# $f : \mathbb{R} \to \mathbb{R}$, $f \in \mathscr{C}^{0}$. Find all $f \not \equiv 0$ satisfying a functional equation

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 :)

• What happens if $x=y=0$? – Coward Aug 1 at 10:55
• 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 $>0$. – Hagen von Eitzen Aug 1 at 10:55
• 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 $x \in \mathbb{R}$. – Ralph Clausen Aug 1 at 10:59
• 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 $k$. – Paramanand Singh Aug 1 at 15:56
• @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 $g(x) = cx \implies f(x) = b^{cx}$ – Ralph Clausen Aug 1 at 16:07