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\centerline{Mathematics 242 -- Principles of Analysis}
\centerline{Selected Solutions -- Problem Set 6, ``B'' Portion}
\centerline{March 25, 2011}
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{\it A General Comment:}  The typical solution that many of you are 
submitting now for many of the proof-type problems is
more like {\it a description of how you (or we :) ) 
came to find the idea for the proof than what the proof itself should look like}.
Some are too wordy, some do not explain enough, and some are  
not organized correctly. From this point on in the semester we will 
be working on moving beyond this level to practice the way mathematicians 
actually present proofs like this.  That style might appear to be terse 
and uninformative at times, but it is the standard, so you need to learn 
how to do it. 

I think one reason for the style that your proofs are ``coming out in'' is that 
some of you are attempting to do the scratch work for the problem set
on the same set of sheets that you eventually hand in.  Unless you 
are a real wiz at this sort of mathematics, that is not going to work for some
problems -- especially the more complicated proofs.
 What this means of course, 
is that you will probably want to start {\it delaying the final writeup on 
problem sets until you have worked out the ideas, and only write things 
up after you understand the ideas.  You may {\bf want to keep both the 
notes you made ``on the way'' to the proof and the final product}.  
That way you can reconstruct how you got to the final state if you need to.}
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\S 3.2/3.  To show:  $\lim_{x\to c} x^n = c^n$ directly from the definition.
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{\bf Proof:}  Given $\varepsilon$, let $\delta = \min\left(1,{\varepsilon\over (n+1)(|c|+1)^{n-1}}\right)$.
Then for all $x$ with $0 < |x - c| < \delta$, we have $|x - c| < 1$, so $|x| < |c| + 1$.  
From the factorization of a difference of $n$th powers and the triangle inequality,
$$\eqalign{
|x^n - c^n| &= |x - c||x^{n-1} + x^{n-2} c + \cdots + c^{n-1}|\cr
           &\le |x - c|\left(|x|^{n-1} + |x|^{n-2} |c| + \cdots + |c|^{n-1}\right).\cr
 &\le |x - c| (n+1)(|c| + 1)^{n-1}\cr}$$
But then for all $x$ with $|x - c| < \delta$, this is:
$$\eqalign{ 
 &< {\varepsilon\over (n+1)(|c| + 1)^{n+1}} (n+1)(|c|+1)^{n-1}\cr
 &= \varepsilon.\cr}$$
This shows that $\lim_{x\to c} x^n = c^n$.  
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7.  We assume that $f(x) \ge 0$ on some deleted neighborhood of $c$.
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(a)  To show:  If $\lim_{x\to c} f(x) = 0$, then $\lim_{x\to c} \sqrt{f(x)} = 0$.  
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{\bf Proof:}  Since $\lim_{x\to c} f(x) = 0$, for all $\varepsilon > 0$, there exist
corresponding $\delta > 0$ such that $|f(x) - 0| = f(x) < \varepsilon^2$ for all
$x$ with $0 < |x - c| < \delta$.  But then for the same $\delta$ and all these $x$,
$|\sqrt{f(x)} - 0| = \sqrt{f(x)} < \varepsilon$ (where we have used \S 1.3 Exercise 8).
This shows $\lim_{x\to c}\sqrt{f(x)} = 0$.  
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Comment:  Notice how this proof is laid out: It is a linear, step-by-step demonstration that
the definition of $\lim_{x\to c} \sqrt{f(x)} = 0$ is satisfied.  The first statement is
OK since the definition of the limit says ``for all positive numbers ... .''  This is why 
we can get $f(x) < \varepsilon^2$ for any given $\varepsilon > 0$.   This style is what
you should be aiming for.  I cut out all of the explanations (except the justifications
for the individual steps) and just worked through to the conclusion $|\sqrt{f(x)} - 0| < \varepsilon$
for the suitable $\delta$.  
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(b) Now we will show that if $\lim_{x\to c} f(x) = L > 0$, then $\lim_{x\to c} \sqrt{f(x)} = \sqrt{L}$.
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{\bf Proof:}  By the assumption on the limit of $f$, given $\varepsilon > 0$, 
there exists $\delta > 0$ such that
$|f(x) - L| < \varepsilon\sqrt{L}$ for all $x$ with $0 < |x - c| < \delta$. For that
same $\delta$ and all $x$ with $0 < |x - c| < \delta$, using the difference of 
squares factorization, 
$$\eqalign{
|\sqrt{f(x)} - \sqrt{L}| &= {|f(x) - L| \over |\sqrt{f(x)} + \sqrt{L}|}\cr
&\le {|f(x) - L| \over |\sqrt{L}|} \quad ({\rm since } f(x) \ge 0)\cr
&\le {\varepsilon\sqrt{L}\over \sqrt{L}}\cr
&=\varepsilon.\cr}$$
This shows $\lim_{x\to c} \sqrt{f(x)} = \sqrt{L}$.  
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\S 3.4/3. (a)  We assume $f$ and $g$ are continuous on an interval $I$ containing $c$.
If $f(c) \ne g(c)$, then we either have $f(c) > g(c)$ or vice versa.  We will show that
$h(x) = \max(f(x),g(x))$ and $k(x) = \min(f(x),g(x))$ are continuous at $c$ under
the assumption $f(c) > g(c)$.  If the reverse inequality is true, the proof is similar
but the maximum and minimum are interchanged.  By problem 2, we know that if $f(c) > g(c)$,
then there exists an interval $I'$ contained in $I$ and containing $c$ such that $f(x) > g(x)$
for all $x\in I'$.  This means that $f(x) = h(x)$ and $g(x) = k(x)$ for all $x \in I'$.  
Since the limits depend only on the values of the function near $c$, this means (using
continuity of $f$ and $g$):
$$\lim_{x\to c} h(x) = \lim_{x\to c} f(x) = f(c) = h(c)$$
and 
$$\lim_{x\to c} k(x) = \lim_{x\to c} g(x) = g(c) = k(c).$$
Hence both $h$ and $k$ are continuous at $c$ by definition.  

