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Chapter 1: Problem 34

Let \(I\) be an interval and \(f: I \rightarrow \mathbb{R}\) be any function. If \(f\) is convex on \(I\), then show that for any \(x_{1}, \ldots, x_{n} \in I\) and any nonnegative real numbers \(t_{1}, \ldots, t_{n}\) with \(t_{1}+\cdots+t_{n}=1\), we have $$ f\left(t_{1} x_{1}+\cdots+t_{n} x_{n}\right) \leq t_{1} f\left(x_{1}\right)+\cdots+t_{n} f\left(x_{n}\right) $$ [Note: The above inequality is sometimes called Jensen's inequality.]

Short Answer

Expert verified
In summary, Jensen's inequality states that for any convex function \(f: I \rightarrow \mathbb{R}\) on interval \(I\), and any points \(x_1, \ldots, x_n \in I\) with nonnegative weights \(t_1, \ldots, t_n\) such that \(t_1 + \cdots + t_n = 1\), the inequality \(f\left(t_{1} x_{1}+\cdots+t_{n} x_{n}\right) \leq t_{1} f\left(x_{1}\right)+\cdots+t_{n} f\left(x_{n}\right)\) holds. We proved this using induction on \(n\), showing that the inequality is true for the base case of \(n = 2\) (by definition of convex functions) and then establishing the inductive step by considering an arbitrary \(n \geq 2\) and proving the inequality for \(n+1\).

Step by step solution

01

Define a convex function

A function f: I 鈫� 鈩� is convex on an interval I if, for all x, y 鈭� I and t 鈭� [0, 1], the following inequality holds: $$ f(tx + (1-t)y) \leq tf(x) + (1-t)f(y) $$
02

Consider the given function f and interval I

Let f: I 鈫� 鈩� be a convex function on the interval I. We are given that for any x鈧�, ..., x鈧� 鈭� I and any nonnegative real numbers t鈧�, ..., t鈧� with t鈧� + ... + t鈧� = 1, we have to show that Jensen's inequality holds: $$ f\left(t_{1} x_{1}+\cdots+t_{n} x_{n}\right) \leq t_{1} f\left(x_{1}\right)+\cdots+t_{n} f\left(x_{n}\right) $$
03

Prove Jensen's inequality using induction

We will prove Jensen's inequality using the method of induction on n. Base case (n=2): For n = 2, we have $$ f\left(t_{1}x_{1} + t_{2}x_{2}\right) \leq t_{1}f(x_{1}) + t_{2}f(x_{2}) $$ Since f is convex, this is true by definition, so the base case is established. Inductive step: Suppose the inequality holds for some n 鈮� 2 (inductive hypothesis). Now, we will prove it for n + 1. Let x鈧� = t鈧亁鈧� + ... + t鈧檟鈧�. Now, define new coefficients s岬� = t岬� / (1 - t鈧欌倞鈧�) for i = 1,...,n. Notice that s鈧� + ... + s鈧� = 1 because t鈧� + ... + t鈧欌倞鈧� = 1. Now, we have $$ t_{1} x_{1}+\cdots+t_{n} x_{n} = (1 - t_{n+1})\left(s_{1} x_{1}+\cdots+s_{n} x_{n}\right) $$ Therefore, $$ f\left(t_{1} x_{1}+\cdots+t_{n+1} x_{n+1}\right) = f\left((1 - t_{n+1})(s_{1} x_{1}+\cdots+s_{n} x_{n}) + t_{n+1}x_{n+1}\right) $$ Applying the definition of convex functions to x鈧� and x鈧欌倞鈧�, we get $$ f\left((1 - t_{n+1})\left(s_{1} x_{1}+\cdots+s_{n} x_{n}\right) + t_{n+1}x_{n+1}\right) \leq (1 - t_{n+1})f\left(s_{1} x_{1}+\cdots+s_{n} x_{n}\right) + t_{n+1}f\left(x_{n+1}\right) $$ By the inductive hypothesis, we know that $$ f\left(s_{1} x_{1}+\cdots+s_{n} x_{n}\right) \leq s_{1} f\left(x_{1}\right)+\cdots+s_{n} f\left(x_{n}\right) $$ Multiplying both sides by (1 - t鈧欌倞鈧�) gives $$ (1 - t_{n+1})f\left(s_{1} x_{1}+\cdots+s_{n} x_{n}\right) \leq (1 - t_{n+1})\left(s_{1} f\left(x_{1}\right)+\cdots+s_{n} f\left(x_{n}\right)\right) $$ Now, combining the two inequalities we derived above, we have $$ f\left(t_{1} x_{1}+\cdots+t_{n+1} x_{n+1}\right) \leq t_{1} f\left(x_{1}\right)+\cdots+t_{n+1} f\left(x_{n+1}\right) $$ Thus, Jensen's inequality holds for n + 1, completing the inductive step. By induction, Jensen's inequality holds for all n 鈮� 2.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Convex Function
Imagine you're taking a peaceful walk along a smoothly curving hill. The path you take is akin to what mathematicians call a convex function. In the realm of mathematics, a function, let's name it \( f \), is deemed convex over an interval if, intuitively speaking, the segment connecting any two points on the graph of \( f \) lies never below the curve itself.

More formally, for a function \( f: I \rightarrow \mathbb{R} \) to be convex on an interval \( I \), it must satisfy a specific condition. If you pick any two points, say \( x \) and \( y \), from this interval and a number \( t \) between 0 and 1, the function satisfies the inequality \( f(tx + (1-t)y) \leq tf(x) + (1-t)f(y) \).

