Proving Strict Increase of Continuous Functions on Intervals

Understanding the behavior of functions, particularly whether they are strictly increasing or not, is fundamental in many areas of mathematics and its applications. This article aims to provide a rigorous proof for the statement: If a function fx is continuous on the interval [a, b] and strictly increasing on (a, b), then it is strictly increasing on [a, b]. We will explore the proof through several cases and provide additional insights into the properties of continuous and strictly increasing functions.

Definition and Key Concepts

A function fx is said to be strictly increasing on an interval [a, b] if for any two points x_1 and x_2 within the interval, with x_1 x_2, it follows that fx_1 fx_2. Additionally, fx is continuous on [a, b] if it has no abrupt changes or jumps within the interval.

Proof Structure

To prove that if fx is continuous on [a, b] and strictly increasing on (a, b), then it is strictly increasing on [a, b], we need to show that for any x_1, x_2 in [a, b] with x_1 x_2, it follows that fx_1 fx_2. We will consider four cases:

Case 1: Both x_1 and x_2 are in (a, b)

Since fx is strictly increasing on (a, b), we have fx_1 fx_2 for x_1 x_2.

Case 2: x_1 a and x_2 isin; (a, b)

Due to the continuity of fx at a, we have the limit as x approaches a from the right equal to fa. Since fx is strictly increasing on (a, b), fx_2 fa. Therefore, fa fx_2.

Case 3: x_1 isin; (a, b) and x_2 b

Similarly, due to the continuity of fx at b, the limit as x approaches b from the left is equal to fb. Since fx is strictly increasing on (a, b), fx_1 fb. Therefore, fx_1 fb.

Case 4: x_1 a and x_2 b

From cases 2 and 3, we have fa fx_2 and fx_1 fb. Therefore, fa fb.

Conclusion

In all cases, we have shown that if x_1 x_2, then fx_1 fx_2. Thus, fx is strictly increasing on [a, b].

Additional Insights

Let c be in [a, b]. Then, there exists a decreasing sequence x_n in (a, b) that converges to a with x_nc for every n. Then fx_n is a decreasing sequence converging to fa, which implies fa le; f(x_n) for every n. Thus fa le; f(x_1) le; c. By a similar argument, we prove that fc le; fb. This confirms the behavior of the function at the boundaries of the interval.

If there were a point x in (a, b) such that fa ge; fx, then since fx is strictly increasing on (a, b), fx ge; f(z) for all z isin; (x, b). Choose one such z and observe that fa le; f(z). Because fx is continuous on [a, z], there must be a point y isin; [a, z] such that fy f(z). This contradicts the fact that fx is strictly increasing on (a, b). Hence, fx is strictly increasing on [a, b]. An analogous argument can be used to verify that fx is strictly increasing on [b, a].