One Sided Range Theorems
I have not seen this theorem elsewhere, but I imagine it is
a re-invention of someoneelse's result.
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 point M.
String on One Side Theorem: Let M > d > 0. Suppose
for 0 < j < k that y(j) are a real numbers
with |y(j) - M| > d, y(0) < M and |y(j+1) - y(j)| < d
then y(j) < M for 0 < j < k.
Proof by Mathematical Induction. Define a statement (q) as
follows:
(q): y(q) < M-d
Observe statement (q) holds when q = 0 since |y(0) - M| > d
and y(0) < M
Assume (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 (q+1) holds. So statement (q) implies statement
(q+1). That completes the proof.
The proof of the following theorem is similar.
String on Other Side Theorem: Let M > d > 0.
Suppose for 0 < j < k that y(j)
are a real numbers with |y(j) - M| > d,and |y(j+1) - y(j)| <
d then y(j) > M for 0 < j < k.
The graph of a function y = f(x) can be compare with that of a string.
One Sided Range Theorem for Lipshitz Continuous Functions: Let d >
0. Suppose f has a Lipschitz continuity constant K on the interval
I. Suppose f(a) =\= M for some point a in the interval I,
and that |f(x) - M| > d for all x in I. Then (i) f(x) <
M for all x in I or (ii) f(x) > M for all x in I
Proof: First assume f(a) < M. The other case f(a) 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 > 0 be a
positive number with
K | b-a| (1/m) < d.
Let
x(j) = a + j(b-a) ( 1/m)
for each integer j. Now put
y(j) = f(x(j)) = f( a + j(b-a) ( 1/m) )
Then |y(j) - M| = | f( x(j) ) - M | > d
The Lipshitz Continuity property implies
|y(j+1) - y(j)| < K | b-a| (1/m) < d.
By the General Barrier Theorem, we conclude y(j) < M for each natural
number j for which y(j) belongs to I. Similarly, we conclude y(j) < M
for each integer j for which y(j) belongs to 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) +(b-a)(1/m)
Therefore f(x) < f(x(j)) + K(b-a)(1/m) < f(x(j)) + d < M-d
+d = M
The case where f(a) > M is treated similarly.
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.
Equi-Continuity, One Side Range Theorem: Let d > 0. Suppose
f is equi-continuous on the interval I . If f(a) =\= M and |f(x) -
M| > d for all x in [a,b]. Then (i) f(x) < M -d. for all x in I or
(ii) f(x) > M. for all x in I
Proof: For e = d >
0, there exist at least a real number d > 0 such
that
whenever x1 and x2 are both in the
interval [a,b] and |x1-x2|
< d. Choose a positive number m > 0 so that d
> (b-a) (1/m) and put
x(j) = a + j(b-a) ( 1/m)
The rest of the proof is like that of the previous proof of the Lipshitz-Continuity
Barrier Theorem.
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Calculus Appetizers
& Lessons
Starter Guide (Views)
Real Player Videos
Limit Properties Algebraically
Pigeon Hole Principle
Bolzano
Constant Difference Thm
Continuous Functions
Rational Functions
Mean Value Theorem
One Side Range Theorem
Range On One Side
From Lipschitz Continuity
YOU are better than YOU think. Show yourself how:
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