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This problem is from an AoPS thread post.


$\blacksquare~$ Problem: Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a continuous function that satisfies the functional relation $$ f(x+y)=a^{x y} f(x) f(y) \quad \forall~ x, y \in \mathbb{R} $$ where $a$ is a positive real constant. Find all functions $f$ that satisfy the above conditions and are not identically zero.


$\blacksquare~$ My approach: At first we have a few claims.

$\bullet~$ Claim: For any $x \in \mathbb{R},$ $f(x) > 0$, i.e., im($f$) $\subseteq$ $\mathbb{R}^{+}$.

$\bullet~$ Proof: Plugging in $x = y = \dfrac{t}{2}$ in the functional equation (for any arbitrary $t \in \mathbb{R}$)
$$ f(t) = a^\frac{t^2}{4} f\bigg( \frac{t}{2} \bigg)^2 > 0 \quad \text{for any arbitrary }~t \in \mathbb{R}$$ Therefore we have that $f(x) > 0$ for any $x \in \mathbb{R}$.

Let's now define a function $g : \mathbb{R} \to \mathbb{R}$ such that $$ g(x) := \log_{a} (f(x)) $$ As $f(x) > 0 $ for any $x \in \mathbb{R},$ then $g$ is well defined.

Now the functional equation becomes $$ g(x + y) = g(x) + g(y) + xy \quad \quad \cdots \cdots (1) $$ Then let's pick $~h:\mathbb{R} \to \mathbb{R}$ such that $$ h(x) = g(x) - \frac{x^2}{2} $$ Therefore the functional equation $(1)$ is reduced to $$ h(x + y) = h(x) + h(y) \quad \quad ~~~~~~~\cdots \cdots (2) $$ And we know that as $f \in \mathscr{C}^{0}(\mathbb{R}) \implies g \in \mathscr{C}^{0}(\mathbb{R}) \implies h \in \mathscr{C}^{0}(\mathbb{R})$. Then, by $~\mathbb{N} \to \mathbb{Z} \to \mathbb{Q} \to \mathbb{R}~$ rule we have that $~h(x) = cx~$ for $x \in \mathbb{R}$.

Then we have that $$ f(x) = a^{g(x)} = a^{h(x) + \frac{x^2}{2}} = a^{cx + \frac{x^2}{2}} $$


Please check the Solutions for glitches and other ideas :)

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  • $\begingroup$ What happens if $x=y=0$? $\endgroup$ – Coward Aug 1 at 10:55
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    $\begingroup$ Your claim should be $\ge0$, not $>0$. But you can show that if $f(x_0)=0$ for some $x_0$, then $f(x)=0$ for all $x$ (and this is excluded). So, with that addition, you do indeed get $>0$. $\endgroup$ – Hagen von Eitzen Aug 1 at 10:55
  • $\begingroup$ But if $f(0) = 0$, then we have that $$ f(x) = f(x) \cdot f(0) = 0 \quad \text{for any } x \in \mathbb{R} $$ hence $f(0) = 0 \implies f(x) \equiv 0 ~$ for any $x \in \mathbb{R}$. $\endgroup$ – Ralph Clausen Aug 1 at 10:59
  • $\begingroup$ Your approach works fine for every positive $a$ except $a=1$ (logarithm to base $1$ can't be defined). But for $a=1$ it is easy to see that $f(x) =k^x$ for some positive $k$. $\endgroup$ – Paramanand Singh Aug 1 at 15:56
  • $\begingroup$ @ParamanandSingh I can deal with the case $f(x + y) = f(x) f(y)$ easily, as the Claim remains same for this case too. i.e., $g(x) := \log_{b}(f(x))$ for $b > 0$ and $g$ makes sense. Then the equation becomes Cauchy's Functional equation $$g(x + y) = g(x) + g(y)$$ having sloution $g(x) = cx \implies f(x) = b^{cx} $ $\endgroup$ – Ralph Clausen Aug 1 at 16:07

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