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\centerline{MATH 242 -- Principles of Analysis}
\centerline{Selected Solutions for Problem Set 7, ``B portion''}
\centerline{April 8, 2011}
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\noindent
\S 3.5/15.  A comment first:  A fairly common idea that people
tried to use in solutions for (a) was this.  Let $0 < c < 1$.
Then $0 < c^2 < 1$ as well.  Since $f(0) < 0$ and $f(1) > 1$,
this means $c^2$ is between $f(0)$ and $f(1)$.  So by the Intermediate
Value Theorem there is
some $x_0 \in [0,1]$ where $f(x_0) = c^2$.  The {\it problem
with this} is that the question asked you to show that there
is some $c$ where $f(c) = c^2$.  The $x_0$ above will not
usually equal $c$.  (Indeed that statement shows 
$\forall c \in [0,1], \exists x_0 \in [0,1]$ s.t. $f(x_0) = c^2$.)  
The correct way to apply the IVT here is to apply it {\it 
to a different function}.  We want to find $c$ where $f(c) = c^2$, 
so $f(c) - c^2 = 0$.

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\noindent
(a)  Let $g(x) = f(x) - x^2$.  By Theorem 3.4.10, this is continuous 
on $[0,1]$ since $f$ is assumed to be continuous and $x^2$ is continuous 
everywhere.  Using the assumptions on $f$, we have
$$\eqalign{g(0) &= f(0) - 0 = f(0) < 0\cr
           g(1) &= f(1) - 1 > 0\cr}$$
Hence since $k = 0$ is a value between $g(0)$ and $g(1)$, there
is a $c \in (0,1)$ where $g(c) = f(c) - c^2 = 0$.  This shows
$f(c) = c^2$.  
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\noindent
(b)  Let $h(x) = f(x) - g(x)$.  By Theorem 3.4.10, this is continuous 
on $[0,1]$ since $f,g$ are assumed to be continuous.  Since $f(0) < 0$, 
$g(0) \ge 0$, $f(1) > 1$, and $g(1) \le 1$, we have
$$\eqalign{h(0) &= f(0) - g(0) < 0\cr
           h(1) &= f(1) - g(1) > 0\cr}$$
Hence since $k = 0$ is a value between $h(0)$ and $h(1)$, there
is a $c \in (0,1)$ where $h(c) = f(c) - g(c) = 0$.  This shows
$f(c) = g(c)$.
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\noindent  
22.  This is FALSE.  The function 
$$f(x) = {\cos(\pi/x)\over x}$$
on $(0,1]$ is one counterexample.  Notice that at $x_n = {1 \over n}$ we have
$$f(1/n) = n \cos(n\pi) = (-1)^n n.$$
Since $x_n \to 0$ as $n \to \infty$, the interval $(0,1]$ contains
points where $f$ takes both positive and negative values that are arbitrarily
large in absolute value:
$$f(1/n) = \cases{n & if $n$ is even\cr
                  -n & if $n$ is odd\cr}$$
It follows that $f$ has no maximum or minimum on $(0,1]$.
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\noindent
23.  (a)  This is FALSE.  There are many non-continuous functions
that have the ``intermediate value property'' (that is, for which the conclusion
of the IVT holds).  For example, on $[0,1]$, the function
$$f(x) = \cases{2x & if $0 \le x \le 1/2$\cr
           2x - 1 & if $1/2 < x \le 1$\cr}$$
has a jump discontinuity at $x = 1/2$.  But it takes every value
between $f(0) = 0$ and $f(1) = 1$ at least once (most of them twice, in fact).  
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\noindent
(b)  This is FALSE.  Here's a counterexample, and an explanation:
Let 
$$f(x) = \cases{\sin(1/x) & if $x \ne 0$ \cr
                     0 & if $x = 0$\cr}$$
and consider the interval $[a,b] = [-1,1]$. First, $f$ fails to be continuous
at $0$ because $\lim_{x\to 0} f(x)$ does not exist.  But $f$ {\it does 
have the ``intermediate value property'' on {\it every} subinterval} $[c_1,c_2] \subset [-1,1]$.
We can see this as follows.  First, if $0 \ne [c_1,c_2]$, then $f$ is actually 
continuous on $[c_1,c_2]$ and the IVT applies.  Next, if $0 \in [c_1,c_2]$, then 
since $f$ oscillates more and more quickly as $x \to 0$, $f$ must take every
value between $-1$ and $1$ on that interval.  This certainly implies
that $f$ takes every value between $f(c_1)$ and $f(c_2)$, since those
are also in the range $[-1,1]$.  
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\noindent
{\it Comment:}  A lot of people tried to construct a counterexample $f$ by somehow
using a pair $c_1$ and $c_2$ in the {\it definition} of $f$.  That {\it cannot work}
here because the property we want is that $f$ is not continuous, but {\it for 
all $a \le c_1 < c_2 \le b$}, the intermediate value property holds for $f$
on $[c_1,c_2]$.  For instance if you tried to make a counterexample with
jump discontinuities at particular $c_1$ and
$c_2$, then on closed subintervals straddling one of those discontinuities
there can be some $y$-values that are omitted by $f$.  So the intermediate
value property would fail for some intervals of $x$'s.

