# digitális matek

### When the hangman is being hanged

In my earlier posts I wrote about equations which play a central role in mathematical research and applications.

Roughly speaking, an equation is of the form where are given functions defined on some set, and the letters represent the „unknowns” — the quantities which we don’t know, which are hidden in the equation. „Solving” the equation means that we can list, or describe in some way all those possible values of these unknowns from the joint domain of the two functions for which the given equation holds true if we replace by the value , by the value , by the value , etc. It is really hopeless to try to define equations of the most general type: it is better to define some special types, like quadratic equations, ordinary differential equations, linear diophantine equations in two variables, and so on. A relatively simple case is the equation where is a given function on some set, and its values belong to a certain set of numbers which includes . In this case we say that we are looking for the zeros of . If, for instance, is a quadratic polynomial on the real numbers, then we have a well-known formula which produces all zeros — by the way, in this case we know that „all” means „at most two”. But what if in the equation we don’t know , the function itself, but we know that the equation holds for each from its domain? What if the hangman is being hanged? Well, in this trivial case what is the problem? We are looking for a function which takes the value at each point of its domain. What is this function? Of course, this is the identically zero function. Let’s make it a bit more complicated. Find a function on the positive reals which maps to for each positive . This problem can be formulated in the following way: for each . To give the answer observe that every positive real number is the square root of some : if and only if . Substitution into the above equation we have . So, the only function satisfying the required property is the function , the squaring function. We solved our problem, we described all solutions of the given equation, which is called a functional equation to emphasize that the function is the unknown.

The theory of functional equations is an amazing field of mathematics. Of course, it is very difficult, maybe impossible to give a general definition for the concept we call a functional equation: again, it is better to specify different types of functional equations, like difference equations, differential equations, etc. For instance, an important classification depends on the number of variables in the equation. In our previous equations we had a single variable, denoted by , but we can consider equations like , , etc., where we usually suppose that the variables may take any value of the domain of the unknown function . Clearly, we may allow more than one unknown functions in the equation, and all those unknown functions may have more than one variables, and so on. For instance, the functional equation on the reals characterizes exactly those functions which we call odd functions: the ones with graph symmetric with respect to the origin in the plane. Similarly, the equation describes those functions whose graph is symmetric with respect to the -axis in the plane; they are called even functions. These are very simple functional equations in a single variable. Above I mentioned the functional equation — this is a functional equation in two variables.

Before we try to solve this equation we must specify the expected domain of the unknown function . Let’s start with the set of all real numbers : in this case we have a great degree of freedom, as we can substitute arbitrary real values for and in the equation. For instance, substituting we have , that is, is the constant function taking the value at each point. Conversely, it is obvious that any constant function satisfies our equation. We express this result by saying that the general solution of our equation on the reals is provided by the family of all constant functions.

We may have the feeling that here the existence of the plays a crucial role: let’s try to solve our equation on the positive reals. There is no , we cannot substitute it. It is clear, that on a smaller set we might expect more solutions — indeed, a larger domain means more conditions on the unknown function, which are more difficult to satisfy. So, what to do on the positive reals? Let be arbitrary and . Then we have

and it is easy to see that every real number can be written in the form with some . Indeed, for we have the quadratic equation

which implies , and both these values are positive. It follows that whenever — our function is constant on the right of . If , then we take and with we have

as . Now we have that our function is constant on the right of . In the next step we take a with , then with the choice and we have

as . Continuing this process we see that for every positive we have , that is, our function is constant. Since every constant function obviously satisfies our original equation we can draw the conclusion again: the general solution on the positive reals is represented by all constant functions.

Cool, right? You can see that the point is to substitute different values for and in a smart way so that we get more and more information about . And we have two „free” variables — fixing one of them we still have another one which can be arbitrary. Yes, but it would be nice to see a functional equation which has non-constant solutions. A famous equation follows: one of the so-called Cauchy equations, the functional equation

where we suppose that is defined on the whole real line and it takes real values. We see immediately that every function of the form is a solution with an arbitrary real number — these are the so-called {\it linear functions}. A reasonable question is if there are any other solutions? Augustin Louis Cauchy was the great French mathematician who systematically studied this functional equation and he proved that if we add certain further conditions on , then must be a linear function. But he was unable to prove linearity without additional assumptions. Here we show that, for instance, if we assume that is increasing, or decreasing on any open interval, then it must be linear.

First we derive some basic consequences of the equation without any additional condition on . Clearly, , which follows from the substitution . Then the substitution gives for each real , that is, is an odd function. Next, applying induction we obtain that

for any positive integer . In particular, holds whenever is arbitrary and is any integer — as a consequence of the odd property. Now we have

which implies . Put here for with some integer , then we have

that means, rational numbers can be factored out of : , whenever is a real number and is a rational number. So, putting here we have that, in fact, is linear on the rational numbers:

with , for each rational number . Unfortunately, „most” real numbers are irrational, and our equation does not say too much about the function values at irrational numbers. That’s the point when algebra is not enough anymore: we have to add some analysis to go on.
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Now we assume that is, say, increasing on the interval . The decreasing case can be treated in a similar manner — or just replace by , which clearly satisfies the same equation. Let be any real number — we take an increasing sequence and a decreasing sequence of rational numbers such that

and ; this is possible, as every nonempty open interval on the real line includes rational numbers. Subtracting from each sides of this chain of inequalities we obtain

hence we can use the increasing property of on the interval , together with the functional equation:

or

Using the linearity of on the rationals we have

|f(x)-x f(1)|\leq (q_n-p_n) f(1)\leq\frac{1}{n} f(1)nf(x)= f(1)\cdot xxffff

y=0f(x)=g(x)+h(0)x=0f(y)=g(0)+h(y)ghf

k=-h(0)-g(0)

x\mapsto f(x)-kg,hf00\$ is not in the domain of the unknown functions, then the above method does not work. In fact, in the theory of functional equations, it is a common feature that changing the domain can make serious troubles. A functional equation, which can be treated nicely on some domain, may be very difficult, or even impossible to solve on another set. So, the field is open, your creativity can be tested on different types of functional equations. Don’t hesitate, go to the Problems and Solutions!