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\begin{center}
{\bf Homework Set 7: Solutions}
\end{center}
\vspace*{.3in}

\noindent{\bf Page 100, \# 17, solution.}
Let $x$ be a point at which $f$ has a simple discontinuity.  Then exactly
one of the following is true.
\begin{enumerate}
\item $f(x+) < f(x-)$;
\item $f(x+) > f(x-)$;
\item $f(x+) = f(x-) < f(x)$;
\item $f(x+) = f(x-) > f(x)$.
\end{enumerate}
It will be enough to show that each of these things, taken separately, 
occurs at no more than countably many values of $x$.

So let us first consider those $x$ as in case (1).  By the density property 
we can choose $p\in\Q$ such that $f(x+) < p < f(x-)$.  Then we set 
$$
\epsilon = \min\{f(x-) - p,p - f(x+)\}
$$
By definition of the left and right hand limits $f(x-)$ and $f(x+)$, there
exists $\delta > 0$ so that 
\begin{itemize}
\item $x-\delta < t < x$ implies $|f(x) - f(x-)| < \epsilon$ (so, 
  in particular, $f(x) < f(x-) + \epsilon \leq p$);
\item $x < t < x + \delta$ implies $|f(x) - f(x+)| < \epsilon$ (so,
  in particular, $f(x) > f(x+) - \epsilon \geq p$).
\end{itemize}
Invoking the density property again, we can choose $q,r\in\Q$ such that
$$
x - \delta < q < x < r < x + \delta,
$$
thereby associating to $x$ a triple $(p,q,r)\in\Q^3$ such that 
\begin{itemize}
\item $q<x<r$,
\item $f(t) < p$ for all $t\in (q,x)$, and 
\item $f(t) > p$ for all $t\in (x,r)$.
\end{itemize}
I claim now that if $x'$ is another point at which $f$ has a simple 
discontinuity of type (1), then (at least one member of) the triple 
$(p',q',r')$ associated to $x'$ differs from the triple $(p,q,r)$ 
associated to $x$.  To see this suppose that $x'< x$ (the case $x < x'$ is 
identical) but $(p',q',r') = (p,q,r)$.  Then $q<x'<x < r$, so if $t$ is
any point between $x'$ and $x$, we have $f(t) > p$ because $x'<t<r$ but
$f(t)<p$ because $q<t<x$.  Since this is impossible, it must be the case
that $(p',q',r') \neq (p,q,r)$.

In summary we have defined an injective function from the set of points
where $f$ has a simple discontinuity of type (1) into $\Q^3$.  Since 
the latter set is countable, we conclude that $f$ has no more than countably
many simple discontinuities of type (1).

The case of simple discontinuities of type (2) is handled in a completely 
analogous fashion.  Dealing with simple discontinuities of type (3)
differs only slightly in that we choose $(p,q,r)\in\Q^3$ such that
$f(t) < p$ for all $t\in (q,x)\cup (x,r)$.  Finally simple discontinuities
of type (4) are dealt with in the same way as those of type (3).
\qed


\newpage

\noindent{\bf Page 101, \# 23, solution.}
Beginning with a convex function $f:(a,b)\to\R$, let us first prove that
$g\circ f$ is also convex whenever $g:(c,d)\to\R$ is a convex increasing 
function whose domain includes the range of $f$.  If 
$x<y$ and $\lambda \in (0,1)$, then by definition of continuity, we have
$$
f(\lambda x + (1-\lambda) y) \leq \lambda f(x) + (1-\lambda) f(y).
$$
Thus since $g$ is increasing, we obtain
$$
g\circ f(\lambda x + (1-\lambda) y) 
\leq 
g(\lambda f(x) + (1-\lambda) f(y)).
$$
Finally, applying the fact that $g$ is convex to the right side of this
inequality gives
$$
g\circ f(\lambda x + (1-\lambda) y) 
\leq 
\lambda g(f(x)) + (1-\lambda)g(f(y)).
$$
We conclude that $g\circ f$ is a convex function.  Ta da!

