# The Lumber Room

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## The functional equation f(x+y) = f(x)f(y)

Suppose $f: \mathbb{R} \to \mathbb{R}$ satisfies $f(x+y) = f(x) f(y)$. What can we say about $f$?

Putting $y = 0$ gives

$\displaystyle f(x) = f(x+0) = f(x)f(0),$

which can happen if either $f(x) = 0$ or $f(0) = 1$. Note that the function $f$ which is identically zero satisfies the functional equation. If $f$ is not this function, i.e., if $f(x) \neq 0$ for at least one value of $x$, then plugging that value of $x$ (say $x^*$) into the equation gives $f(0) = 1$. Also, for any $x$, the equation $f(x^*) = f(x +x^* - x) = f(x)f(x^* - x)$ forces $f(x) \neq 0$ as well. Further, $f(x) = f(x/2 + x/2) = f(x/2)^2$ so $f(x) > 0$ for all $x$.

Next, putting $y = x$ gives $f(2x) = f(x)^2$, and by induction $f(nx) = f(x)^n$. Putting $\frac{x}{n}$ in place of $x$ in this gives $f(n\frac{x}{n}) = f(\frac{x}{n})^n$ which means $f(\frac{x}{n}) = f(x)^{\frac1n}$ (note we’re using $f(x) > 0$ here). And again, $f(\frac{m}{n}x) = f(x)^{m/n}$. So $f(\frac{m}{n}) = f(1)^{m/n}$, which completely defines the function at rational points.

[As $f(1) > 0$, it can be written as $f(1) = e^k$ for some constant $k$, which gives $f(x) = e^{kx}$ for rational $x$.]

To extend this function to irrational numbers, we need some further assumptions on $f$, such as continuity. It turns out that being continuous at any point is enough (and implies the function is $f(x) = f(1)^x$ everywhere): note that $f(x + m/n) = f(x)f(m/n) = f(x)f(1)^{m/n}$. Even being Lebesgue-integrable/measurable will do.

Else, there are discontinuous functions satisfying the functional equation. (Basically, we can define the value of the function separately on each “independent” part. That is, define the equivalence class where $x$ and $y$ are related if $y = r_1x + r_2$ for rationals $r_1$ and $r_2$, pick a representative for each class using the axiom of choice (this is something like picking a basis for $\mathbb{R}/\mathbb{Q}$, which corresponds to the equivalence class defined by the relation $y = r_1x$), define the value of the function independently for each representative, and this fixes the value of $f$ on $\mathbb{R}$. See this article for more details.)

To step back a bit: what the functional equation says is that $f$ is a homorphism from $(\mathbb{R}, +)$, the additive group of real numbers, to $(\mathbb{R}, \times)$, the multiplicative monoid of real numbers. If $f$ is not the trivial identically-zero function, then (as we saw above) $f$ is in fact a homomorphism from $(\mathbb{R}, +)$, the additive group of real numbers, to $(\mathbb{R_+^*}, \times)$, the multiplicative group of positive real numbers. What we proved is that the exponential functions $e^{kx}$ are precisely all such functions that are nice (nice here meaning either measurable or continuous at least one point). (Note that this set includes the trivial homomorphism corresponding to $k = 0$: the function $f(x) = 1$ identically everywhere. If $f$ is not this trivial map, then it is in fact an isomorphism.)

Written by S

Mon, 2013-04-08 at 11:24:08

Posted in mathematics