\documentclass{article}
\input{18100bpreamble}

\begin{document}
\pset{8}{Dan Ports}{2003/05/02}{drkp@mit.edu}

\begin{ppart}{Problem 1}
\paragraph{Part a}
Note that since $\frac{1}{p} + \frac{1}{q} = 1$ and $p$ and $q$ are both positive, both $p$ and $q$ must be greater than $1$ (else their reciprocals could not add to $1$). Let $x$ = $\frac{u^p}{v^q}$ (we assume both $u$ and $v$ are non-zero so $x$ is defined). Assume without loss of generality that $u^p > v^q$ (by the symmetry of the problem, we can exchange $u^p$ and $v^q$; we will consider the $u^p = v^q$ case later). Then $x > 1$. Consider the function $f(x) = \frac{x}{p} - x^{\frac{1}{p}}$. Its derivative is $f'(x) = \frac{1}{p} - \frac{1}{p}x^{\frac{1}{p}-1} = \frac{1}{p}-\frac{1}{p}x^{-q} = \frac{1}{p}(1-x^{-q})$. Certainly $\frac{1}{p} > 0$. Since $q > 0$ and $x > 1$, $x^{-q} > 1$ and thus $(1-x^{-q}) > 0$. So $f'(x) > 0$ and $f(x)$ is strictly increasing on $(1,\infty)$. Note that $f(1) = \frac{1}{p} - 1 = -\frac{1}{q}$. Thus on $(1,\infty)$, $f(x) > -\frac{1}{q}$, so $f(x) + \frac{1}{q} > 0$: 
\begin{eqnarray*}
\frac{x}{p} - x^{\frac{1}{p}} + \frac{1}{q} &>& 0\\
\frac{1}{p}\frac{u^p}{v^q} - \frac{u}{v^{\frac{q}{p}}} + \frac{1}{q} &>& 0\\
\frac{1}{p} u^p - u v^{-\frac{q}{p} + q} + \frac{1}{q}v^q &>& 0 \\
\frac{u^p}{p} + \frac{v^q}{q} &>& uv^{-\frac{q}{p} +q} = uv^{-(q-1)+q} = uv \\
\end{eqnarray*}
We need also to consider the cases where $x$ is undefined. If $u \neq v = 0$, then the inequality is simply $0 < \frac{u^p}{p}$, which is clearly true, and similarly if $v \neq u = 0$. If $u = v = 0$, then both sides of the inequality are zero and equality holds.
\paragraph{}
Next we consider the possibility that $u^p = v^q$. In this case, $x = 1$. Dividing the inequality by $v^q$ and substituting $x$ as before gives $\frac{x}{p} + \frac{1}{q} \ge x^{\frac{1}{p}}$.  With $x= 1$, the inequality becomes an equality: $\frac{1}{p} + (1 - \frac{1}{p}) = 1$. Thus equality holds if $x=1$, in which case $u^p = v^q$. If $x\neq 1$, we showed above that the strict inequality $\frac{u^p}{p} + \frac{v^q}{q} > uv$ holds. Thus, equality holds if and only if $u^p = v^q$.
\paragraph{Part b}
Since $f,g \in \riemann$, $fg \in \riemann$. From part a, $\forall x \in [a,b] ~ fg \le \frac{f^p}{p} + \frac {g^q}{q}$. Thus, applying theorem 6.12(b), \[\int_a^b fg\,dx \le \frac{1}{p} \int_a^b f^p\,dx + \frac{1}{q} \int_a^b g^q\,dx = \frac{1}{p} + \frac{1}{q} = 1\]

