Let f : X → R be a differentiable function with X ⊆ R being a convex
set. Prove that f′ is nondeccreasing iff
f(x2) ≥ f(x1) + f′(x1)(x2 − x1), ∀x1, x2 ∈ X

To prove the equivalence between the two conditions for a differentiable function \( f: X \to \mathbb{R} \), where \( X \subseteq \mathbb{R} \) is a convex set, we will establish the two implications:

1. **\( f'(x) \) is nondecreasing implies \( f(x_2) \geq f(x_1) + f'(x_1)(x_2 - x_1) \) for all \( x_1, x_2 \in X \).**

2. **\( f(x_2) \geq f(x_1) + f'(x_1)(x_2 - x_1) \) for all \( x_1, x_2 \in X \) implies \( f'(x) \) is nondecreasing.**

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### Part 1: \( f'(x) \) is nondecreasing implies the inequality

Assume that \( f'(x) \) is nondecreasing on the convex set \( X \), meaning that for any \( x_1, x_2 \in X \) with \( x_1 \leq x_2 \), we have:
\[
f'(x_1) \leq f'(x_2).
\]
Now, we want to show that:
\[
f(x_2) \geq f(x_1) + f'(x_1)(x_2 - x_1).
\]

Let \( g(x) = f(x) - f(x_1) - f'(x_1)(x - x_1) \). The goal is to show that \( g(x_2) \geq 0 \). Note that:
\[
g(x_1) = f(x_1) - f(x_1) - f'(x_1)(x_1 - x_1) = 0.
\]
The derivative of \( g(x) \) is:
\[
g'(x) = f'(x) - f'(x_1).
\]
Since \( f'(x) \) is nondecreasing, for \( x \geq x_1 \), we have \( f'(x) \geq f'(x_1) \), which implies that \( g'(x) \geq 0 \) for \( x \geq x_1 \). Therefore, \( g(x) \) is nondecreasing for \( x \geq x_1 \), and since \( g(x_1) = 0 \), it follows that \( g(x_2) \geq 0 \) for \( x_2 \geq x_1 \). Thus, we have:
\[
f(x_2) \geq f(x_1) + f'(x_1)(x_2 - x_1),
\]
as required.

### Part 2: The inequality implies \( f'(x) \) is nondecreasing

Now assume that for all \( x_1, x_2 \in X \), we have:
\[
f(x_2) \geq f(x_1) + f'(x_1)(x_2 - x_1).
\]
We want to show that \( f'(x) \) is nondecreasing.

Let \( x_1 < x_2 \) be any two points in \( X \), and define the function:
\[
\phi(x) = \frac{f(x) - f(x_1)}{x - x_1}.
\]
The function \( \phi(x) \) represents the slope of the secant line connecting \( (x_1, f(x_1)) \) and \( (x, f(x)) \). By the assumption, we have:
\[
f(x_2) \geq f(x_1) + f'(x_1)(x_2 - x_1),
\]
which can be rearranged as:
\[
\frac{f(x_2) - f(x_1)}{x_2 - x_1} \geq f'(x_1).
\]
This means that \( \phi(x_2) \geq f'(x_1) \).

Now, consider the limit as \( x_2 \to x_1^+ \) (i.e., approaching \( x_1 \) from the right):
\[
\lim_{x_2 \to x_1^+} \frac{f(x_2) - f(x_1)}{x_2 - x_1} = f'(x_1).
\]
Thus, \( \phi(x_2) \geq f'(x_1) \) for \( x_2 > x_1 \), and in the limit, the inequality holds with equality. This implies that the derivative \( f'(x) \) is nondecreasing.

### Conclusion

We have shown that:

- If \( f'(x) \) is nondecreasing, then \( f(x_2) \geq f(x_1) + f'(x_1)(x_2 - x_1) \) for all \( x_1, x_2 \in X \).
- If \( f(x_2) \geq f(x_1) + f'(x_1)(x_2 - x_1) \) for all \( x_1, x_2 \in X \), then \( f'(x) \) is nondecreasing.

Thus, the two conditions are equivalent.

---

Would you like more details on any part of the proof or have any questions?

Here are 5 related questions you can explore:
1. How does convexity of the set \( X \) play a role in this proof?
2. What other equivalent conditions can describe a function's convexity?
3. Can the inequality in the problem statement be used to define convex functions?
4. How can we use second-order derivatives to verify that \( f'(x) \) is nondecreasing?
5. What is the geometric interpretation of the inequality involving \( f'(x_1) \)?

**Tip**: When proving implications between statements, breaking the problem into two separate cases (if and only if) can clarify the structure of the proof.

Let f : X → R be a differentiable function with X ⊆ R being a convex set. Prove that f′ is nondecreasing iff f(x2) ≥ f(x1) + f′(x1)(x2 − x1), ∀x1, x2 ∈ X

Explore the proof that a function's derivative is nondecreasing if and only if it satisfies the inequality f(x2) ≥ f(x1) + f′(x1)(x2 − x1). Learn how differentiation, convexity, and inequalities come together in this advanced calculus topic. Suitable for undergraduate-level students.

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