Now we establish the `slope inequality' given in the problem for $f$, because it
will be useful in proving that $f$ is continuous.  That is, if $s<t<u$ are 
numbers in $(a,b)$, we will prove that 
$$
\frac{f(t) - f(s)}{t-s} 
\leq 
\frac{f(u) - f(s)}{u-s} 
\leq 
\frac{f(u) - f(t)}{u-t}
$$
Since $t$ is between $s$ and $u$, there exists $\lambda\in(0,1)$ such that
$$
t = \lambda s + (1-\lambda) t,
$$
which implies that
$$
t - s = (1-\lambda)(u-s)
$$
Moreover, by definition of convexity, we have
$$
f(t) \leq \lambda f(s) + (1-\lambda) f(u),
$$
which, after subtracting $f(s)$ from both sides becomes
$$
f(t) - f(s) \leq (1-\lambda)(f(u) - f(s)) \leq (t-s)\frac{f(u) - f(s)}{u-s}.
$$
Dividing through by $t-s$ then gives the first inequality above.  The proof
of the second inequality is similar.  Cha-ching!

Now let $x\in(a,b)$ be any given point.  We will show that $f$ is continuous
at $x$.  Note that by the inequalities we just proved, the function
$$
m(t) := \frac{f(t) - f(x)}{t-x}
$$
is increasing in $t\neq x$ (verifying that $m(t_1) \leq m(t_2)$ for $t_1 < t_2$
requires applying our inequality for $s<t<u$ to each of the three cases 
$x < t_1 < t_2$, $t_1 < x < t_2$, and $x < t_1 < t_2$).

So fix numbers $A<x<B$
in $(a,b)$.  Then $m(A) \leq m(t) \leq m(B)$ for all $t\in (A,B)$ not equal to
$x$.  In particular, taking $C = \max\{|m(A)|,|m(B)|\}$, we see that 
$|m(t)| \leq C$ for all $t\in (A,B)$.  That is,
$$
|f(t) - f(x)| \leq C|t-x|.
$$

Now if $\epsilon > 0$ is given, we take $\delta = \min\{\epsilon/C,x-A,B-x\}$.
Then $0< |t-x| < \delta$ implies that $A<t<B$, and therefore
$$
|f(t) - f(x)| \leq C|t-x| < C\delta \leq \epsilon.
$$
This shows that $\lim_{t\to x} f(t) = f(x)$---i.e. $f$ is continuous at $x$.
As $x$ is arbitrary $f$ is continuous on $(a,b)$.
\qed
\newpage

\noindent{\bf Page 114, \# 6, solution.}
Since $f$ is differentiable, so is $g$.  We will show that $g'(x)\geq 0$ for 
every $x > 0$.  Then if $0 \leq x_1 < x_2$, the mean value theorem gives us a 
number $c\in (x_1,x_2)$ such that 
$$
g(x_2) - g(x_1) = g'(c)(x_2 - x_1) \geq 0,
$$ 
so that $g$ is increasing, as desired.

Now
$$
g'(x) = \frac{x f'(x) - f(x)}{x^2}.
$$
Moreover, another application of the mean value theorem gives us $c\in(0,x)$
such that
$$
\frac{f(x)}{x} = \frac{f(x) - f(0)}{x - 0} = f'(c).
$$
But $f'$ is an increasing function, so 
$$
\frac{f(x)}{x} \leq f'(x).
$$
Rearranging this and using the fact that $x>0$ gives
$$
\frac{x f'(x) - f(x)}{x^2} \geq 0.
$$
We conclude that $g'(x) \geq 0$, and therefore $g$ is increasing.
\qed
\vspace*{.3in}

\noindent{\bf Page 114, \# 9, solution.}
Yes, it does.  Let $\{x_n\}\subset\R-\{0\}$ be any sequence of points 
converging to $0$.
The mean value theorem gives us a second sequence $\{c_n\}$ such that
for every $n\in\N$, $c_n$ is between $x_n$ and $0$ (so by the Squeeze
Theorem $c_n \to 0$) and
$$
\frac{f(x_n) - f(0)}{x_n - 0} = f'(c_n).
$$
It follows that
$$
\lim_{n\to\infty}\frac{f(x_n) - f(0)}{x_n - 0} = \lim_{n\to\infty} f'(c_n) = 3.
$$
Since the sequence $\{x_n\}$ was arbitrary, we conclude from Theorem 4.2
that 
$$
f'(0) = \lim_{x\to 0} \frac{f(x) - f(0)}{x - 0} = 3.
$$
\qed

