Solving the equation $f(x+t)=f(x)+f(t)+2\sqrt{f(x)}\sqrt{f(t)}$

Question

I am trying to solve the equation $f(x+t)=f(x)+f(t)+2\sqrt{f(x)}\sqrt{f(t)}$ - as in find a function that satisfies this equation. I notice that the RHS is $({\sqrt{f(x)}+\sqrt{f(t)}})^2$ but I am stuck after this.

Answer 1

The presence of $\sqrt{f(x)}$ in your functional equation implies that the range of $f$ is nonnegative.

If you are looking for a continuous function, Simon's comment shows that $g$ must be linear. (With real functions, continuity and additivity imply full linearity. This is mentioned here although I'd prefer to show a link to an actual proof.)

Therefore $$\sqrt{f(x)} = ax$$

for some $a$.

1. If $a=0$, then $f$ is the zero function.
2. If $a>0$, then $f$ is only defined for $x\geq 0$, and $f(x)=a^2x^2$.
3. If $a<0$, then $f$ is only defined for $x\leq 0$, and $f(x)=a^2x^2$.

That is, there are three families of solutions. They all have the same form $f(x)=a^2x^2$, but either you must restrict the domain to non-negatives, non-positives, or $a $ is necessarily $0$.

Answer 2

Use the hint and solve in the usual step by step way for $g$ first. Assume wlog $g(1)= 1$.

1. solve for $x, t$ natural numbers and $0$ (domain can't be negative) to see $g(n) = n$.

2. solve for fractions with numerator being $1$ to see $n\cdot g(1/n) = 1$.

3. now for rationals to see $q/p \cdot g(p/q) = 1$

4. use limit argument for irrationals to get $g(r) = r$

5. plug back into $f$ to conclude one solution is $f(n) = n^2$.

Answer 3

solution: $f(x) = a^2 x^2$ where $a$ is any real number