Question

Let \(f\) be a differentiable real function defined in \((a, b)\). Prove that \(f\) is convex if and only if \(f^{\prime}\) is monotonically increasing. Assume next that \(f^{\prime \prime}(x)\) exists for every \(x \in(a, b)\), and prove that \(f\) is convex if and only if \(f^{\prime \prime}(x) \geq 0\) for all \(x \in(a, b)\).

Short answer

Question: Prove that (1) a function \(f\) is convex if and only if its first derivative \(f'\) is monotonically increasing, and (2) assuming the second derivative \(f''(x)\) exists for all \(x \in (a, b)\), the function is convex if and only if \(f''(x) \geq 0\) for all \(x \in (a, b)\). Answer: We used the definition of convexity, the Mean Value Theorem, and the properties of derivatives to prove both statements. 1. To prove the first statement, we showed that if \(f\) is convex, then \(f'\) is monotonically increasing by applying the Mean Value Theorem to \(f\). Then, we proved that if \(f'\) is monotonically increasing, then \(f\) is convex by applying the Mean Value Theorem to \(f'\). 2. To prove the second statement, we showed that if \(f''(x) \geq 0\), then \(f\) is convex because \(f'\) is monotonically increasing (which we proved in the first part of the problem), and if \(f'\) is monotonically increasing, then \(f\) is convex. Then, we proved that if \(f\) is convex, then \(f''(x) \geq 0\) by showing that if \(f\) is convex, then \(f'\) is monotonically increasing, and applying the Mean Value Theorem to \(f'\).

Step-by-step solution

Review the definition of convexity

A function \(f: (a, b) \rightarrow \mathbb{R}\) is convex if, for any \(x, y \in (a, b)\) and \(t \in (0, 1)\), we have that: \[f(tx + (1-t)y) \leq tf(x) + (1-t)f(y)\] Now let's proceed to prove the first statement, which states that a function \(f\) is convex if and only if its first derivative \(f^\prime\) is monotonically increasing.

Prove that if \(f\) is convex, then \(f'\) is monotonically increasing

Using the definition of convexity, we need to show that if \(f\) is convex, then for any \(x, y \in (a, b)\) with \(x < y\): \[f'(x) \leq f'(y)\] By the Mean Value Theorem, we have that there exists a point \(c \in (x,y)\) such that: \[f'(c) = \frac{f(y) - f(x)}{y - x}\] Note that, since \(f\) is convex, we have: \[f(y) \leq f(x) + (y - x)f'(c)\] Rearranging the inequality, we get: \[f'(c) \geq \frac{f(y) - f(x)}{y - x}\] Now, let's take another point \(d \in (x, y)\). Similarly, by the Mean Value Theorem, we have that there exists a point \(e \in (x, d)\) such that: \[f'(e) = \frac{f(d) - f(x)}{d - x}\] Since \(f\) is convex, we have: \[f(d) \leq f(x) + (d - x)f'(e)\] Rearranging the inequality, we get: \[f'(e) \geq \frac{f(d) - f(x)}{d - x}\] Now, since \(d < y\), we can conclude that: \[f'(x) \leq f'(e) \leq f'(c) \leq f'(y)\] Thus, \(f'\) is monotonically increasing.

Prove that if \(f'\) is monotonically increasing, then \(f\) is convex

We need to show that if \(f'\) is monotonically increasing, then \(f\) is convex, which means for any \(x, y \in (a, b)\) and \(t \in (0, 1)\), we have: \[f(tx + (1-t)y) \leq tf(x) + (1-t)f(y)\] By the Mean Value Theorem, there exists points \(c_1 \in (x, tx + (1-t)y)\) and \(c_2 \in (tx + (1-t)y, y)\) such that: \[f'(c_1) = \frac{f(tx + (1-t)y) - f(x)}{tx + (1-t)y - x}\] and \[f'(c_2) = \frac{f(y) - f(tx + (1-t)y)}{y - tx - (1-t)y}\] Now, since \(f'\) is monotonically increasing and \(x < c_1 < tx + (1-t)y < c_2 < y\), we have: \[f'(x) \leq f'(c_1) \leq f'(c_2) \leq f'(y)\] Using the definitions of \(f'(c_1)\) and \(f'(c_2)\), we can rewrite the inequality as: \[f(tx + (1-t)y) - f(x) \leq t(y-x)f'(c_1)\] and \[f(y) - f(tx + (1-t)y) \leq (1-t)(y-x)f'(c_2)\] Summing up these inequalities, we obtain: \[f(tx + (1-t)y) \leq tf(x) + (1-t)f(y)\] Thus, \(f\) is convex. Now let's proceed with the second statement, which states that assuming the second derivative \(f^{\prime\prime}(x)\) exists for all \(x \in (a, b)\), the function is convex if and only if \(f^{\prime\prime}(x) \geq 0\) for all \(x \in (a, b)\).