\paragraph{Part c}
If $f$ and $g$ are complex-valued Riemann-integrable functions, the functions $\abs{f}$ and $\abs{g}$ are positive real-valued Riemann-integrable functions, and by theorem 6.13(b), $\abs{\int_a^b fg\,dx} \leq \int_a^b \abs{fg}\,dx \leq \int_a^b \abs{f}\abs{g}\,dx$. Let $\alpha = (\int_a^b \abs{f}^p\,dx)^{\frac{1}{p}}$ and $\beta = (\int_a^b \abs{g}^q\,dx)^{\frac{1}{q}}$. Suppose first that $\alpha$ and $\beta$ are both non-zero. Then let $h = \frac{\abs{f}}{\alpha}$ and $j = \frac{\abs{g}}{\beta}$. Then $\int_a^b h^p\,dx = \frac{1}{\int_a^b \abs{f}^p\,dx}\int_a^b \abs{f}^p\,dx = 1$, and by the same reasoning $\int_a^b j^q\,dx = 1$. By part b:
\begin{eqnarray*}
\int_a^b hj\,dx = \int_a^b \frac{\abs{f}}{\alpha} \frac{\abs{g}}{\beta}\,dx &\leq& 1 \\
\int_a^b \abs{f}\abs{g}\,dx &\le& \alpha \beta\\
\int_a^b \abs{f}\abs{g}\,dx &\le& \left(\int_a^b \abs{f}^p\,dx\right)^{\frac{1}{p}} \left(\int_a^b \abs{g}^q\,dx\right)^{\frac{1}{q}} \\
\abs{\int_a^b fg\,dx} &\le& \left(\int_a^b \abs{f}^p\,dx\right)^{\frac{1}{p}} \left(\int_a^b \abs{g}^q\,dx\right)^{\frac{1}{q}}
\end{eqnarray*}
We also need to consider the case where either $\alpha$ or $\beta$ is zero. Suppose $\alpha = 0$. Then $\int_a^b \abs{f}^p\,dx = 0$. So $\forall x~\abs{f(x)}^p = 0$ because $\abs{f(x)}^p$ is everywhere non-negative and its integral is zero (this is a result from the last problem set). Thus $\abs{f} = 0$, so $\abs{\int_a^b fg\,dx} \le \int_a^b \abs{f}\abs{g}\,dx = 0$. This means $\abs{\int_a^b fg\,dx}$ is zero (since it certainly cannot be negative). Since $\alpha= (\int_a^b \abs{f}^p\,dx)^{\frac{1}{p}}$ is also zero, the inequality reduces to $0 \le 0$, which is obviously true. The same argument can be applied if $\beta = 0$. So the desired inequality $\abs{\int_a^b fg\,dx} \le \left(\int_a^b \abs{f}^p\,dx\right)^{\frac{1}{p}} \left(\int_a^b \abs{g}^q\,dx\right)^{\frac{1}{q}}$ always holds.
\end{ppart}


\begin{ppart}{Problem 2}
\paragraph{Part a}
The $n=0$ case reduces to $f(x) = f(a) + \int_a^x f'(x)\,dx$. This follows from the fundamental theorem of calculus, which tells us $\int_a^x f'(x)\,dx = f(x) - f(a)$. Applying this, \[f(x) = f(a) + f(x) - f(a) = f(a) + \int_a^x f'(x)\,dx\].

\paragraph{Part b}
Assume by induction that $f(x) = \sum_{k=0}^{n-1} \frac{f^{(k)}(a)}{k!}(x-a)^k+R_n(x)$. Let $g(t) = -\frac{(x-t)^n}{n}$ and $h(t) = f^{(n)}(t)$. Then $g'(t) = (x-t)^{n-1}$ and $h'(t) = f^{(n+1)}(t)$. Then we can write $R_n(x) = \frac{1}{(n-1)!} \int_a^x g'(t) h(t)\,dt$. Applying integration by parts, 
\begin{eqnarray*}
R_n(x) &=& \frac{1}{(n-1)!} \left[ g(x)h(x) - g(a)h(a) - \int_a^x g(t) h'(t)\, dt \right] \\
 &=& \frac{1}{(n-1)!} \left[-\frac{(x-x)^n}{n} f^{(n)}(x) - \left(-\frac{(x-a)^n}{n} f^{(n)}(a)\right) - \int_a^x \left(-\frac{(x-t)^n}{n} f^{(n+1)}(t) \right)\, dt \right] \\
&=& \frac{1}{(n-1)!} \left[\frac{(x-a)^n}{n} f^{(n)}(a) + \int_a^x \frac{(x-t)^n}{n} f^{(n+1)}(t) \, dt \right] \\
&=& \frac{1}{n!} (x-a)^n f^{(n)}(a) + R_{n+1}(x)
\end{eqnarray*}
So we have shown that
\begin{eqnarray*}
f(x) &=& \sum_{k=0}^{n-1} \frac{f^{(k)}(a)}{k!}(x-a)^k + \frac{1}{n!}\frac{(x-a)^n}{n} f^{(n)}(a) + R_{n+1}(x) \\
&=&\sum_{k=0}^{n} \frac{f^{(k)}(a)}{k!}(x-a)^k+R_{n+1}(x) \\
\end{eqnarray*}
Thus we have proven the inductive hypothesis for $n$, and by induction the theorem holds for all $n \in \N$.
\end{ppart}

