0.1 Cauchy’s FE and Its Variants
Unlike the other problems, if there is one thing that you must know to solve a FE problem, it would be Cauchy FE. This could be a very nice problem, but could also be a very mean one. Before we look at what Cauchy FE is, let’s discuss some very introductory topics from real analysis.
Theorem 0.1.1: Density of Rational Numbers
For any real numbers \(a\) and \(b\), it is possible to choose a rational number \(q\) such that \(a < q < b\).
Proof.
Consider the interval \((na, nb)\) for some integer \(n\). Choose a sufficiently large positive integer \(n\) such that \(nb - na > 1\). Because the length of the interval is greater than \(1\), we can choose another integer \(m\) such that \(na < m < nb\). Therefore, \(a < \frac {m}{n} < b\) and \(\frac {m}{n} \in \mathbb {Q}\) since \(m, n \in \mathbb {Z}\).
This seems quite obvious when you think about it. However, it is important to consider and question the very seemingly obvious properties since conditions are very important for Cauchy FE. Continuing, we could try the same for irrational numbers.
Theorem 0.1.2: Density of Irrational Numbers
Given arbitrary real numbers \(a\) and \(b\), it is possible to choose a irrational number \(q\) such that \(a < q < b\).
Proof.
First, consider the inequality \(\frac {a}{\sqrt {2}} < r < \frac {b}{\sqrt {2}}\) for some rational number \(r\) by Theorem 0.1.1 . Multiplying \(\sqrt {2}\), \(a < \sqrt {2} r = q < b\) holds.
Using these two, we can notice that given two reals, we can choose a real number between them. Now before moving on to Cauchy FE, consider the following theorem.
Theorem 0.1.3: Bolzano-Weierstrass Theorem
Any real number \(x\) can be written as the limit of bounded sequence of rational numbers and of irrational numbers.
Proof.
Let \(r_n\) be an increasing sequence of rational numbers defined as \(x - \frac {1}{n} < r_n < x - \frac {1}{n+1}\) by Theorem 0.1.1 . Therefore, \[ x - \frac {1}{n} < r_n < x - \frac {1}{n + 1} < r_{n+1} \] holds and \(x = \lim _{n \to \infty } r_n\) by Squeeze Theorem. If \(r_n\) is a decreasing sequence defined as \(x + \frac {1}{n+1} < r_n < x + \frac {1}{n}\), then \(r_{n+1} < x + \frac {1}{n+1} < r_n < x + \frac {1}{n}\). Therefore \(x = \lim _{n \to \infty } r_n\) by Squeeze Theorem. The case of irrational numbers are proven in analogous way.
Now with these ideas in mind, let’s finally discuss Cauchy’s Functional Equations!
Definition 0.1.4
Equations with the following form is known as Cauchy’s Functional Equations. \[ f(x + y) = f(x) + f(y) \]
Now this is pretty straight forward for certain input domain. Let’s start by what we can notice just by looking at the equations. Substituting \(x = y = 0\), we get \(f(0) = 0\). Moreover, substituting \(y = -x\) gives \(f(x) = -f(-x)\) and we can see that \(f\) is an odd function. With these in mind and having \(k = f(1)\), let’s start looking into different input domains.
I. \(\mathbf {f: \mathbb {N} \to \mathbb {R}}\)
This case is pretty straightforward. Consider the following equations for some integer \(n\). \[ f(n) = f(n - 1) + f(1) = f(n - 2) + f(1) + f(1) = \cdots = n f(1) \] Therefore, \(f(x) = kx\) if \(x\) is defined under natural numbers. An extension would be using integers.
II. \(\mathbf {f: \mathbb {Z} \to \mathbb {R}}\)
For this case, we will be using the fact that \(f\) is an odd function. For \(x > 0\), we have showed that \(f(x) = kx\). For \(x < 0\), consider the following equations. \[ f(x) = -f(-x) = -k(-x) = kx \] Therefore with \(f(0) = 0\), \(f\) is still linear.
