User:Jotomicron/Functional Equations

Solve the functional equation $f(x+y)=f(x)+f(y)+2{\sqrt {f(x)f(y)}}\quad \forall {x,y\in \mathbb {R} }$ (taken from ...)

If this equation is true for any y, then it is true for the particular case y = 0. Thus:

$f(x)=f(x)+f(0)+2{\sqrt {f(x)f(0)}}$

$\implies -f(0)=2{\sqrt {f(x)f(0)}}$

$\implies f(0)^{2}=4f(x)f(0)\implies f(0)=0\lor 4f(x)=f(0)\quad \forall {x\in \mathbb {R} }$

Now, if $4f(x)=f(0)$ for all x, then it is also true for x = 0, which means that 4 f(0) = f(0), thus f(0) = 0 and $f(x)=0$, which is a solution. From the two different conclusions above, this eliminates at once the need to analyse the second. Let's now focus on the first: f(0) = 0.

Let a>0 be a real number with the value f(1). We now prove that any solution must be of the form $f(x)=ax^{2}$: We know that $f(0)=0$ and that $f(1)=a$. Now let's assume $f(x)=ax^{2}$ for the first n integers. Then, $f(n+1)=f(n)+f(1)+2{\sqrt {f(n)f(1)}}=an^{2}+a+2{\sqrt {an^{2}a}}=a(n^{2}+1+2n)=a(n+1)^{2}$.

Now we simplify the original equation into $f(x+y)=\left({\sqrt {f(x)}}+{\sqrt {f(y)}}\right)^{2}$. Then:

$a=f(1)=f\left({\frac {p-1}{p}}+{\frac {1}{p}}\right)=\left({\sqrt {f\left({\frac {p-1}{p}}\right)}}+{\sqrt {f\left({\frac {1}{p}}\right)}}\right)^{2}$ $=\left({\sqrt {\left({\sqrt {f\left({\frac {p-2}{p}}\right)}}+{\sqrt {f\left({\frac {1}{p}}\right)}}\right)^{2}}}+{\sqrt {f\left({\frac {1}{p}}\right)}}\right)^{2}$ $=\left({\sqrt {f\left({\frac {p-2}{p}}\right)}}+{\sqrt {f\left({\frac {1}{p}}\right)}}+{\sqrt {f\left({\frac {1}{p}}\right)}}\right)^{2}$$=\cdots$$=\left(p{\sqrt {f\left({\frac {1}{p}}\right)}}\right)^{2}$ which means that $f\left({\frac {1}{p}}\right)=a\ {\frac {1}{p^{2}}}$