\begin{ppart}{Problem 3}
Let $\{f_n\}$ and $\{g_n\}$ be sequences of bounded functions that converge uniformly on $E$. We will show that their product $\{f_n g_n\}$ converges via the Cauchy criterion. The sequences are bounded, so we can find some $M_f$ such that $\forall n\in \J~ \forall x \in E~ \abs{f_n(x)} < M_f$, and similarly $M_g$ such that $\forall n\in \J~ \forall x \in E~ \abs{g_n(x)} < M_g$; let $M = \max\{M_f, M_g\}$. Let $\epsilon > 0$ be given. Then by the uniform convergence of $\{f_n\}$, we can find a $N_f$ such that $\forall n,m \ge N_f~ \forall x \in E~ \abs{f_n(x) - f_m(x)} \le \frac{\epsilon}{2M}$, and a $N_g$ similarly chosen by the same criteria on $g$. Let $N = \max\{N_f, N_g\}$. Then for any $n,m \ge N$ and any $x \in E$, consider $\abs{f_n(x) g_n(x) - f_m(x) g_m(x)}$. By the triangle inequality, this is less than or equal to $\abs{f_n(x) g_n(x) - f_n(x) g_m(x)} + \abs{f_n(x) g_m(x) - f_m(x) g_m(x)} = \abs{f_n(x)}\abs{g_n(x) - g_m(x)} + \abs{f_n(x) - f_m(x)}\abs{g_m(x)}$. But $\abs{f_n(x)}$ and $\abs{g_m(x)}$ are both less than $M$ by their boundedness, and  $\abs{f_n(x) - f_m(x)}$ and $\abs{g_n(x) - g_m(x)}$ are less than or equal to $\frac{\epsilon}{2M}$ by our choice of $N$. So this quantity is less than $M\frac{\epsilon}{2M} + M\frac{\epsilon}{2M} = \epsilon$. Thus the sequence $\{f_n g_n\}$ satisfies the Cauchy criterion for uniform convergence.
\paragraph{}
We now show that this is not true in the unbounded case. Consider two sequences of functions defined on the interval $[0,\infty)$. Define $f_n(x) = x^2$ for all $n$. Clearly, $\{f_n\}$ is uniformly convergent to $f(x) = x^2$ (since it does not even depend on $n$). Define $g_n(x)$ to be zero for $x\in[0,n]$ and $\frac{1}{x}$ for $x\in(n,\infty)$. Then $\{g_n\}$ converges uniformly to $g(x) = 0$: given any $\epsilon > 0$, we let $N = \lceil \frac{1}{\epsilon} \rceil$. Then for any $n \ge N$, $\abs{g_n(x)}$ is everywhere less than $\epsilon$ because $g_n$ is defined to be zero for $x \le n$, and for $x > n$ $g_n(x) = \frac{1}{x} < \frac{1}{\frac{1}{\epsilon}} = \epsilon$. Next consider the product of the sequences:
\[ f_n g_n = \begin{cases} 0  & 0 \le x \le n \\ x^2\frac{1}{x} = x & x>n \end{cases} \]
The sequence $\{f_n g_n\}$ certainly converges to $f(x)g(x) = 0$, but this convergence is not uniform. Given any $\epsilon > 0$ and any $n \in \J$, we will find a $x$ such that $f_n(x)g_n(x) - f(x)g(x) > \epsilon$. Specifically, consider $x = \max\{\epsilon + 1,n+1\}$. Then $f_n(x)g_n(x) = x > \epsilon$. So $f_n(x)g_n(x) - f(x)g(x) = x - 0 > \epsilon$. Thus $\{f_ng_n\}$ cannot satisfy the definition of uniform convergence.
\end{ppart}

