Question 6.T.7: Let f : I → R, where I is an interval. If f is continuous an...
Let where I is an interval. If f is continuous and injective, then it is strictly monotonic.
We shall prove the theorem for the case when I = [a, b], and leave the other cases as an exercise. Without loss of generality, we can assume f(a) < f(b) and we have to show that f is strictly increasing (otherwise we consider −f ).
First we shall prove that
f(a) < f(x) < f(b) for all x ∈ (a, b). (6.6)
Since f is injective, f(x) ≠ f(a) and f(x) ≠ f(b), and we have to exclude the two possibilities
f(x) < f(a) < f(b), (6.7)
f(a) < f(b) < f(x). (6.8)
If the inequality (6.7) were valid, then, by the Intermediate Value Theorem, there would exist a point c ∈ (x, b) where f(c) = f(a), which contradicts the injective property of f. Similarly, inequality (6.8) implies that there is a c ∈ (a, x) where f(c) = f(b), again contradicting the injectivity of f. This proves (6.6).
To complete the proof, we have to show that, if x, y ∈ (a, b) and x < y, then f(x) < f(y). From (6.6) we have
f(a) < f(y) < f(b).
If f(y) < f(x), then
f(a) < f(y) < f(x),
in which case there would have to be a point c ∈ (a, x) where f(c) = f(y), thereby contradicting the injectivity of f. Therefore f(y) > f(x).