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
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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