ but $L(P,f)=0$, which implies that the lower
and upper integrals or $f$ are not equal and hence $f\not\in\mathcal{R}$.

Fix a partition $P=\{a=x_0<x_1<\dots<x_n=b\}$. (Note: the definition
on p.\ 120 allows $x_0\leq x_1\leq \dots\leq x_n$, but if $x_i=x_{i+1}$ then
$\Delta x_i=0$, so we can discard any point that appears more than once).
Each interval $[x_{i-1},x_i]$ contains a rational number (by density of the
rational numbers, Theorem 1.20(b)), so $M_i=1$ for all $i$. On the other
hand, each interval $[x_{i-1},x_i]$ also contains irrational numbers, so
$m_i=0$ for all $i$. This implies that
$L(P,f)=\sum_{i=1}^n m_i\,\Delta x_i=0$ and
$U(P,f)=\sum_{i=1}^n M_i\,\Delta x_i=\sum_{i=1}^n (x_i-x_{i-1})=b-a,$
which implies that $f$ is not Riemann-integrable.%
\medskip

{\bf Problem 2.}

Consider the partition $P_n=\{1,2,\dots,n\}$ of the interval $[1,n]$.
Since $f$ is monotonically decreasing, $\inf_{[i-1,i]} f=f(i)$ and
$\sup_{[i-1,i]} f=f(i-1)$, while $\Delta x_i=1$ for all $i$, so we have
$L(P_n,f)=\sum_{i=2}^n f(i)$ and $U(P_n,f)=\sum_{i=2}^n
f(i-1)=\sum_{i=1}^{n-1} f(i)$. Since $L(P_n,f)\le \int_1^n f\,dx\le
U(P_n,f)$, we have the inequalities $\sum_{i=2}^n f(i)\le \int_1^n f\,dx
\le \sum_{i=1}^{n-1} f(i)$. By comparison, we get that the integral 
converges if and only if the series converges.

More precisely, assume that $\sum_{i=1}^\infty f(i)$ converges: then the 
integrals $I_n=\int_1^n f\,dx\le \sum_{i=1}^{n-1} f(i)\le \sum_{i=1}^\infty 
f(i)$ are bounded, and so $\int_1^\infty f\,dx$ is convergent (remark that 
$\{I_n\}$ is a bounded monotonically increasing sequence, and that any
integral of the form $\int_1^A f\,dx$ can be bounded between $I_n$ and
$I_{n+1}$ for some value of $n$).
Conversely, assume that $\int_1^\infty f\,dx$ converges: then the
sums $\sum_{i=2}^n f(i)\le \int_1^n f\,dx\le \int_1^\infty f\,dx$ are
bounded, so the series $\sum_{i=2}^\infty f(i)$ is convergent (it is a
series of non-negative terms and its partial sums are bounded); adding the
single term $f(1)$ to the series, $\sum_{i=1}^\infty f(i)$ is also
convergent.
\medskip

{\bf Problem 3.}

(a) First observe that, since $\frac{1}{p}+\frac{1}{q}=1$ and $p,q$ are
positive, we must have $\frac{1}{p},\frac{1}{q}<1$, i.e.\ $p,q>1$. For a
fixed value of $v\ge 0$, let $\phi:[0,+\infty)\to\R$ be the
function defined by $\phi(u)=\frac{1}{p}u^p-uv$. The function $\phi$ is
differentiable and $\phi'(u)=u^{p-1}-v$. Let $\alpha=v^{1/(p-1)}$: since
$p-1>0$, $\phi'$ is a strictly increasing function, so $\phi'$
takes negative values over $[0,\alpha)$ and positive values over
$(\alpha,+\infty)$. Hence, by the mean value
theorem, $\phi$ is strictly decreasing over the interval $[0,\alpha]$ and 
strictly increasing over the interval $[\alpha,+\infty)$. In particular, 
$\forall u\ge 0$ we have $\phi(u)\ge \phi(\alpha)$, with equality if and
only if $u=\alpha$. Observe
that $\frac{p-1}{p}=1-\frac{1}{p}=\frac{1}{q}$; therefore 
$\phi(\alpha)=\frac{1}{p}\alpha^p-\alpha v=\frac{1}{p}v^{p/(p-1)}-
v^{1/(p-1)+1}=-\frac{p-1}{p}v^{p/(p-1)}=-\frac{1}{q}v^q$. We conclude that, 
for every $u,v\ge 0$, $\frac{1}{p}u^p-uv\ge -\frac{1}{q}v^q$, or
equivalenty, 
$uv\le \frac{1}{p}u^p+\frac{1}{q}v^q$, with equality if and only if
$u=v^{1/(p-1)}$, i.e.\ $u^p=v^{p/(p-1)}=v^q$.