This peculiar property has numerous implications on how the function behaves, including how its values can be squeezed in between certain bounds, a fact extensively leveraged in calculus and optimization. Consequently, Jensen's inequality captures this concept, allowing us to compare the function's value at a blend of points to a blend of the function's values at those points鈥攁 truly invaluable tool for mathematicians and economists alike.
Induction in Mathematics
Induction in mathematics is akin to a domino effect; once the first piece falls, the rest follow seamlessly. It is a method used to prove that a statement is true for all natural numbers. The technique involves two critical steps: establishing the base case and proving the inductive step.

First, the base case: it's like proof for the first domino. You show that your statement holds for the initial number, commonly the number 1 or 2. Once the base case is verified, you proceed to the inductive step. Here, you assume, as a hypothesis, that the statement holds for some number \( n \), and then you need to prove that this truth cascades to the next number, \( n+1 \).

When the inductive step is proven, you have essentially set up a reaction that proves your statement for all subsequent numbers, just as the fall of one domino causes the fall of the next. This method is exactly what breathes life into the proof of Jensen's inequality for any number of variables, guaranteeing its truth for a grand ensemble of scenarios.
Inequality in Calculus
In calculus, inequality is not about who gets more pie, but it is a fundamental concept where we determine the relationships between different functions or values. We often encounter expressions like \( a < b \) or \( f(x) \leq g(x) \), and these comparisons can tell us which function is larger or if a particular function always lies below another.

Understanding inequalities is crucial because they help us describe the behavior of functions and set constraints in optimization problems. They are tools enabling us to confine our solutions within certain bounds, adamant boundaries from which our answers cannot escape.

Moreover, inequalities like Jensen's are particularly potent as they encapsulate properties of functions鈥攕uch as convexity鈥攁nd allow us to derive vital estimates, approximate solutions, and sometimes grasp the nature of complex systems. It鈥檚 this establishment of bounds and estimates that is often the first step in tackling the more intricate problems one may find in the magical land of calculus.

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Most popular questions from this chapter

Given any \(\ell, m \in \mathbb{Z}\) with \(\ell \neq 0\), prove that there are unique integers \(q\) and \(r\) such that \(m=\ell q+r\) and \(0 \leq r<|\ell| .\) (Hint: Consider the least element of the subset \(\\{m-\ell n: n \in \mathbb{Z}\) with \(m-\ell n \geq 0\\}\) of \(\mathbb{Z} .\) )

Show that if \(n \in \mathbb{N}\) and \(a_{1}, \ldots, a_{n}\) are nonnegative real numbers, then \(\left(a_{1}+\cdots+a_{n}\right)^{2} \leq n\left(a_{1}^{2}+\cdots+a_{n}^{2}\right) .\) (Hint: Write \(\left(a_{1}+\cdots+a_{n}\right)^{2}\) as \(t_{1}+\cdots+t_{n}\) where \(\left.t_{k}:=a_{1} a_{k}+a_{2} a_{k+1}+\cdots+a_{n-k+1} a_{n}+a_{n-k+2} a_{1}+\cdots+a_{n} a_{k-1} .\right)\) [Note: Exercise 35 gives an alternative approach to this inequality.]

Let \(I\) be an interval and \(f: I \rightarrow \mathbb{R}\) be any function. (i) If \(f\) is monotonically increasing as well as monotonically decreasing on \(I\), then show that \(f\) is constant on \(I\). (ii) If \(f\) is convex as well as concave on \(I\), then show that \(f\) is given by a linear polynomial (that is, there are \(a, b \in \mathbb{R}\) such that \(f(x)=a x+b\) for all \(x \in I)\).

The Fundamental Theorem of Algebra states that if \(p(x)\) is a polynomial in \(\mathbb{C}[x]\) of positive degree, then \(p(x)\) has at least one root in \(\mathbb{C} .\) (i) Assuming the Fundamental Theorem of Algebra, show that if \(p(x)\) is a polynomial in \(\mathbb{C}[x]\) of positive degree \(n \in \mathbb{N}\), then we can write $$ p(x)=c\left(x-\alpha_{1}\right) \cdots\left(x-\alpha_{n}\right) $$ where \(c\) is the leading coefficient of \(p(x)\) and \(\alpha_{1}, \ldots, \alpha_{n}\) are (not necessarily distinct) complex numbers. (ii) Show that if \(p(x) \in \mathbb{R}[x]\) and if a complex number \(\alpha=a+i b\) is a root of \(p(x)\), then its conjugate \(\bar{\alpha}:=a-i b\) is also a root of \(p(x)\). (iii) Show that the Fundamental Theorem of Algebra, as stated above, and the Real Fundamental Theorem of Algebra, as stated in this chapter, are equivalent to each other, that is, assuming one of them, we can deduce the other.

If \(m, n, n^{\prime} \in \mathbb{Z}\) are such that \(m\) and \(n\) are relatively prime and \(m \mid n n^{\prime}\), then show that \(m \mid n^{\prime}\). Deduce that if \(p\) is a prime (which means that \(p\) is an integer \(>1\) and the only positive integers that divide \(p\) are 1 and \(p\) ) and if \(p\) divides a product of two integers, then it divides one of them. (Hint: Exercise 38.)
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