Crack the Code: The Surprising Solution to Jensen’s Equation Revealed!

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**Proposition 31**.: _Suppose \(f\) is a continuous function on \([0,1]\). Then_

\[f(x)=cx+a\]

_is a solution of Jensen’s equation._

Proof

: Setting \(f(0)=a\), \(f(1)=b\) and using

\[f\left(\frac{x+y}{2}\right)=\frac{f(x)+f(y)}{2},\]

we obtain

\[f\left(\frac{1}{2}\right)=f\left(\frac{0+1}{2}\right)=\frac{f(0)+f(1)}{2}=a+ \frac{1}{2}(b-a).\]

Next, we look at \(f\left(\frac{1}{4}\right)\) and \(f\left(\frac{3}{4}\right)\):

\[f\left(\frac{1}{4}\right)=\frac{f(0)+f(1/2)}{2}=a+\frac{1}{4}(b-a),\]

\[f\left(\frac{3}{4}\right)=\frac{f(1/2)+f(1)}{2}=a+\frac{3}{4}(b-a).\]

Thus we claim that

\[f(x)=a+(b-a)x\]

holds for all dyadic \(x\). For dyadic fractions with denominator \(2^{n}\), we have:

\[f\left(\frac{2m}{2^{n+1}}\right)=f\left(\frac{m}{2^{n}}\right)=c+\frac{m}{2^{ n}}+a=c+\frac{2m}{2^{n+1}}+a\]

and

\[f\left(\frac{2m+1}{2^{n+1}}\right)=\frac{f(m/2^{n})+f(m+1/2^{n})}{2}=c+\frac{ 2m+1}{2^{n+1}}+a,\]

which proves the claim. Since \(f\) is continuous we have already shown that \(f(x)=cx+a\) for all \(x\in[0,1]\), where \(c=b-a\)

**Remark 56**.: Let \(A\subset\mathbb{R}^{n}\) be an open set. A function \(f:A\to\mathbb{R}\) is called _convex_ if and only if it satisfies Jensen’s functional inequality:

\[f\left(\frac{x+y}{2}\right)\leq\frac{f(x)+f(y)}{2}\]

for all \(x\in A\). For example \(f(x)=x^{2}\), \(f(x)=e^{x}\) and \(f(x)=|x|\) are all convex functions on any open interval of \(\mathbb{R}\). It turns out that the notion of convex functions is just as important as positive or increasing functions. That is probably why the treatment of convex functions can be found in many calculus books. In the following we just mention the relationship of convex functions and additive functions.

**Theorem 106**.: _Every additive function is convex._

Proof

: For any additive function \(f:\mathbb{R}^{n}\to\mathbb{R}\), we have

\[f\left(\frac{x+y}{2}\right)=\frac{1}{2}f(x+y)=\frac{f(x)+f(y)}{2}.\qed\]

Note that every additive function is also concave, i.e., it satisfies

\[f\left(\frac{x+y}{2}\right)\geq\frac{f(x)+f(y)}{2}.\]

Thus additive functions are both convex and concave. The converse is not true, since functions of the form \(f(x)=h(x)+b\) with \(h\) an additive function and \(b\) any constant are concave and convex, but not additive if \(b\neq 0\). If a function \(f:\mathbb{R}^{n}\to\mathbb{R}\) is at the same time convex and concave, then it differs from an additive function by a constant (for details see [36], Chapter 13).

### Exercises

1. Find all functions \(f\) satisfying

\[f(x+y)^{2}=f(x)^{2}+f(y)^{2}\]

for all \(x,y\in\mathbb{R}\).

2. Let \(f:\mathbb{R}\to\mathbb{R}\) be a function satisfying Jensen’s functional equation

\[f\left(\frac{x+y}{2}\right)=\frac{f(x)+f(y)}{2}\]

for every \(x,y\in\mathbb{R}\). Show that there is a constant \(c\) such that \(f(x)-c\) is an additive function.

3. If the function \(f:\mathbb{R}\to\mathbb{R}\) satisfies

\[x+f(x)=f(f(x))\ \ \text{for every}\ \ \ x\in\mathbb{R},\]

find all solutions of the equation \(f(f(x))\).

4. Prove that if \(u=(a,b)\), \(v=(c,d)\) form a linearly independent set in \(\mathbb{R}^{2}\), then \(W=\text{span}_{\mathbb{Q}}\{u,v\}=\{ru+sv:\ r,s\in\mathbb{Q}\}\) forms a dense subset of \(\mathbb{R}^{2}\).

5. Define and solve Jensen’s functional equation for continuous functions defined on any interval \([a,b]\), where \(0

6. Let \(H\) be a Hamel basis and let \(f:H\to\mathbb{R}\) be an arbitrary function. Prove that \(f\) can be extended uniquely to \(\mathbb{R}\) as an additive function.

7. The Gamma Function is defined by

\[\Gamma(p+1)=\int_{0}^{\infty}e^{-x}x^{p}dx.\]

1. Show that for \(p>0\), \(\Gamma(p+1)=p\Gamma(p)\).
2. Show that \(\Gamma(1)=1\).
3. If \(p\) is a positive integer \(n\), show that \(\Gamma(n+1)=n!\).

8. Suppose that \(f(x)\) is convex on an interval \(I\) in the sense of Jensen. Show that

\[f\left(\frac{x_{1}+x_{2}+\cdots+x_{n}}{n}\right)\leq\frac{f(x_{1})+f(x_{2})+ \cdots+f(x_{n})}{n}\]

for arbitrary \(n\)-points \(x_{1},x_{2},\ldots,x_{n}\) in \(I\).

9. Show that \(f(x)\) is convex on an interval \(I\) if and only if \(e^{\lambda f(x)}\) is convex on \(I\) for any positive \(\lambda\).

10. Show that \(f\in C[a,b]\) is convex if and only if \(\frac{1}{t-s}\int_{s}^{t}f(x)dx\leq\frac{f(x)+f(y)}{2}\) for any \(s\neq t\) in \([a,b]\).

### 2.6 Fourier Series

In this section we focus on an introduction to Fourier Series. It is well known that in analysis, power series expansions of functions play an important role. Recall that to determine a Taylor series expansion of a function, we need to find the coefficients and the coefficients are determined by differentiation. But what if the function is not differentiable or what if it has points of discontinuity? Can we still have another expansion where the coefficients in this representation are determined by integration? (Since we have not developed measure theory or Lebesgue integration, what we mean by integration here is Riemann integrable functions). The answer to this question lies in the theory of Fourier Analysis. Since the time of Fourier (1768-1830), an extensive mathematical theory of Fourier series has emerged. Today, besides being a vibrant area of research of pure mathematics, it has broad applications to fields such as mathematical physics, medical imaging, data storage and retrieval, and coding theory. A Fourier series is an expansion of a periodic function \(f(x)\) in terms of an infinite sum of sines and cosines. Fourier series make use of the orthogonality relationships of the sine and cosine functions, as we shall explain shortly. The computation and study of Fourier series is known as harmonic analysis and is extremely useful as a way to break up an arbitrary periodic function into a set of simple terms that can be plugged in, solved individually, and then recombined to obtain the solution to the original problem.

### 2.6 Historical Background