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{\large {\bf Cheap Set-valued Analysis}}
{\bf 1. The All-or-Nothing Filter}
Most of the time when we want to prove something about inequalities we
make use of the trick of `worsening the estimates': we may know that,
say, $x$ lies in $A$ and $y$ lies in $B$, but we drop this in favour
of weaker facts, $x × A'$ and $y × B'$, for a certain $A' \supset A$
and a certain $B' \supset B$; with these new conditions some
calculations become easier to do, and, after maybe a series of steps
like these we get to the conclusion that we desire, that is something
like $f(x, y) \in W$ (think in $A$, $B$, $A'$, $B'$ and $W$ as being
intervals to get a good mental picture). Here we'll show how a
technique very close to the one in section ... lets us do all
calculations with the variables assuming many values at the same time
(they'll be ``set-valued''), while still retaining all the properties
of the familiar, ``point-valued'' situation.
As in the previous section, we'll start with a simple example and then
proceed to translate it into the usual language. Everything in this
example will be happening over the reals.
Suppose that we have $p(x) := x^4 + a(x) x^3 + b(x) x^2 + c(x) x +
d(x)$, where each of the functions $a$, $b$, $c$ and $d$ is bounded,
and that we want to show formally that for all $|x|$ sufficiently
large (i.e., $ìx \;\; |x| > R$) the sign of $p(x)$ is the same as the
sign of its leading term, $x^4$; moreover, suppose that we want to
obtain an estimate for $R$. Then we can reason like this, if we are
allowed to use set-valued variables:
(start indentation...)
take a value $k$ such that $ìx \;\; |a(x)| \le k, \ldots, |d(x)| \le
k$; without loss of generality we can take $k \ge 1$. We'll show that
$R$ can be taken as $2k$. Let's denote by $s$ a variable that assumes
exactly the values in the interval $[-1,1]$, and by $u$ the
``unsigned'' version of $s$, that assumes all values in the interval
$[0,1]$. We'll ``worsen'' $a(x)$ by replacing it by $ku$, and we'll do
the same for $b$, $c$ and $d$; then we'll show that for any $x$ such
that $|x| \ge 2k$ all values assumed by $ks x^3 + ks x^2 + ks x + ks$
have module smaller than $|x^4|$.
Fix an $x$ with $|x| \ge 2k$, and define $h := \frac x 2$. Then $ks$
``fits into'' $hs$, in the sense the set of values that $ks$ assume is
contained in the set of values of $hs$. It is easy to check that
$a(x) x^3 + b(x) x^2 + c(x) x + d(x)$ \qquad fits into
$ks x^3 + ks x^2 + ks x + ks$ \qquad that fits into
$hs x^3 + hs x^2 + hs x + hs$ \qquad that fits into, and is equal to
$hs (2h)^3 + hs (2h)^2 + hs (2h) + hs$ \qquad that fits into, and is equal to
$8 h^4 s + 4 h^3 s + 2 h^2 s + hs$ \qquad that fits into, and is equal to
$8 h^4 s + 4 h^4 s + 2 h^4 s + h^4 s$ \qquad that fits into, and is equal to
$15 h^4 s$,
\noindent but all the values that $|15 h^4 s|$ take are less than $|16
h^4| = |(2h)^4| = |x^4|$, and so we're done: $a(x) x^3 + b(x) x^2 +
c(x) x + d(x) \in (-|x^4|, |x^4|).$
(...end indentation).
