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.

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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.