We have $f:R\rightarrow R$ continuous with $f(x+y)=e^{3xy}f(x)f(y)$ and $f(1)=e$. Find f.

Question

We have $f:\mathbb{R}\rightarrow \mathbb{R}$ a continuous function with $f(x+y)=e^{3xy}f(x)f(y)$ and $f(1)=e$, Find $f$.

I showed by induction that f is strictly positive on $\mathbb{N}$, then I showed again with induction that f is strictly positive on $\mathbb{Z}$. I don't know how to show f is strictly positive on $\mathbb{Q}$. If I can show that, by the density of $\mathbb{Q}$ in $\mathbb{R}$ we can say that $f\geq0$. But if f is 0 at some point then f is 0 at all points by the relation so we get a contradiction. So f is strictly positive on $\mathbb{R}$.

Now I will take the function $g=\ln(f)$, which is continuous. We get the relation

$g(x+y)=3xy+g(x)+g(y)$

From here we get $g(x+1)=3x+1+g(x)$

We have $g(x+1)-g(x)=3x+1$

By iteration we get $g(x+1)-g(x-(n+1))=\frac{(n+2)(3x-3n-2)}{2}$

The iteration I took looks very bad, so I don't have an idea right now. Can anyone help me with this problem?

Answer 1

I do not see why you struggled so much to prove that $f$ is positive on $\mathbb R$, unless you are missing a hypothesis. If $f(x)=0$ for some $x$, then for any $z \in \mathbb R$, simply taking $y=z-x$ in the equation gives $$ f(z)=f(y+x)=e^{3xy}f(x)f(y) = 0 $$ so that $f(z)=0$ identically. By the intermediate value theorem, $f$ is either positive or negative on $\mathbb R$. However, $f(1)>0$ so $f>0$ on $\mathbb R$, and $g = \ln f$ is well defined and satisfies $$ g(x+y) = 3xy+g(x)+g(y) $$

Note : If you want to alternately show that $f(z)$ is positive without using the intermediate value theorem (which uses continuity of $f$), then for any $z$, let $x=y=\frac z2$ in the functional equation to get $$ f(z) = e^{3z^2/4}f(z/2)^2 $$ must be non-negative for all $z$. Then the logic from my first proof above tells you that $f(z) \neq 0$ for any $z$, so the above automatically implies that $f(z)>0$ for all $z$,and hence $g(z) = \ln f(z)$ is well defined.

The advantage of the above approach is that continuity of $f$ is not required. In particular,

Suppose that $f$ satisfies $f(x+y)=f(x)f(y)e^{3xy}$ and $f$ is merely Lebesgue measurable. Then, $f(x) = \exp\left(\frac{3x^2-x}{2}\right)$ for all $x \in \mathbb R$.

The idea is that we will reduce this question to the Cauchy functional equation, and that has a solution just conditioned the function being Lebesgue measurable.

---

The quadratic guess was a good one. Basically, you need to know how to proceed from $$ g(x+y) = g(x)+g(y)+3xy $$

A quick idea will tell you that if $g(a)=a^2$ was the case, then $g(a+b) = g(a)+g(b)+2ab$ would have been true. However, we have $3ab$ here which is a slight change, so what do we do?

The answer is to see that if $g(a) = ka^2$ for some $k$, then $$ g(a+b) = k(a+b)^2 = ka^2+kb^2+2kab = g(a)+g(b) + 2kab $$

and our equation corresponds to $k = \frac 32$.

Since we worked backward to find this out, we haven't got a "proper" proof yet. To get that kind of a proof, start with $$ g(x+y) = g(x)+g(y)+3xy $$ and let $h(a) = g(a) - \frac{3}{2}a^2$. Then, substituting this into this equation gives $$ h(x+y) = g(x+y) - \frac{3}{2}(x+y)^2 = g(x)+g(y) + 3xy - \frac{3}{2}(x+y)^2 \\ = g(x) - \frac{3}{2}x^2+g(y) - \frac{3}{2}y^2 = h(x)+h(y) $$

which implies that $h(x+y) = h(x)+h(y)$ is a solution of the Cauchy functional equation, and since $f$ is continuous(or Lebesgue measurable), so is $h$. It follows that $h(x) = \alpha x$ for some $\alpha \in \mathbb R$. Going back, $$ g(x) = \frac{3}{2}x^2+\alpha x \implies f(x) = \exp\left(\frac{3}{2}x^2+\alpha x\right) $$

However, we know that $f(1)=e = \exp(1)$, therefore $\frac 32 + \alpha = 1$ and hence $\alpha = -\frac{1}{2}$. All in all $$ f(x) = \exp\left(\frac{3x^2-x}{2}\right) $$ is the answer.

---

There are two key parts of this answer. One is the identification of how one goes from the functional equation $g(x+y)=g(x)+g(y)+2xy$ to $g(x+y) = g(x)+g(y)+3xy$ using a "perturbation" , and the other is how rigorously this is brought back to the Cauchy functional equation via the function $h$, whose solutions are completely characterized under continuity. In isolation these are good tricks to know.

Answer 2

$\textbf{Aliter:}$

$f(x+y+z)=f(x)f(y+z)e^{3{x(y+z)}}=f(x)f(y)f(z)e^{3xy+3xz+3yz}$.

Induct to get $$f(x_1+x_2\dots +x_k)=e^{3(\sum\limits_{i\ne j} x_ix_j)}f(x_1)\dots f(x_k)$$

Set $k=n$, $x_i=\frac1 n \forall i$ to get $$f(1)=e^{3\frac{n(n-1)}{2n^2}}f\big(\frac{1}{n}\big).$$

Once you've found $f(1/n)$, we do this for $k=m$ and $x_i=\frac1 n\forall i$ to get $f(\frac m n)$ and conclude for the reals by continuity.