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?

  • $\begingroup$ 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 $f(x) = e^{x + x^2}$? $\endgroup$ – 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 $f(x) = e^{x + x^2}$.


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