(b) Note that, if $f,g\in\mathcal{R}$, then by Theorems 6.11 and 6.13,
$f^p$, $g^q$ and $fg$ are also integrable. By part (a), we have the
inequality
$f(x)g(x)\le \frac{1}{p}f(x)^p+\frac{1}{q}f(x)^q$ for every $x\in [a,b]$.
Therefore, by Theorem 6.12, we have $\int_a^b fg\,dx\le \int_a^b
(\frac{1}{p} f^p+\frac{1}{q}f^q)\,dx=\frac{1}{p}\int_a^b f^p\,dx+\frac{1}{q}
\int_a^b g^q\,dx=\frac{1}{p}+\frac{1}{q}=1$, which is the desired result.

(c) Let $I=(\int_a^b |f|^p\,dx)^{1/p}$ and $J=(\int_a^b |g|^q\,dx)^{1/q}$.
Observe that $\int_a^b (\frac{1}{I}|f|)^p\,dx=\frac{1}{I^p} \int_a^b
|f|^p\,dx=1$, and similarly $\int_a^b (\frac{1}{J}|g|)^q\,dx=\frac{1}{J^q}
\int_a^b |g|^q\,dx=1$. Therefore, applying the result of (b) to the
functions $\frac{1}{I}|f|$ and $\frac{1}{J}|g|$, we get
$\int_a^b \frac{1}{IJ}|fg|\,dx\le 1$, or equivalently $\int_a^b |fg|\,dx\le
IJ$. Using Theorem 6.13, which remains true for complex-valued functions
(cf.\ Theorem 6.25), we have $|\int_a^b fg\,dx|\le \int_a^b |fg|\,dx\le IJ$,
which completes the proof.
\medskip

{\bf Problem 4.}

(a) For $n=0$, the formula becomes $f(x)=f(a)+\int_a^x f'(t)\,dt$, which
is the fundamental theorem of calculus (Theorem 6.21).

(b) By comparing the given expressions for $n-1$ and $n$, one sees that it
is sufficient to prove that
$R_n(x)=\frac{1}{n!}f^{(n)}(a)\,(x-a)^n+R_{n+1}(x)$.
We use integration by parts (Theorem 6.22) with $F(t)=f^{(n)}(t)$ and
$G(t)=-\frac{1}{n}(x-t)^n$; we have
\begin{eqnarray*}
R_n(x)&=&\frac{1}{(n-1)!}\int_a^x (x-t)^{n-1}f^{(n)}(t)\,dt=
\frac{1}{(n-1)!}\int_a^x F(t) G'(t)\,dt\\
&=&\frac{F(x)G(x)-F(a)G(a)}{(n-1)!}-
\frac{1}{(n-1)!}\int_a^x F'(t)G(t)\,dt\\
&=&\frac{0+f^{(n)}(a)(x-a)^n}{n\,(n-1)!}+\frac{1}{n\,(n-1)!}\int_a^x 
f^{(n+1)}(t)(x-t)^n\,dt=\frac{f^{(n)}(a)}{n!}(x-a)^n+R_{n+1}(x).
\end{eqnarray*}
The result follows by induction on $n$.
\medskip

{\bf Problem 5.}

Assume that $\{f_n\}$ converges uniformly to $f$ and $\{g_n\}$ converges
uniformly to $g$. First observe that $f$ is bounded. Indeed, there exists
$N$ such that if $n\ge N$ then $|f_n(x)-f(x)|\le 1$ $\forall x\in E$.
Therefore, setting $M=1+\sup_{x\in E} |f_N(x)|$, for every $x\in E$ we have
$|f(x)|\le |f_N(x)|+1\le M$. Similarly, $g$ is bounded by a constant $M'$.

