Question about a Functional Equation.

Let $g(x)= \sin(\frac{\pi}{6}x).$ Then it is easy to see that $g(x)$ satisfies the given functional equation. Now, let us define, $f(x)= \begin{cases} g(x) & \text{if} \; x\text{ is irrational } \\ 2g(x) & \text{if} \; x\; \text{is rational, but not integer} \\ g(x-1) & \text{if}\; x \;\text{ is any integer } \end{cases}$ Then it can be proved that $f(x)$ satisfies the functional equation and is nowhere continuous.

The proof of the fact that $f(x)$ satisfies the functional equation is very simple and can be derived from the fact that $g(x-1)+g(x+1)=\sqrt{3} g(x)$ for all real values of $x,$ but there is a detail that we should not ignore when writing a detailed proof. This important point is the fact that if $x$ is irrational, so are $x-1$ and $x+1$ . Additionally, if $x$ is a non-integer rational number, so are $x-1$ and $x+1,$ and something similar in the case that $x$ is just an integer.

Now, lets prove that $f(x)$ cannot be continuous at any point.

Case 1. (When $x$ is irrational). We can find a sequence ${x_n}$ formed by non-integer rational numbers, such that $\lim_{n\rightarrow \infty }x_n =x.$ Then $\lim_{n\rightarrow \infty }f(x_n)=\lim_{n\rightarrow \infty }2g(x_n)=2g(x)\neq g(x)= f(x).$ Notice that $g(x)\neq2g(x)$ for any irrational value of $x,$ because $2g(x)=g(x)$ if and only if $x$ is an integer number multiple of 6.

Case 2. (When $x$ is a non-integer rational number). We can find a sequence ${x_n}$ formed by
irrational numbers, such that $\lim_{n\rightarrow \infty }x_n =x.$ Then $\lim_{n\rightarrow \infty }f(x_n)=\lim_{n\rightarrow \infty }g(x_n)=g(x)\neq 2g(x)= f(x).$ The reason why the inequality takes place is similar to the one in the Case 1.

Case 3. (When $x$ is an integer). We can find a sequence ${x_n}$ formed by
non-integers rational numbers, such that $\lim_{n\rightarrow \infty }x_n =x.$ Then $\lim_{n\rightarrow \infty }f(x_n)=\lim_{n\rightarrow \infty }g(x_n)=g(x)\neq g(x-1)= f(x).$ The inequality $g(x)\neq g(x-1)$ takes place for any integer value of $x,$ because the solutions of the equation $g(x)= g(x-1)$ are the numbers of the form $6 n - 5/2,$ where $n$ is an arbitrary integer.

Therefore, this proves that the functions $f$ is not continuous at any point.

Remark. The function $f(x)= \begin{cases} g(x) & \text{if} \; x \; \text{is irrational } \\ 2g(x) & \text{if} \; x \; \text{ is rational} \end{cases}$ would not be nowhere continuous. Can you find out why?

We have $f(x-1) + f(x+1) = \sqrt{3}f(x)$ So in easy language sum of two f's is nothing but equal to the root 3 times the mean of the argument of f's.

So plugging $x=0, 1, -1$ we get. $f(-1)+f(1)=\sqrt{3}f(0) \\ f(0)+f(2)=\sqrt{3}f(1) \\ f(-2)+f(0)=\sqrt{3}f(-1) \\$ Adding all we get $(2-\sqrt{3})f(0)+f(2)+f(-2)=(\sqrt{3}-1)(f(-1)+f(1)) \\ (2-\sqrt{3})f(0)+\sqrt{3}f(0)=(3-\sqrt{3})f(0)$ So there may be an equation such that f(0) is not 0 proving it as a nowhere continuous function.