Page 168 - DMTH401_REAL ANALYSIS

Real Analysis

Notes
1
Example: Show that the function f defined by f(x) = V x  ] 0, 1[ is continuous but not
x
bounded.
Solution: The function f is continuous being the quotient of continuous functions F(x) = 1 and G(x)
= x with
G(x)  0, x  ]0, 1[

Domain of f is bounded but not a closed interval. The function is not bounded since its range is
(1/x : x ] 0, l [ ] = ]1, [ which is not a bounded set.

Example: Show that the function f such that f(x) = x " x ]0, 1[ is continuous but does not
attain its bounds.

Solution: As mentioned the identity function f is continuous in ]0, 1[. Here the domain of f is
bounded but is not a closed interval. The function f is bounded with least upper bound (1.u.b) =
1 and greatest lower bound (g.l.b) = 0 and both the bounds are not attained by the function, since
range of f = ]0, 1[.

Example: Show that the function f such that

1

f(x) = 2 " x [0, 1[.
x
is continuous but does not attain its g.l.b.
2
Solution: The function G given by G(x) = x " x ]0, 1[ is continuous and G(x)  0 " x ]0, 1[
therefore its reciprocal function f(x) = 1/x is continuous in ]0, 1[. Here the domain f is bounded
2
but is not a closed interval.
Further l.u.b. of f does not exist whereas its g.l.b. is 1 which is not attained by f.

Task Show that the function f given by f(x) = sin x, x  ]0, /2[ is continuous but does not
attain any of its bounds.

Task Prove that the function f given by f(x) = x " x ] –, 0[ is continuous but does not
2
attain its g.l.b.
We next prove another important property known as the intermediate value property of a
continuous function on an interval I. We do not need the assumption that I is bounded and losed.
This property justifies our intuitive idea of a continuous function namely as a function f which
cannot jump from one value to another since it takes on between any two values f(a) and f(b) all
values lying between f(a) and f(b).
Theorem 3: (Intermediate Value Theorem). Let f be a continuous function on an interval
containing a and b. If K is any number between f(a) and f(b) then there is a number c, a  c S b
such that f(c) = K.
Proof: Either f(a) = f(b) or f(a) < f(b) or f(b) < f(a). If f(a) = f(b) then K = f(a) = f(b) and so c can be
taken to be either a or b. We will assume that f(a) < f(b). (The other case can be dealt with
similarly.) We can, therefore, assume that f(a) < K < f(b).

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