Prove that if \(f''(x) \geq 0\), then \(f\) is convex

Since \(f^{\prime\prime}(x) \geq 0\) for all \(x \in (a, b)\), we have that \(f'\) is monotonically increasing (by the first part of this problem). And, from Step 3, we know that if \(f'\) is monotonically increasing, then \(f\) is convex. Thus, we have that \(f\) is convex.

Prove that if \(f\) is convex, then \(f''(x) \geq 0\)

From Step 2, we know that if \(f\) is convex, then \(f'\) is monotonically increasing. So, for any \(x, y \in (a, b)\) with \(x < y\), we have: \[f'(x) \leq f'(y)\] Rearranging the inequality and dividing by \((y - x)\), we obtain: \[\frac{f'(y) - f'(x)}{y - x} \geq 0\] Now, since \(f^{\prime\prime}(x)\) exists for all \(x \in (a, b)\), we can use the Mean Value Theorem once again for \(f'\). There exists a point \(c \in (x, y)\) such that: \[f^{\prime\prime}(c) = \frac{f'(y) - f'(x)}{y - x}\] From the inequality above, we conclude that \(f^{\prime\prime}(c) \geq 0\) for all \(x, y \in (a, b)\) and \(x < y\), which means that \(f^{\prime\prime}(x) \geq 0\) for all \(x \in (a, b)\). We have now completed the proof for both statements, showing that a function \(f\) is convex if and only if its first derivative \(f'\) is monotonically increasing, and assuming the second derivative \(f^{\prime\prime}(x)\) exists for all \(x \in (a, b)\), the function is convex if and only if \(f^{\prime\prime}(x) \geq 0\) for all \(x \in (a, b)\).

Key concepts

Properties of Convex Functions

Convex functions have distinct characteristics that are crucial in various branches of mathematics and optimization. A function, say, f, is defined to be convex over an interval if the line segment connecting any two points on the graph of the function lies above or on the graph. In mathematical terms, for any points x and y within the domain and any t between 0 and 1, the following inequality must hold:
\[f(tx + (1-t)y) \leq tf(x) + (1-t)f(y)\]
This property implies that convex functions always form an 'upward' curve, and there are no 'dips' in the graph within the interval of convexity. Another important aspect of convex functions is related to their first derivatives. Specifically, if a function f is convex on an interval, then its first derivative f' is monotonically increasing on that interval. It signifies that as x increases, f'(x) does not decrease, which is typically demonstrated in the slope of the tangent lines to the graph of f, ever becoming steeper.
Understanding these properties allows us to analyze the behavior of functions and the optimization problems involving such functions more efficiently.

Monotonicity and Derivatives

The concept of monotonicity in functions is directly connected to the behavior of their derivatives. For differentiable functions, monotonicity of the derivatives holds valuable insights. A function f whose derivative f' is monotonically increasing means that the slope of the tangent to the function at any point is lesser than or equal to the slope of the tangent at a point to the right of it. Formally, for any two points x and y in the domain such that x < y, it must be the case that:
\[f'(x) \leq f'(y)\]
This condition is integral in proving that a function is convex. If the derivative of a function increases (or at least, does not decrease) as we move along the interval, this indicates that the function is becoming steeper and does not turn downwards, a quality inherent to convex functions. Consequently, the monotonicity of the first derivative is a powerful tool in the analysis of convexity.

Second Derivative Test for Convexity

The second derivative test for convexity offers a more straightforward approach to assessing whether a function is convex over an interval. If a function f has a continuous second derivative, f'', over an interval then f is convex over that interval if and only if:
\[f''(x) \geq 0\]
for every x in the interval. What the second derivative tells us is that if it is positive, the slope of the function's tangent is increasing, hence, the function is bending upwards. This test simplifies the process of determining convexity as it avoids the need to directly verify the inequality involved in the definition of convex functions. By checking the sign of the second derivative, one can readily conclude about the convexity of functions that are twice differentiable over the interval in question.

Mean Value Theorem

The Mean Value Theorem (MVT) is a pivotal concept in calculus that relates to the increment of the function to the increment of its derivative. If a function f is continuous on the closed interval \[[a, b]\] and differentiable on the open interval \[(a, b)\], then there exists at least one point \[c\] in \[(a, b)\] such that the derivative f' at \[c\] equals the average rate of change of f over \[[a, b]\]:
\[f'(c) = \frac{f(b) - f(a)}{b - a}\]
The MVT not only holds significance in theoretical mathematics but also in practical applications, like in the proofs relating to the properties of convex functions. When we consider the convexity of a function, the MVT assures us that somewhere between any two points on the function, there's a point where the tangent has the same slope as the secant connecting those two points. Implementing the MVT becomes especially useful in demonstrating the link between the monotonicity of the first derivative and the convexity of the function. The MVT bridges local and average properties of functions and, hence, is a cornerstone in furthering understanding of more complex calculus concepts.