One Sided Range Theorems
Appendices, Volume 3, Why Slopes and More Math.
The graph of a function $y = f(x)$ can be compare with that of a
string.
Preliminary Step: A One Sided String or Chain Theorems
Imagine a vertical rope with knots separated by a distance d or less.
Suppose the distance of all knots from the point of height M is greater
than d. Then all the knots and the string itself lie on one side of the
height M - are all above or below. A proof by mathematical induction
follows. The point or height M is acting as barrier or divided in the
line.
A String Below Theorem: Let M > d > 0. Suppose for 0
< j < k that $y(j)$ are a real numbers with
\[|y(j) - M| >d,\]
\[|y(j+1) - y(j)| < d\]
and $y(0) < M$
then $y(j) < M$ for all integers j where 0 < j <
k.
Proof:
Define a statement $P(q)$ as follows:
\[P(q): y(q) < M-d \]
Observe statement $P(q)$ holds when q = 0 since $|y(0) - M| > d$ and
$y(0) < M $
Assume $P(q)$ holds for some $q < k.$ Then $y(q)< M - d.$
Therefore
\[y(q+1) < y(q)+ d < (M - d) + d < M.\]
Now $|y(q+1) - M| > d$ and $y(q+1) < M$ implies statement $y(q+1)
< M-d$ and hence statement $P(q+1)$ holds. So statement $P(q)$
implies statement $P(q+1).$ That completes the proof by mathematical
induction.
The proof of the following theorem is similar.
A String Above Theorem: Let M > d > 0. Suppose for 0
< j < k that $y(j)$ are a real numbers with
\[|y(j) - M| > d,\]
\[|y(j+1) - y(j)| < d\]
and $y(0) > M$
then $y(j) > M$ for all integers j with 0 < j <
k.
Intermediate Value Theorem for Lipshitz Continuous Functions:
Theorem: Suppose f has a Lipschitz continuity constant K on the
interval [a,b]. If a point M belongs to the closed interval with end
points $f(a)$ and $f(b)$ then there is a point c in the interval [a,b]
such that $f(c) = M.$
Proof of the Contrapositive : If M is not in the image
$f([a,b])$ of the closed interval [a,b] then there is a $d(M, f([a,b])) =
d > 0$ because the image $f([a,b])$ is closed - contains all its limit
points. Let K be the Lipshitz constant for $f$ on the interval [a,b]. Now
choose an positive integer n such that $\frac{b-a}n \lt d$ and put
$x_j=a+\frac{j(b-a)}n$ for $0 \le j \le n$. The above theorem implies the
points $x_j$ all lie on one side of M - are all greater or less than M.
That implies the two function values $f(a)= f(x_0)$ and $f(b) = f(x_n)$
must both be larger than M+d or both be less than M-d. Hence M does not
belongs to the closed interval with end points $f(a)$ and $f(b)$.
Corollary: If M = 0 is not in the range of a Lipshitz Continuous
function $f(x)$ on an interval $[a,b]$ then the image of $[a,b]$ lies in an closed interval $[A,B]$ where both endpoints
are positive or both endpoints $A$ and $B$ are negative. Moreover, the function $|f(x)| $, the absolute value of $f$ function
values, is also Lipshitz continuous on
$[a,b]$ and in consequence the absolute value of $f$ function has image in an interval $[C, D]$ where both endpoints $C$ and
$D$ have postive values with $ 0 \lt C \le D.$
Note: The error control inequalities used to establish the continuity of employs
the Corollary to show that the quotient and reciprocal of functions in the case where the numerator and denominator
are Lipshitz continuous on an interval [a,b] with the denominator nonzero for then the denominator is bounded
away from zero. The latter condition bounded away from zero can also be applied on unbounded interval I to imply
the Lipshitz continuity of quotients and reciprocals
of Lipshitz continuous functions.
One Sided Range Theorems for Lipschitz Continuous Functions
Theorem: Let d > 0. Suppose f has a Lipschitz continuity
constant K on the interval I. Suppose $f(a) \ne M$ for some point a in
the interval I, and that $|f(x) - M| > d$ for all x in I. Then
-
$f(x) < M$ for all x in I, or
-
$f(x) > M$for all x in I.
Proof: First assume $f(a) < M.$ The other case $f(a) > M$
is similar.
