can we just check an obvious solution of a functional equation?

Question

In a functional equation, if a solution is obvious, can we just check it and say that this is the solution?

Answer 1

In general, that method will work fine for picking out particular solutions to functional equations, but it may not get you every solution to the equation. Consider the functional equation

$$ f(x)\cdot f(-x) = 1. $$

One solution is $f(x) = 1$. Another solution is $f(x) = e^x$. In fact, there is an entire family of solutions of the form $f(x) = c^x$ for $c \neq 0$. So, to answer your question, it is OK to specify a particular solution, but you should be careful when claiming you have found the solution when infinitely many solutions may exist.

If you want to claim that you have found the solution, say $f$, to a functional equation, you will have to show that if there is any other solution $g$, that $g = f$. This says that $f$ is the unique solution of the functional equation in consideration. As I have already exemplified, this is not always possible to show.

I am going to give an example of a situation where one can show that the solution $f$ of a functional equation is the solution. Incidentally, this is almost coming verbatim from the Wikipedia page for "functional equation." However, the way they illustrate uniqueness is not the most general way, so I am taking the time to demonstrate the general method here.

Consider the functional equation $f(x + y)^2 = f(x)^2 + f(y)^2$. Putting $x = y = 0$ shows that $f(0) = 0$. Putting $y = -x$ shows that $0 = f(x)^2 + f(-x)^2$. Squares are always nonnegative, so the sum of squares is nonnegative, and hence $f(x)^2=0$ implies that $f(x) = 0$, identically.

Suppose there was a distinct function $g$ that also satisfied $g(x+y)^2 = g(x)^2 + g(y)^2$. Since $g$ is distinct from $f$, there exists $x^\ast \in \Bbb R$ such that $g(x^\ast) \neq 0$. Then, by applying the same procedure of putting $y = -x^\ast$, we see that $0 = g(x^\ast)^2 + g(-x^\ast)^2$. However, the square of a positive number is itself positive, and squares are nonnegative, so $g(-x^\ast)^2 \geq 0$. The sum of a number that is greater than zero and a number that is at least zero is greater than zero. We see that no such $g$ can exist, and we conclude that $f = 0$ is the unique solution to $f(x+y)^2 = f(x)^2 + f(y)^2$.

Answer 2

No, it is not enough. If you have a functional equation, "solving" the functional equation means finding every solution. That is, your solution set $S$ must be such that a function $f$ is a solution to the equation if and only if $f \in S$.

In practice, what this means is "guessing and checking" will practically never work, because you probably don't have every solution.