\noindent{\bf Page 115, \# 13abcd.  Solution}
In all cases, it is only necessary to verify statements at $x=0$ (or in part
(c), in a neighborhood of $x=0$).
\begin{description}
\item{(a)} 
If $a = 0$, then we have already observed in class that 
$\lim_{x\to 0} f(x)$ does not exist, so $f$ cannot be continuous at $x=0$.
if $a < 0$, then $f$ is not even bounded near $0$, so $f$ is not continuous
at $x=0$.  If $a >0$, on the other hand, then $|f(x)| < |x|^a \to 0$ as
$x\to 0$.  So the squeeze theorem implies that $\lim_{x\to 0} f(x) = 0 =
f(0)$, and $f$ is continuous at $0$.

\item{(b)}
By definition of the derivative, we have
$$
f'(0) = \lim_{x\to 0} x^{a-1}\sin(x^{-c})
$$
which, as we observed in part (a), exists if and only if $a-1>0$.

\item{(c)}
Note that by part (b) we must have $a>1$.  
For $x\neq 0$, we have
$$
f'(x) = ax^{a-1}\sin(x^{-c}) + cx^{a-c-1}\cos(x^{-c}).
$$
The first term is bounded since $a > 1$, and the second is bounded
if and only if $a \geq c+1$.

\item{(d)}
This follows immediately from parts (a) (which works for the cosine as well
as the sine function) and (c).
\end{description}
\vspace*{.3in}

\noindent{\bf Page 115, \# 14.  Solution}
Let $f:(a,b)\to\R$ be a differentiable a function.  Supposing first that
$f'$ is monotonically increasing, we will show that $f$ is convex.  Given
numbers $x < y$ in $(a,b)$ and $\lambda\in(0,1)$, let
$$
z = \lambda x + (1-\lambda) y.
$$
Then the mean value theorem gives us $c_1\in(x,z)$, $c_2\in(z,y)$ such that
$$
f(z) - f(x) = f'(c_1)(z-x), \qquad 
f(y) - f(z) = f'(c_2)(y-z).
$$
Rewriting $z$ in terms of $x$, $y$, and $\lambda$ on the right sides of
both equations, and using the fact that $f'(c_2) \geq f'(c_1)$ gives
$$
\lambda(f(z) - f(x)) \leq (1-\lambda)(f(y) - f(z)),
$$
which, upon solving for $f(z)$, yields
$$
f(z) \leq \lambda f(x) + (1-\lambda) f(y).
$$
This shows that $f$ is convex.

Now let us begin again supposing that $f$ is convex, and trying to show that 
$f'$ is monotonically increasing.  That is, if $x<y$ are two points in
$(a,b)$, we seek to prove that $f'(x)\leq f'(y)$.  To do this let $t$ be any
point strictly between $x$ and $y$.  Then our `slope inequality' from
page 101/\# 23 tells us that
$$
\frac{f(t) - f(x)}{t-x} \leq 
\frac{f(y) - f(x)}{y-x} \leq 
\frac{f(y) - f(t)}{y-t}.
$$
Letting $t\to x$ in the left and middle expressions gives
$$
f'(x) \leq \frac{f(y) - f(x)}{y-x}.
$$
Letting $t\to y$ in the middle and right expressions gives
$$
\frac{f(y) - f(x)}{y-x}\leq f'(y).
$$
Combining the two inequalities gives $f'(x)\leq f'(y)$, as desired.

Finally, if $f''$ exists on $(a,b)$, we note that $f'$ is increasing if and
only if $f''(x)\geq 0$ at every $x$.  So by our work above,
$f$ is convex if and only if $f''$ is non-negative on $(a,b)$.
\qed
\vspace*{.3in}

