One Sided Range 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 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

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

|f(x2)-f(x1)| < e

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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Here are the Appendices from  Volume 3, Why Slopes and More Math,  Chapters 14 to 19 in Vol 3 are related. Here is a  reference for college or university mathematics, electrical engineering and physics.

A. What's Next
B. Pigeon Hole Principle
B. Bolzano-Weierstrass
C1. Triangle Inequality
C2. Triangle Inequality
C. More T.Inequality
D. Sets & Sequences
D. Monotone Sequences
E. Limits,  Properties
E Limits & Error Control
F. Continuous Functions
F. Closed Range Thm
F. Intermediate Val. Thm
F. Compactness Thm
F. Equicontinuity Thm
F Extreme Value Thm
G. Rolle's Theorem etc
G. Mean Val. Thm.
G. Constant Difference Thm
G. Lipschitz Continuity I
PS: One Sided Range Theorems
G. Velocity Revisited
G. Sufficient Conditions
H. Riemann Sums Conv
H. Lipschitz Continuity II

The site area More Calculus contains a one-sided theorem with proof that should be of interest too.

Vol 1A Logic Postscripts
online only:-

Proof by Absurdity alias proof by contradiction

How the demand for consistency supports the law of the excluded middle

Reality versus or with the aid of Imagination

Science, Engineering & Math Students: Have you seen a simpler  geometric introduction to complex numbers? ( java applet included) . Can you explain what is a variable without using a symbol? Can you derive trig expression for dot & cross &  cosine law from complex number properties? For truth tables and indirect methods of reason, see  chapters 19-24 & postscripts in  Pattern Based Reason  and visit Volume 1A, Pattern Based Reason, striving for objectivity, the empirical challenge & limits.  

Vol 1A Postscripts
online only

Proof by Absurdity alias proof by contradiction

How the demand for consistency supports the law of the excluded middle

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