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:

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

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

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

Now, if $$4 f(x) = f(0)$$ 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 $$f(x) = a x^2$$: We know that $$f(0) = 0$$ and that $$f(1) = a$$. Now let's assume $$f(x) = a x^2$$ for the first n integers. Then, $$f(n + 1) = f(n) + f(1) + 2\sqrt{f(n)f(1)} = a n^2 + a + 2\sqrt{a n^2 a} = a (n^2 + 1 + 2n) = a(n+1)^2$$.

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 $$f\left( \frac{1}{p} \right) = a\ \frac{1}{p^2} $$