# Find all function $f$ : $f(x+y)=e^{2xy}f(x)f(y)$

Problem:

Suppose $$f\colon\mathbb{R}\to\mathbb{R^+}$$ is differentiable function which satisfies $$f'(0)=1$$ and $$\forall x, y \in \mathbb{R}, \quad f(x+y)=e^{2xy}f(x)f(y)$$

Where $$\mathbb{R^+}$$ is set of positive real numbers.

Find all function $$f$$.

I found $$f(0)=1$$ and one of $$f(x) = e^{x-x^2}$$ (Am I correct?)

from the problem "Find all function $$f$$", are there more function $$f$$ which satisfies condition?

• Substituting $f(x) = e^{x - x^2}$ and taking logarithm of both sides, we get $x + y - x^2 - y^2 - 2xy = 2xy + x - x^2 + y - y^2$, which is not tautology. Is it a typo and you meant $f(x) = e^{x + x^2}$? – mihaild May 29 '19 at 17:51

As $$f(a)$$ is positive, we can let $$f(a) = e^{g(a)}$$ and get equation: $$g(x + y) = 2xy + g(x) + g(y)$$. Making one more substitution, $$g(a) = h(a) + a^2$$ (so $$f(a) = \exp(h(a) + a^2)$$, we get $$h(x + y) = h(x) + h(y)$$, which is Cauchy's functional equation, and only continuous solutions are $$h(x) = \alpha x$$.
Substituting it back, we get $$f(x) = e^{\alpha x + x^2}$$. $$f'(0) = \alpha$$, so we have the only solution $$f(x) = e^{x + x^2}$$.