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\centerline{Mathematics 242 -- Principles of Analysis}
\centerline{Selected Solutions -- Problem Set 8}
\centerline{November 22, 2004}
\bigskip
\noindent
Section 25/16 a)  We assume that $f$ is differentiable
at $c$, so according to the definition used in the
book, $\lim_{x \to c} {f(x) - f(c)\over x - c} = f'(c)$
exists.  Introduce the new quantity $h = x - c$.
then $x = c + h$, and the limit as $x \to c$ is 
the same as the limit as $h \to 0$.  Hence
the above limit becomes
$$f'(c) = \lim_{h\to 0} {f(c+h) - f(c) \over h}.$$

b)  From part a, we have since $f$ is differentiable
at $c$,
$f'(c) = \lim_{h\to 0} {f(c+h) - f(c) \over h}$.
Replacing $h$ by $-h$ in the above, we also have
$f'(c) = \lim_{h\to 0} {f(c-h) - f(c) \over -h}
= \lim_{h\to 0} {f(c) - f(c-h)\over h}$.
Hence by the limit sum rule,
$$\eqalign{f'(c) &= {1\over 2} 
\left(\lim_{h\to 0}{f(c+h) - f(c)\over h} + 
\lim_{h\to 0} {f(c) - f(c-h)\over h}\right)\cr
&=\lim_{h\to 0} {f(c+h) - f(c) + f(c) - f(c-h)\over 2h}\cr
&=\lim_{h\to 0} {f(c+h) - f(c-h)\over 2h},\cr}$$
which is what we wanted to show.

c)  Since $f$ is differentiable at $c$, by part a,
$f'(c) = \lim_{h\to 0}{f(c + h) - f(c)\over h}$.
By Theorem 20.8, if we take a sequence $h_n$ of $h$-values
converging to 0, with $h_n \ne 0$ for all $n$, 
then 
$f'(c) = \lim_{n\to \infty}{f(c + h_n) - f(c)\over h_n}$.
So let $h_n = 1/n$.  We then have 
$$f'(c) = \lim_{n\to \infty}{f(c + 1/n) - f(c)\over 1/n}
= \lim_{n\to \infty} n(f(c + 1/n) - f(c)),$$
which is what we wanted to show.

d)  The converse of part b would say: If
$\lim_{h\to 0} {f(c+h) - f(c-h)\over 2h}$ exists,
then $f$ is differentiable and this limit gives 
the derivative of $f$ at $c$.  To see that this
is false, consider an {\it even function} that
is not differentiable at $c = 0$, for example,
$f(x) = |x|$.  We have
$$\lim_{h\to 0} {|0+h| - |0-h|\over 2h} 
=\lim_{h\to 0} {|h| - |h|\over 2h} = 0.$$
(Geometrically, the secant lines through equally
spaced points on either side of zero are always
horizontal, so this limit is zero.)
But we know from problem 25.7 that $f'(0)$ 
does not exist.

The converse of part c would say: If   
$\lim_{n\to \infty} n(f(c+1/n) - f(c))$ exists,
then $f$ is differentiable and this limit gives 
the derivative of $f$ at $c$.  To see that this
is false, consider a function that
is not differentiable at $c = 0$, but
whose difference quotient {\it does have a 
well-defined one-sided limit} as $h\to 0^+$
(note that $1/n > 0$, so this limit basically
only takes into account what is happening
to the right of $c$ on the graph of $f$). 
For example, consider 
$f(x) = |x|$ again.  We have
$$\lim_{h\to 0+} {|0+h| - |0|\over h} 
=\lim_{h\to 0^+} 1 = 1.$$
But we know from problem 25.7 that $f'(0)$ 
does not exist because the other one-sided
limit gives $-1$.

\bigskip
\noindent
Section 25/6. First a comment about a proposed solution
that a number of people tried to use.  If you assume
that $f$ is increasing on $I$ and take $x_1 < x_2$, 
then it is true that the Mean Value Theorem
gives $f'(c) = {f(x_2) - f(x_1) \over x_2 - x_1} \ge 0$
for some $c$ between $x_1$ and $x_2$.  Unfortunately,
there is no reasonable way, given what we know, to use this
to show that $f'(x) \ge 0$ for all $x \in I$.  The problem
is that we have no ``control'' over {\it which} $c$'s 
appear in these equations.  There's nothing that says 
every $x\in I$ has to appear in one of them.   

So a complete solution for part a has to proceed a different 
way.  First we will show that if $f$ is increasing 
on $I$ then $f'(x) \ge 0$ for all $x \in I$ by a 
direct argument.  Take any $x \in I$.  Then the 
definition of $f'(x)$ says 
$f'(x) = \lim_{x_1 \to x} {f(x_1) - f(x)\over x_1 - x}$.
Since we assume $f$ is increasing, the difference
quotient ${f(x_1) - f(x)\over x_1 - x} \ge 0$ for
all $x_1$.  Any limit of a nonnegative function 
is nonnegative (see Exercise 20.12, for instance).
Hence, $f'(x) = \lim_{x_1\to x}{f(x_1) - f(x)\over x_1 - x} \ge 0$.
Next, we show the reverse implication, and this is where
the Mean Value Theorem comes in.  Assume $f'(x) \ge 0$
for all $x \in I$, and let $x_1 < x_2$ in $I$.  Applying
the MVT to $f$ on the interval $[x_1,x_2]$, we get that 
there exists some $c \in (x_1,x_2)$ such that
$0 \le f'(c) = {f(x_2) - f(x_1)\over x_2 - x_1}$.
Since $x_2 > x_1$, this implies $f(x_2) \ge f(x_1)$. 

Part b is parallel to this and will be omitted here.

\bigskip
\noindent
Section 26/8.  Since $f$ is assumed differentiable
on $(0,1)$, by the quotient rule for derivatives,
$g(x) = f(x)/x$ is also differentiable on $(0,1)$,
and 
$$g'(x) = {xf'(x) - f(x) \over x^2}$$
for all $x \in (0,1)$. By Exercise 26.6 above, it suffices
to show that $g'(x) \ge 0$ for all $x\in (0,1)$. 
By the MVT, for all $x\in (0,1)$, there exists
a $c \in (0,x)$ such that 
$f'(c) = {f(x) - f(0) \over x - 0} = {f(x)\over x}$.
Since $f'$ is assumed to be increasing, since $x > c$,
we have $f'(x) \ge f'(c)$.  Hence
$f'(x) \ge {f(x)\over x}$ for all $x \in (0,1)$. 
But then 
$${x f'(x)\over x^2} \ge {f(x)\over x^2} \Rightarrow
{x f'(x) - f(x) \over x^2} \ge 0$$
which shows that $g'(x) \ge 0$ for all $x \in (0,1)$.

\bye