Fix $\epsilon>0$; decreasing $\epsilon$ if necessary we can assume that
$\frac{\epsilon}{2M'}<1$. Since $f_n\to f$ uniformly, there exists $N_1$ such that
$\forall n\ge N_1$, $\forall x\in E$, $|f_n(x)-f(x)|\le
\frac{\epsilon}{2M'}$. Similarly there exists $N_2$ such that $\forall n\ge
N_2$, $\forall x\in E$, $|g_n(x)-g(x)|\le \frac{\epsilon}{2(M+1)}$.
Let $n\ge N=\max(N_1,N_2)$ and $x\in E$: we have 
\begin{eqnarray*}
|f_n(x)g_n(x)-f(x)g(x)|&=& |f_n(x)(g_n(x)-g(x))+(f_n(x)-f(x))g(x)|\\
&\le& |f_n(x)||g_n(x)-g(x)|+ |f_n(x)-f(x)||g(x)|\\
&\le& (|f(x)|+\tfrac{\epsilon}{2M'})\tfrac{\epsilon}{2(M+1)}+
\tfrac{\epsilon}{2M'}|g(x)|\\
&\le&
(M+1)\tfrac{\epsilon}{2(M+1)}+M'\tfrac{\epsilon}{2M'}=\epsilon.
\end{eqnarray*} So we have
proved that $f_ng_n$ converges uniformly to $fg$.

The assumption that $f_n$ and $g_n$ are bounded is necessary: consider e.g.\
$E=(0,+\infty)$, $f_n(x)=x$, and $g_n(x)=\frac{1}{x}+\frac{1}{n}$.
It is clear that $f_n$ converges uniformly to $f(x)=x$ and $g_n$ converges
uniformly to $g(x)=\frac{1}{x}$. We have $f_n(x)g_n(x)=1+\frac{x}{n}$, which 
converges to $1$ as $n\to \infty$ for every value of $x$. However, $f_n g_n$
does not converge uniformly to $1$ over $(0,+\infty)$, in fact $(f_n g_n-1)$
is not even bounded (consider $x\to\infty$).
\medskip

\newpage

{\bf Problem 6.}

If $x>0$ then $|\frac{1}{1+n^2x}|\le \frac{1}{n^2x}=\frac{x^{-1}}{n^2}$.
Since $\sum \frac{1}{n^2}$ converges, by the comparison criterion $\sum
\frac{1}{1+n^2x}$ converges absolutely for all $x>0$. If $x=0$ the series
is divergent since $\frac{1}{1+n^2x}=1$.

Given any constant $a>0$, we have the inequality $|\frac{1}{1+n^2x}|
\le \frac{1}{n^2x}\le \frac{a^{-1}}{n^2}$ for all $x\in
[a,+\infty)$. Observe that the series $\sum \frac{a^{-1}}{n^2}$
is convergent; therefore, by Theorem 7.10, the series $\sum
\frac{1}{1+n^2x}$
converges uniformly on the interval $[a,+\infty)$ for all $a>0$.

We now consider an interval of the form $(0,a)$: fix an integer $N$, and
let $x\in (0,a)$ be such that $x<\frac{1}{N^2}$. Then $\frac{1}{1+N^2x}>
\frac{1}{1+1}=\frac{1}{2}$. This proves that $\frac{1}{1+n^2x}$ does not
converge uniformly to $0$ over $(0,a)$, which is enough to derive a
contradiction to the Cauchy criterion (Theorem 7.8): 
therefore the series does not converge uniformly on any of the intervals
$(0,a)$ for $a>0$.\\
Alternatively: letting $f_n(x)=\sum_{k=1}^n \frac{1}{1+k^2n}$, we set
$A_n=\lim_{t\to 0} f_n(t)=n$. If one assumes the sequence $\{f_n\}$ to 
converge uniformly over $(0,a)$, by Theorem 7.11 the sequence $\{A_n\}$ must 
be convergent (and its limit equals $\lim_{t\to 0} f(t)$). This gives a
contradiction, so the convergence is not uniform over $(0,a)$.

Since the individual terms in the series are continuous functions and
since the convergence is uniform over $[a,+\infty)$, by Theorem 7.12 the
function $f$ is continuous over $[a,+\infty)$ for all $a>0$. Therefore
$f$ is continuous at every point of $(0,+\infty)$.

The function $f$ is not bounded: given any integer $N$, if $x\in
(0,\frac{1}{N^2})$ then for every $n\le N$ we have $\frac{1}{1+n^2x}>
\frac{1}{1+(n/N)^2}\ge \frac{1}{2}$, so $\sum_{n=1}^N \frac{1}{1+n^2x}>
\frac{N}{2}$, and therefore $f(x)>\frac{N}{2}$.



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