Page 120 - DMTH401_REAL ANALYSIS

Real Analysis

Notes (ii) It is said to be a monotonically decreasing function on S if
x < x f(x )  f(x ) for any x,, x S.
1 2 1 2 2
(iii) The function f is said to be a monotonic function on S if it is either monotonically increasing
or monotonically decreasing.
(iv) The function f is said to be strictly increasing on S if

x < x  f(X ) < f(x ), for x,, x S,
2 1 2 2
(v) It is said to be strictly decreasing on S if
x < x  f(x ) > f(X ), for x,, x S.
1 2 1 2 2
You can notice immediately that if f is monotonically increasing then –f i.e. –f: R  R defined by
(–f)(x) = –f(x), " x R

is monotonically decreasing and vice-versa.

Example: Test the monotonic character of the function f: R  R defined as
2
ì x , x 0

f(x) = í 2
>
î  x , x 0
Solution: For any X , x R, X  0; X  0
1 2 1 2
2
X < X  X > X  f(x ) > f(X )
2
1 2 1 2 1 2
which shows that f is strictly decreasing.
Again if X > 0, X > 0, then
1 2
2
X < X  X < x  –X > –X  f(X ) > f(X )
2
2
2
1 2 1 2 1 2 1 2
which shows that i is strictly decreasing for x > 0. Thus f is strictly decreasing for every x  R.
Now, we discuss an interesting property of a strictly increasing function in the form of the
following theorem:
Theorem 1: Prove that a strictly increasing function is always one-one.
Proof: Let f : S  T be a strictly increasing function. Since f is strictly increasing, therefore,

X < X  f(x ) < f(x ) for any X , x  S.
1 2 1 2 1 2
Now to show that f : S T is one-one, it is enough to show that
f(x ) = f(x )  X = X .
1 2 1 2
Equivalently, it is enough to show that distinct elements in S have distinct images in T
i.e. X  X  f(x )  f(x ), for X , X  S.
1 2 1 2 1 2
Indeed,
x  x  x < x or x > x
1 2 1 1 1 2
 f(x ) < f(x ) or f(x ) > f(x )
1 2 1 2
 f(x )  f(x )
l 2
which proves the theorem.

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