\noindent{\bf Page 116,\# 19.  Solution:}
For parts (a) and (b) it is useful to note that
$$
\frac{f(\beta_n) - f(\alpha_n)}{\beta_n - \alpha_n}
= 
\frac{f(\beta_n) - f(0)}{\beta_n - 0}\frac{\beta_n}{\beta_n - \alpha_n}
-
\frac{f(\alpha_n) - f(0)}{\alpha_n - 0}\frac{\alpha_n}{\beta_n - \alpha_n}
$$
Therefore, if we compare with the derivative of $f$ at zero, we can
use the fact that
$$
f'(0) = f'(0) \frac{\beta_n}{\beta_n - \alpha_n} - 
        f'(0) \frac{\alpha_n}{\beta_n - \alpha_n}
$$
together with the triangle inequality to obtain
$$
\left|\frac{f(\beta_n) - f(\alpha_n)}{\beta_n - \alpha_n} - f'(0)\right|
\leq
\left|\frac{f(\beta_n) - f(0)}{\beta_n - 0} - f'(0)\right|
\left|\frac{\beta_n}{\beta_n - \alpha_n}\right|
+
\left|\frac{f(\alpha_n) - f(0)}{\alpha_n - 0} - f'(0)\right|
\left|\frac{\alpha_n}{\beta_n - \alpha_n}\right|
$$
Therefore if both $\{\beta_n/(\beta_n - \alpha_n)\}$ and
$\{\alpha_n/(\beta_n - \alpha_n)\}$ are bounded, we conclude that
$$
\lim_{n\to\infty} 
\left|\frac{f(\beta_n) - f(\alpha_n)}{\beta_n - \alpha_n} - f'(0)\right|
= 
0.
$$
We now deal with parts (a) and (b) in light of this discussion.
\vspace*{.1in}

\noindent{\bf(a)} If $\alpha_n < 0 < \beta_n$, then $\beta_n - \alpha_n$ is
larger than both $|\beta_n|$ and $|\alpha_n|$.  Hence,
$$ 
\left|\frac{\beta_n}{\beta_n - \alpha_n}\right|, 
\left|\frac{\alpha_n}{\beta_n - \alpha_n}\right| < 1
$$
for every $n\in\N$.  It follows immediately, then, from the discussion 
above that
$$
\lim_{n\to\infty} 
\frac{f(\beta_n) - f(\alpha_n)}{\beta_n - \alpha_n} = f'(0).
$$
\qed

\vspace*{.1in}
\noindent{\bf(b)}
By assumption, there exists $C\in\R$ such that
$|\beta_n/(\beta_n - \alpha_n)| \leq C$ for all $n\in\N$. Thus
$$
\left|\frac{\alpha_n}{\beta_n - \alpha_n}\right| 
\leq 
\left|\frac{\alpha_n - \beta_n}{\beta_n - \alpha_n}\right| 
+
\left|\frac{\beta_n}{\beta_n - \alpha_n}\right|
\leq 1 + C
$$
for every $n\in\N$.
It therefore follows again from the discussion preceding part (a) that
$$
\lim_{n\to\infty} 
\frac{f(\beta_n) - f(\alpha_n)}{\beta_n - \alpha_n} = f'(0).
$$
\qed
\vspace*{.1in}

\noindent{\bf(b)}
By the mean value theorem, we have for every $n\in\N$ a number
$c_n \in (a_n,b_n)$ such that
$$
f'(c_n) = \frac{f(\beta_n) - f(\alpha_n)}{\beta_n - \alpha_n}
$$
Since $a_n,b_n\to 0$, it follows from the Squeeze Theorem that $c_n\to 0$.
And since we are assuming now that $f'$ is continuous at 0, we have
$$
\lim_{n\to\infty} \frac{f(\beta_n) - f(\alpha_n)}{\beta_n - \alpha_n}
= 
\lim_{n\to\infty} f'(c_n)
= 
f'( \lim_{n\to\infty} c_n )
= 
f'(0),
$$
as advertised.
\qed

\newpage
\noindent{\bf Page 117,\# 22abc. Solution:}
\vspace*{.1in}

\noindent{\bf(a)}
Suppose by way of contradiction that $x<y$ are distinct fixed points
of $f$.  Then the mean value theorem gives us $c\in(x,y)$ such that
$$
f'(c) = \frac{f(y) - f(x)}{y-x} = \frac{y-x}{y-x} = 1
$$
contrary to our assumption that $f'$ is never equal to $1$.  Therefore,
$f$ has at most one fixed point.
\qed
\vspace*{.1in}