\begin{ppart}{Problem 4}
We consider the series $\sum_{n=1}^\infty \frac{1}{1+n^2x}$, with $x \ge 0$.  Fix any $x>0$, and note that the series is termwise less than $\sum_{n=1}^\infty \frac{1}{n^2x}$ = $\frac{1}{x} \sum_{n=1}^\infty \frac{1}{n^2}$. Thus by the comparison test, since $\sum \frac{1}{n^2}$ converges, the series must converge for any $x > 0$. Since we are restricted to considering non-negative $x$, every term of the series is always positive, so the series converges absolutely for $x>0$. The series obviously diverges for $x = 0$.
\paragraph{}
We next show that the series converges uniformly on any interval of the form $[x,\infty)$ (for $x>0$). Write $f_n(x) = \frac{1}{1+n^2x}$; then $f(x) = \sum_{n=1}^\infty f_n(x)$. Let $x>0$ be given, and consider the set $E = [x,\infty)$. For any $t\in E$, $t \ge x$, and $\abs{f_n(t)} = \abs{\frac{1}{1+n^2t}} \le \abs{\frac{1}{1+n^2x}} = \abs{f_n(x)}$ By the previous result, the series $\sum_{n=1}^\infty f_n(x)$ converges. Therefore, by the Weierstrass m-test, $\sum_{n=1}^\infty f_n(t) = f(t)$ converges uniformly on $E$. So for any $x>0$, $f$ converges uniformly on $[x,\infty)$.
\paragraph{}
The series fails to converge uniformly on any interval $(0,x]$ (with $x>0$). Let $x>0$ be given and consider $E = (0,x]$. We will show that $\sum f_n(t)$ fails to satisfy the Cauchy criterion for uniform convergence on $E$. Let $p,q \in \J$ be given ($p \neq q$), and assume without loss of generality that $p < q$. Then consider $\abs{\sum_{n=1}^p f_n(t) - \sum_{n=1}^q f_n(t)} = \sum_{n=p}^{q} f_n(t)$. The first term of this summation is $f_p(t) = \frac{1}{1+p^2t}$. Since we are considering $t \in (0,x]$, we can take $u = \min\{x,\frac{1}{p^2}\}$ and we will have a $u$ that satisfies both $u \in E$ and $u \le \frac{1}{p^2}$. Then $f_p(u) \ge \frac{1}{1+p^2\frac{1}{p^2}} = \frac{1}{2}$. So $\sum_{n=p}^{q} f_n(u)$ is certainly bounded below by its first term, $f_p(u) = \frac{1}{2}$. So we have shown that for any $p,q$ we can find a $u \in E$ such that $\abs{\sum_{n=1}^p f_n(y) - \sum_{n=1}^q f_n(y)} > \frac{1}{2}$. Thus $\sum f_n(t)$ fails to satisfy the Cauchy criterion for uniform convergence, and therefore $f$ does not converge uniformly on $(0,x]$ for any $x>0$.
\paragraph{}
Next we show $f$ is continuous on the domain $(0,\infty)$ where it converges. Consider any $x > 0$. We will show that $f$ is continuous at $x$. Note that by the density of the real numbers, we can find some $a$ satisfying $0 < a < x$. Then $x \in (a,\infty)$. From above, $f$ is uniformly convergent on $(a,\infty)$. Also, for each $n$, $f_n(x) = \frac{1}{1+n^2x}$ is clearly continuous on $(a,\infty)$. Thus, since $f = \sum f_n$ is a uniformly convergent sum of continuous functions on $(a,\infty)$, it is continuous on $(a,\infty)$. Thus it is continuous at $x$. Since $x$ was arbitrary, $f$ is continuous on $(0,\infty)$.
\paragraph{}
Finally, we show that $f$ is unbounded. For any $M > 0$, we will find a point $x \in (0,\infty)$ such that $f(x) > M$. Let $k = \lceil 2M \rceil$. Then choose $x = \frac{1}{k^2}$. Thus for $n \le k$, $f_n(x) = \frac{1}{1+n^2\frac{1}{k^2}} \ge \frac{1}{2}$. So $f(x) = \sum_{n=1}^\infty f_n(x) > \sum_{n=1}^k f_n(x) \ge \frac{1}{2}k \ge M$. Thus $f$ is unbounded.
\end{ppart}