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\noindent
\S 3.6/4  {\it Comment:}  Many people had the correct idea on this one
after talking to me in office hours.  However, in order to give a complete 
proof that
$$f(x) = \sin(2\pi[x]x)$$
is not uniformly continuous on ${\bf R}$, you have to supply more details
than many people did.  Here is a complete argument.
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\noindent
Aiming for a contradiction, suppose $f$ is uniformly continuous
on ${\bf R}$.  Then for all $\varepsilon > 0$, there exists $\delta > 0$
such that $|f(x) - f(x')| < \varepsilon$ for all $x,x'$ with $|x - x'| < \delta$.
However, take $\varepsilon = 1$.  Assuming there is a $\delta > 0$
that works here, we can find $n \in {\bf N}$ such that ${1\over 4n} < \delta$.  
Consider $x = n$ and $x' = n + {1\over 4n}$.  Then 
$$|x - x'| = {1\over 4n} < \delta,$$
but 
$f(n) = \sin(2\pi n^2) = 0$ and 
$$\eqalign{f\left(n + {1\over 4n}\right) 
&= \sin\left(2\pi \left[n + {1\over 4n}\right]\left(n + {1\over 4n}\right)\right)\cr
&= \sin\left(2\pi n\left(n + {1\over 4n}\right)\right)\cr
&= \sin\left(2\pi n^2 + {\pi\over 2}\right)\cr
&= 1\cr}$$
Hence $|f(x) - f(x')| = 1$, which is not less than $\varepsilon = 1$.  This
contradiction shows that $f$ is not uniformly continuous on ${\bf R}$.    
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\noindent
\S 4.1/11.  {\it Comment:}  Several people overlooked the fact that the statement
to be proved is an ``if and only if'' statement, so both implications must
be shown.
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\noindent
$\Rightarrow$:  Suppose $h(x)$ is differentiable at $c$.  Then the limit of the difference
quotient of $h$:
$$h'(c) = \lim_{x \to c} {h(x) - h(c)\over x - c}$$
must exist.  This is equivalent to saying the two one-sided limits
$$\lim_{x \to c^-} {h(x) - h(c)\over x - c}, \quad \lim_{x \to c^+} {h(x) - h(c)\over x - c}$$
exist and {\it are equal}.  By the definition of $h$, though, 
$$\lim_{x \to c^-} {h(x) - h(c)\over x - c} = \lim_{x \to c^-} {f(x) - f(c)\over x - c},$$
and that limit exists and equals $f'(c)$ since we assume $f$ is differentiable at $c$.
Similarly, 
$$\lim_{x \to c^+} {h(x) - h(c)\over x - c} = \lim_{x \to c^+} {g(x) - g(c)\over x - c},$$
and that limit exists and equals $g'(c)$ since we assume $g$ is differentiable at $c$.
It follows that $f'(c) = g'(c)$.  
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\noindent
$\Leftarrow$:  Conversely, suppose $f'(c) = g'(c)$.  Then taking the limits
of the difference quotients on only one side, we can say:
$$\lim_{x \to c^-} {f(x) - f(c)\over x - c} = \lim_{x \to c^+} {g(x) - g(c)\over x - c}.$$
Because of the way $h$ is defined piecewise, the first limit here is the same as 
$$\lim_{x\to c^-}{h(x) - h(c)\over x - c}$$
and the second is 
$$\lim_{x \to c^+}{h(x) - h(c)\over x - c}.$$
This shows that 
the two one-sided limits
$$\lim_{x \to c^-} {h(x) - h(c)\over x - c}, \quad \lim_{x \to c^+} {h(x) - h(c)\over x - c}$$
exist and {\it are equal}.  Therefore, 
$$h'(c) = \lim_{x \to c} {h(x) - h(c)\over x - c}$$
must exist, so $h$ is differentiable at $c$.  
\bye
 
  