The trick to translate this argument to standard terms is to consider
each $s$ as being a point-valued function of $\ii$; the value of each
$s$ becomes a set when we ``forget at which $\ii$ we were looking at''
and we move on to the full image of $\I$ by the function $s$. However,
each $s$ is a different function of $\ii$, and to get the right
interpretation for
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$$f(s,s) \qquad \text{fits into} \qquad g(s,s)$$
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we have to translate it to
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$$f(s_1,s_2) \qquad = \qquad g(s_3,s_4),$$
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where we omit the parameters $\ii$ from $s_1(\ii), s_2(\ii), s_3(\ii),
s_4(\ii)$; to convey the right meaning we have to choose a function
$\ii \to (s_1, s_2)$ whose image is exactly $[-1,1]^2$, and the
affirmative ``$f(s,s)$ fits into $g(s,s)$'' is then equivalent to
``$ì\ii \; í(s_3, s_4) \ldots$'', or, equivalently, ``$ì(s_1, s_2) \;
í(s_3, s_4) \ldots$''.
There are universal ways to choose an index set and its projections
into the `$s$'s; for the case $f(s_1, s_2) = g(s_3, s_4)$ we may take
$\E_\ii = \setofst{(s_1, s_2, s_3, s_4)}{f(s_1, s_2) = g(s_3, s_4)}$;
if then the natural projection $\ii \to (s_1, s_2)$ does not cover all
of the $[-1, 1]^2$ that will mean that there are pairs $(s_1, s_2)$
for which no good corresponding $(s_3, s_4)$ exist, and so ``$f(s,s)$
fits into $g(s,s)$'' is false. For the big example about $p(x)$ and
$|x^4|$ for all of the steps we have more obvious ways to choose the
new `$s$'s: in ``$a(x) x^3 + b(x) x^2 + c(x) x + d(x)$ fits into $ks_1
x^3 + ks_2 x^2 + ks_3 x + ks_4$'' we can take $s_1 = \frac{a(x)}{k},
\ldots, s_4 = \frac{d(x)}{k}$; for ``$ks_1 x^3 + ks_2 x^2 + ks_3 x +
ks_4$ fits into $hs_5 x^3 + hs_6 x^2 + hs_7 x + hs_8$'' we can take
$s_5 = \frac kh s_1, \ldots, s_8 = \frac kh s_4$, and so on, and this
will be enough to prove that there's an adequate $s_{21}$ such that
$a(x) x^3 + b(x) x^2 + c(x) x + d(x) = 15 h^4 s_{21}$.
\medskip
{\bf 2. Corsets and Continuity}
Still on the context of functions from the reals to the reals let's
define what we'll call a {\sl corset}. Take any non-decreasing
function $h: [0, \infty) \to [0, \infty]$ whose limit when its
parameter goes to zero is zero; the {\sl symmetric corset generated by
$h$} is the set $\setofst{(x,y)}{|y| \le h(x)}$, and we say that a
subset of $\R^2$ is a symmetric corset when there is a function $h$
satisfying the conditions above that generates it. Some examples of
symmetric corsets are the horizontal line, the set
$\setofst{(x,y)}{|y| \le x}$ and sets like the one drawn below:
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$$\eps{corset-small}$$
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It is trivial but boring to define corsets without the symmetry
requirement; they'll be generated by four monotonic functions. We can
also define a {\sl corset over $(x_0, y_0)$} as being a set that
becomes a corset when translated by $-(x_0, y_0)$, a {\sl Lipschitz
corset} as a corset whose $h$ is of the form $\aa x$ for some $\aa \in
[0, \infty)$, a $x\o(x)$ corset as one in which $h(x) = x\o(x)$ for
some function $\o(x)$ from $[0, \infty)$ to $[0, \infty]$ that goes to
zero when its parameter goes to zero, and so on; note that every
corsets (without further attributes) is a $\o(x)$-corsets and
vice-versa.
A function $f: \R \to \R$ that has $f(0)=0$ is continuous at 0 iff its
graph is contained in a corset; a function $f: \R \to \R$ is
continuous iff for every $x$ its graph is contained in a corset over
$(x,f(x))$; it is Lipschitz iff for every $x$ its graph is inside a
Lipschitz corset over $(x,f(x))$, and it is uniformly continuous iff
there's a single corset $C$ such that for every $x$ the graph of $f$
is inside $C+(x,f(x))$, that is $C$ translated to make it into a
$(x,f(x))$-corset.
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