Second Functional Equations Game

First, $f(x) + g(x) = 4f(x)g(0)$ implies that $g(x) = C f(x),$ where $C = 4g(0)-1$ is constant. So the equation becomes $f(x+y) + Cf(x-y) = 4Cf(x)f(y).$ Next, plug in $(y,0)$ and $(0,y)$ to get $\begin{aligned} f(y) + Cf(y) &= 4Cf(0)f(y) \\ f(y) + Cf(-y) &= 4Cf(0)f(y) \end{aligned}$ so the left sides of these equations are equal. So either $f(y) = f(-y)$ for all $y,$ or $C=0.$ Well, if $C=0$ then we get $f(y) = 0$ for all $y.$ We can ignore this case.

So $f$ is even. Now plug in $(x/2, x/2)$ and $(x/2,-x/2)$ to get $\begin{aligned} f(x) + Cf(0) &= 4Cf(x/2)f(x/2) \\ f(0) + Cf(x) &= 4Cf(x/2)f(-x/2) = 4Cf(x/2)f(x/2) \end{aligned}$ and setting the left sides of these two equations equal yields $(C-1)f(0) = (C-1)f(x)$ for all $x.$ If $C \ne 1$ then $f$ is constant, and equals $5/2$ everywhere, so we get $5/2(1+C) = 25C,$ so $C = 1/9$ and $g(2019) = 5/18.$

So the last case to consider is $C=1$ (and hence $f(x) = g(x)$ ), which gives the equation $f(x+y) + f(x-y) = 4f(x)f(y).$ Note that $C=1$ implies $f(0) = g(0) = 1/2,$ and $f(1) = 5/2.$ Now plug in $(n,1)$ for any positive integer to get $f(n+1) + f(n-1) = 4f(n)f(1) = 10f(n).$ This is an easy recurrence to solve, which yields $f(n) = \frac{(5+2\sqrt{6})^n + (5-2\sqrt{6})^n}4.$ It's not hard to compute that $\lfloor \ln(g(2019)) \rfloor = \lfloor \ln(f(2019)) \rfloor = \fbox{4627},$ so this is the answer.

Note that I still need to show that $f$ actually extends to a continuous function on the real numbers in this case. This is actually pretty easy: simply replace the $n$ 's in the above formula with $x$ 's and show that the resulting function satisfies the functional equation. I will leave this to the reader.

Using these $(x, y)$ values, we find the following equations:

and with $f(1) = \frac{5}{2}$ , this system of equations can be simplified to $792(g(0))^3 - 796(g(0))^2 + 250g(0) - 25 = 0$ , which has solutions $g(0) = \frac{5}{18}$ , $g(0) = \frac{5}{22}$ , and $g(0) = \frac{1}{2}$ .

Now from $(x, 0)$ we have $f(x) + g(x) = 4f(x)g(0)$ and from $(x, 1)$ we have $f(x + 1) + g(x - 1) = 4f(x)g(1)$ . The first gives us $g(x) = (4g(0) - 1)f(x)$ , and both combine to make the recursive equation $g(x + 1) = (4g(0) - 1)(10g(x) - g(x - 1))$ .

If $g(0) = \frac{5}{18}$ , then we have $g(x) = \frac{1}{9}f(x)$ and $g(x + 1) = \frac{1}{9}(10g(x) - g(x - 1))$ . Using $g(0) = \frac{5}{18}$ and $g(1) = \frac{1}{9} \cdot \frac{5}{2} = \frac{5}{18}$ , this recursive equation gives $g(x) = \frac{5}{18}$ for all $x$ . Therefore $g(2019) = \frac{5}{18}$ and $\lfloor \ln(g(2019)) \rfloor = -2$ .

If $g(0) = \frac{5}{22}$ , then we have $g(x) = -\frac{1}{11}f(x)$ and $g(x + 1) = -\frac{1}{11}(10g(x) - g(x - 1))$ . Using $g(0) = \frac{5}{22}$ and $g(1) = -\frac{1}{11} \cdot \frac{5}{2} = -\frac{5}{22}$ , this recursive equation gives $g(x) = -\frac{5}{22}$ for all odd $x$ . Therefore $g(2019) = -\frac{5}{22}$ and is undefined for $\lfloor \ln(g(2019)) \rfloor$ .

If $g(0) = \frac{1}{2}$ , then we have $g(x) = f(x)$ and $g(x + 1) = 10g(x) - g(x - 1)$ . Using $g(0) = \frac{1}{2}$ and $g(1) = f(1) = \frac{5}{2}$ , (and with some help from a computer for the large numbers), this recursive equation gives $\lfloor \ln(g(2019)) \rfloor = 4627$ .

Therefore, the maximum value of $\lfloor \ln(g(2019)) \rfloor$ for all possible functions $g$ is $\boxed{4627}$ .