III. \(\mathbf {f: \mathbb {Q} \to \mathbb {R}}\)
The next extension that we can have is defining \(x\) over the rational numbers. Let \(p\) and \(q\) be relatively prime integers. First consider the following equation. \[ f \left ( \frac {p}{q} \right ) = f \left ( \frac {1}{q} + \cdots + \frac {1}{q} \right ) = f \left ( \frac {1}{q} \right ) \cdot p \] Therefore, \(f(1) = f \left ( \frac {1}{q} \right ) \cdot q\) and substituting back to the equation gives the following. \[ f \left ( \frac {p}{q} \right ) = \frac {f(1)}{q} \cdot p = \frac {p}{q} k \] Therefore, \(f(x) = kx\) also holds in this case. The final case that we can consider is when \(f: \mathbb {R} \to \mathbb {R}\).
IV. \(\mathbf {f: \mathbb {R} \to \mathbb {R}}\)
This is the part where it gets quite tricky and where conditions becomes very important. There are many different conditions that we could consider for this case, but let’s focus on two major conditions for now.
IV.I \(\mathbf {f}\) is an increasing or decreasing function.
First, let’s consider the case when \(f\) is increasing. Assume for the sake of contradiction that \(f(x) \neq kx\). If \(f(x) > kx\), then \(x < \frac {f(x)}{k}\) and we can choose a rational \(q\) such that \(x < q < \frac {f(x)}{k}\) by Theorem 0.1.1 . Therefore, \(f(x) \leq f(q) = kq\) since \(q \in \mathbb {Q}\). Continuing, \(\frac {f(x)}{k} \leq q < \frac {f(x)}{k}\) is obtained, which is a contradiction.
This time, assume that \(f(x) < kx\). We can choose a rational \(q\) such that \(\frac {f(x)}{k} < q < x\). Continuing, we get \(kq = f(q) \leq f(x) < kq\) which again is a contradiction. This proves that if \(f\) is an increasing function, then \(f(x) = kx\) for \(x \in \mathbb {R}\).
The case when \(f\) is decreasing is quite simple. Let \(g(x) = -f(x)\). By construction, \(g\) is an increasing function and the following equations hold. \[ -f(x + y) = g(x + y) = g(x) + g(y) = -f(x) - f(y) = -(f(x) + f(y)) \] Therefore, \(f(x + y) = f(x) + f(y)\). This completes the proof!
Note that these conditions are necessary. We cannot blindly assume that \(f(x) = kx\) for \(x \in \mathbb {R}\) without the condition. One very, very popular way of knowing if a function is increasing is \(f(xy) = f(x) f(y)\).
If \(f(xy) = f(x) f(y)\), then \(f(x^2) = f(x)^2 \geq 0\). Therefore for positive \(y\), \(f(x) = f(x - y) + f(y) \geq f(x - y)\) and \(f\) is increasing.
If \(f(xy) = f(x) f(y)\), then \(f\) is an increasing function.
This is very popular, and will make our lives greatly easier to use \(f(x) = kx\).
IV.II \(\mathbf {f}\) is continuous.
The second very popular condition is when \(f\) is continuous. Note that by Bolzano-Weierstrass Theorem, there exists an increasing rational sequence \(a_n\) and decreasing rational sequence \(b_n\) such that \(x = \lim _{n \to \infty } a_n = \lim _{n \to \infty } b_n\) for any real \(x\). Therefore, the following equations hold. \begin{align*} f(x) &= f \left ( \lim _{n \to \infty } a_n \right ) = \lim _{n \to \infty } f(a_n) = \lim _{n \to \infty } ka_n = kx \\ &= f \left ( \lim _{n \to \infty } b_n \right ) = \lim _{n \to \infty } f(b_n) = \lim _{n \to \infty } kb_n = kx \end{align*}
Therefore, \(f(x) = kx\) holds if \(f\) is continuous.
This is the basic idea of Cauchy FE! With this in mind, let’s solve few practice problems to see how they are used.