By the definition of Lipshitz continuity
\[|f(s) - f(t)| < K|s-t| \]
for all points s and t in the interval [a,b]. Let m be a positive
number with
\[ K \frac{| b-a|}{m} < d.\]
Let
\[ x(j) = a + j\frac{(b-a)}{m} \]
for each integer j. Now put
\[y(j) = f(x(j)) = f\left(a + j\frac{(b-a)}{m}\right)\]
Then
\[|y(j) - M| = | f( x(j) ) - M | > d\]
The Lipshitz Continuity property implies
\[|y(j+1) - y(j)| < K \frac{|b-a|}{m}< d.\]
By the General Barrier Theorem above, we conclude $y(j) < M$
for each natural number j for which $y(j) \in I.$ Similarly, we
conclude $y(j) < M$ for each integer j for which $y(j) \in I.$
Finally, for each number x in the interval [a, b], there is an index j
such that
\[x(j) < x < x(j+1) = x(j) +\frac{(b-a)}{m}\]
Therefore
\[f(x) < f(x(j)) + K\frac{(b-a)}{m} < f(x(j)) + d < M-d +d =
M\]
The case where \[f(a) > M\] is treated similarly.
One Sided Range Theorems for Equi-Continuous Functions
Definition: A real-valued function f: I --> R on an
closed interval I is continuous on that interval I if and only if for
every real number c in the interval I, the function f is continuous at
c.
Definition: A function $f(x)$ is said to be
equicontinuous on an interval I if and only if for each
e > 0, there exist at least one d > 0
such that
whenever x1 and x2 are both in the
interval I and |x1-x2| <
d.
Remember by the
EquiContinuity theorem in [F.5a] that a function $f(x)$ continuous on
an interval [a, b] is also equicontinous on that interval.
Theorem: Let d > 0. Suppose f is equi-continuous on the
interval I=[a,b]. If $f(a) \ne M$ and $|f(x) - M| > d$ for all x in
the interval [a,b]. Then
-
$f(x) < M$ for all x in I, or
-
$f(x) > M$for all x in I.
Proof: For each positive number $E,$ there exist at least a real
number d > 0 such that
\[ |f(x_2)-f(x_1)| < E \]
whenever x1 and x2 are both in the
interval [a,b] and |x1-x2| <
d. Choose a positive integer m so that $d
> \frac{(b-a)}{m}$ and put
\[x(j) = a + j\frac{(b-a)}{m}\]
The rest of the proof is like that of the previous proof of the
Lipshitz-Continuity Barrier Theorem.
Intermediate Value Theorem for EquiContinuous Functons
Theorem: Suppose f is equicontinous on the interval [a,b]. If a
point M belongs to the closed interval with end points f(a) and f(b).
Then there is a point c in the interval [a,b] such that f(c) = M.
Proof : If M is not in the image f([a,b]) of the interval
[a,b] then there is a d(M, f([a,b])) = d > 0. Let K be the Lipshitz
constant for f on the interval [a,b]. Now choose a whole n > 0 such
that $|u -v| \lt \frac{b-a}n $ implies $ |(f(u)- f(v) | \lt d$ for pairs
of numbers u and v in the given interval [a,b]. Now put
$x_j=a+\frac{j(b-a)}n$ for $0 \le j \ne n$. The above theorem implies the
points $x_j$ all lie on one side of M - are all greater or less than.
That implies the two function values $f(a)= f(x_0)$ and $f(b) = f(x_n)$
must both be larger than M+d or both be less than M-d.
Note: Lipshitz Continuous functions are equicontinuous. However the
identification of Lipshitz continuous functions and the restriction
of Riemann Integration to Lipshitz continuous functions removes the
need for the proof of equicontinuity of continuous functions over closed
intervals [a, b]
Further Applications: Optional Extension for now to Higher Dimension
Let M be a subset of an n-dimensional set X in IR^n or more
generally a metric space X. Let $d(x,M)$ denote the shortest distance of
a point in X to M. Further suppose X\M has a connected components
M0.
A Membership Theorem: Let M > d > 0. Suppose for 0
< j < k that y(j) are a real numbers with
\[d(y(j), M) > d,\]
\[d(y(j+1), y(j)) < d\]
and $y(0)$ belongs to M0
then $y(j)$ belongs to M0 for 0 < j < k.
Further topological generalization would allow the set of $y(j)$ to be
infinite, and the variable j to be belong to a multidimensional lattice.
The generalization combined with Lipshitz or Equicontinuity would imply
multi-dimensional all one side or all in one component theorems
for functions of say Rn into Rm. That might support
or aid the common know-how in the description and handling of surfaces
and they may separate space, or constrain function values.
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