Now suppose $f(c) = g(c) = b = h(c) = k(c)$.  Since $f$ and $g$ are continuous at $c$, given 
$\varepsilon > 0$, there exist $\delta_1$ and $\delta_2$ such that 
$|f(x) - b| < \varepsilon$ for all $x$ with $|x - c| < \delta_1$ and 
$|g(x) - b| < \varepsilon$ for all $x$ with $|x - c| < \delta_2$.  If $\delta = \min(\delta_1,\delta_2)$,
then for all $x$ with $|x - c| < \delta$, we have both
$$|f(x) - b| < \varepsilon \quad {\rm and} \quad |g(x) - b| < \varepsilon.$$
This implies 
$$|h(x) - b| < \varepsilon \quad {\rm and} \quad |k(x) - b| < \varepsilon,$$
since for every $x$, $h(x)$ is either $f(x)$ or $g(x)$ and similarly for $k(x)$.  
This shows that both $h$ and $k$ are continuous at $c$.  

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{\it Comment:}
Although it is tempting to try to do something like
what we did above in the other part of (a), 
to get a general proof in the case $f(c) = g(c)$, we {\it really do need} to proceed by the 
different argument used here.
The reason is that {\it there need not be any intervals containing $c$ on which the 
maximum or minimum is given by the same function between the two possibilities
$f$ or $g$.  In particular, the graphs of $f$ and 
$g$ can cross at $x$'s arbitrarily close to $c$ so the max and min can be changing infinitely often
in all intervals containing $c$}.   For an example of this type, consider $f(x) = 0$
and $g(x) = x\sin(1/x)$ if $x \ne 0$, with $g(0)$ defined as $g(0) = 0$.  Then 
the minimum and maximum alternate between $f$ and $g$ infinitely often in every
interval containing $c = 0$.  
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(b)  For this part, we will use the problem in Chapter 1 to give different formulas
for the maximum and minimum:
$$h(x) = \max(f(x),g(x)) = {f(x) + g(x)\over 2} + {|f(x) - g(x)|\over 2}$$
and 
$$k(x) = \min(f(x),g(x)) = {f(x) + g(x)\over 2} - {|f(x) - g(x)|\over 2}.$$
We want to show this implies $h$ and $k$ are continuous at $c$, under the assumption
that $f$ and $g$ are continuous at $c$.  This follows from Theorems 3.4.10 and 3.4.11,
since we are taking sums, differences, and compositions of continuous functions to get these
formulas for the maximum and minimum.
\bye
 