\noindent{\bf(b)}
Setting $f(t) = t$ for this particular function $f$ gives us that
$$
e^t = -1,
$$
which is impossible.  Therefore, $f$ has no fixed points.  On the other
hand
$$
f'(t) = 1 - \frac{e^t}{(1+e^t)^2}.
$$
Moreover, since $e^t > 0$ for every $t\in\R$,
$$
0 < \frac{e^t}{(1+e^t)^2} < \frac{e^t+1}{(1+e^t)^2} = \frac{1}{1+e^t} < 1.
$$
That is, $1 > f'(t) > 0$ for all $t \in \R$.
\qed
\vspace*{.1in}

\noindent{\bf(c)}
Let $x_1\in\R$ be any point and $\{x_n\}$ be the sequence determined by
setting $x_{n+1} = f(x_n)$ for every $n\in\N$.  Then by the mean value
theorem 
$$
|x_{n+1} - x_n| = |f(x_n) - f(x_{n-1})| = |f'(c)||x_n - x_{n-1}|
$$
for some $c$ between $x_n$ and $x_{n-1}$.  In particular, 
$$
|x_{n+1} - x_n| \leq A|x_n - x_{n-1}|,
$$
where $A$ is the constant given in the problem.  Applying this inequality
inductively gives
$$
|x_{n+1} - x_n| \leq A^{n-1}|x_2 - x_1|.
$$
We will use this to show that $\{x_n\}$ is a Cauchy sequence.  Let 
$\epsilon > 0$ be given and choose $N\in\N$ large enough that
$\frac{A^{N}}{1 - A} < \epsilon/C$, where $C:= |x_2 - x_1|$.  Then if $m>n\geq N$ we have
$$
|x_m - x_n|
= 
\left|\sum_{j=n}^{m-1} (x_{j+1} - x_j) \right| 
\leq 
\sum_{j=n}^{m-1} |x_{j+1} - x_j|
\leq
C\sum_{j=n}^{m-1} A^j
\leq 
C\sum_{j=n}^\infty A^j
=
C\frac{A^n}{1-A} < \epsilon.
$$
This proves that $\{x_n\}$ is a Cauchy sequence.  Since $\R$ is complete,
we have $x\in\R$ such that $\lim x_n = x$.  Since $f$ is continuous, we
also have
$$
f(x) = f(\lim_{n\to\infty} x_n) = \lim_{n\to\infty} f(x_n)
= \lim_{n\to\infty} x_{n+1} = x.
$$
So $x$ is a fixed point for $f$.  By part (a) there can be no other fixed 
point for $f$, so $x$ does not depend at all on our initial choice of $x_1$
\qed

\vspace*{.1in}
\noindent{\bf Alternative Proof (taken shamelessly from Jenista's hwk):}
Note that by hypothesis, we have
$$
|f(x) - f(0)| \leq A|x - 0| = A|x|
$$
for every $x\in\R$.  So suppose, for instance that $f(0) > 0$ (the
case $f(0) < 0$ is similar, and if $f(0) = 0$, then 0 itself is
the fixed point).  Then the above inequality gives us for $x>0$
that 
$$
f(x) - f(0) \leq Ax 
$$
(i.e. the graph of $f$ stays below the line $y= f(0) + Ax$ which
has slop less than $1$).  Therefore,
$$
f(x) - x \leq f(0) + (A-1)x
$$
for $x > 0$.  In particular, since $A-1<0$, we have $f(b) - b < 0$
for $b$ large.  But $f(x) - x$ is continous and $f(0) - 0 > 0$.
So by the intermediate value theorem, there exists $a\in (0,b)$
such that $f(a) - a = 0$.  We conclude that there exists a fixed
point $x=a$ (which is unique by part (a)).  

Now if $x = x_1$ is some other point and $x_{n+1} = f(x_n)$,
we have
$$
|x_{n+1} - a| = |f(x_n) - f(a)| = |f'(c)||x_n - a|
$$
for some $c$ between $x_n$ and $a$ by the Mean Value Theorem.  
But this means that
$$
|x_{n+1} - a| \leq A|x_n - a| \leq \dots \leq A^n|x_1 - a|
$$
for all $n\in\N$.  Since $\lim_{n\to\infty} A^n = 0$, we
conclude that
$$
\lim_{n\to\infty} x_n = \lim_{n\to\infty} x_{n+1} = a.
$$






\end{document}
\end