\begin{ppart}{Problem 5}
We first show that if $\{f_n\} = \{g_n + h_n\}$ and $\{g_n\}$ and $\{h_n\}$ are uniformly convergent sequences of functions, then $\{f_n\}$ is also uniformly convergent. Let $g$ and $h$ be the limit functions of $\{g_n\}$ and $\{h_n\}$ respectively. We will show that $\{f_n\}$ converges uniformly to $g+h$. Let $\epsilon > 0$ be given. By the uniform convergence of $\{g_n\}$ we can find a $N_g$ such that $\forall n \ge N_g~\forall x ~ \abs{g_n(x)} \le \frac{\epsilon}{2}$. We can find a $N_h$ defined equivalently for $h_n$. Then let $N = \max\{N_g, N_h\}$ and $\forall n \ge N$,  consider $\abs{f_n(x) - (g(x) + h(x))} = \abs{g_n(x) + h_n(x) - g(x) - h(x)} \le \abs{g_n(x) - g(x)} + \abs{h_n(x) - h(x)} \le \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon$. So $\{f_n\}$ converges uniformly to $g+h$. By considering the sequences of partial sums, we easily see that this theorem applies to series as well: if $\sum g_n$ and $\sum h_n$ converge uniformly, and $f_n = g_n + h_n$ then $\sum f_n$ converges uniformly.

\paragraph{}
We use this result to show that $\sum_{n=1}^\infty (-1)^n \frac{x^2+n}{n^2}$ converges uniformly in every bounded interval. Observe that this series can be written as the sum of two series: $\sum_{n=1}^\infty (-1)^n\frac{x^2}{n^2} + \sum_{n=1}^\infty (-1)^n\frac{n}{n^2}$. To show that it converges uniformly in every bounded interval, we need to show that both of these two series converge uniformly. The first series is $\sum_{n=1}^\infty (-1)^n\frac{x^2}{n^2}$. Since we are considering a bounded interval, $\exists M \in \R$ such that $\forall x ~ \abs{x}\le M$. Thus $\abs{(-1)^n\frac{x^2}{n^2}} \le \frac{M^2}{n^2}$. Since $M$ is constant, $\sum \frac{M^2}{n^2}$ converges, so by the Weierstrass m-test $\sum_{n=1}^\infty (-1)^n\frac{x^2}{n^2}$ converges uniformly on any bounded interval.
\paragraph{}
The second series is $\sum_{n=1}^\infty \frac{(-1)^n}{n}$, which is independent of $x$. As a series in $n$, it converges (it is the alternating harmonic series). Since it is independent of $x$, it certainly converges uniformly on any interval. By our result above, since $\sum_{n=1}^\infty (-1)^n \frac{x^2+n}{n^2}$ is the sum of two series that converge uniformly on any bounded interval, it converges uniformly on any bounded interval.
\paragraph{}
$\sum_{n=1}^\infty (-1)^n \frac{x^2+n}{n^2}$ does not converge absolutely for any value of $x$. Observe that $\abs{(-1)^n \frac{x^2+n}{n^2}} = \frac{x^2+n}{n^2} = \frac{\frac{x^2}{n}+1}{n}$. Thus for any $x$ the resulting series is termwise greater than or equal to $\frac{1}{n}$, which diverges, so by the comparison test it diverges.
\end{ppart}
\end{document}