## Cauchy Functional Equation

Over a month ago, I wrote about Steinhaus’ theorem, which states that if $A,B\subset\mathbb{R}$ are sets of positive Lebesgue measure (we will use the notation $\left|\cdot\right|$ to denote the Lebesgue measure of a set), then the set

$\displaystyle A+B:=\left\{x \in \mathbb{R} : x=a+b,a\in A,b\in B\right\}$

contains an open interval. I want now to use the Steinhaus theorem to prove the uniqueness of a measurable solutions to the functional equation $f(t+s)=f(t)+f(s)$, where $f: [0,\infty) \rightarrow\mathbb{R}$. Later, we will use this result to show that the only continuous probability distribution with the so-called “memorylessness” property is the exponential distribution.

We say that a function $f: \mathbb{R}^{> 0} \rightarrow \mathbb{R}$ satisfies the Cauchy (functional) equation if

$\displaystyle f(s+t)=f(s)+f(t),\indent s,t>0$

Note that the extension of $f$ to the domain $\mathbb{R}$ is uniquely defined. Since $f$ is additive, we see that $f(0) = 0$ and therefore $f(-x+x)=-f(x)$. We now prove the reverse direction. Clearly, $f(x+y)=f(x)+f(y)$, for $x,y \geq 0$. If $x \geq 0, y < 0$ and $x+y \geq 0$, then

$\displaystyle f(x)=f(x+y+-y)=f(x+y)+f(-y)\Longleftrightarrow f(x)+f(y)=f(x+y)$

The argument for the case $x < 0$, $y \geq 0$, and $x+y < 0$ is completely analogous. Lastly, for the case $x < 0, y < 0$,

$\displaystyle f(x+y)=-f(-x+-y)=-f(-x)-f(-y)=f(x)+f(y)$

Lemma 1. If $f: \mathbb{R}^{> 0}\rightarrow \mathbb{R}$ satisfies the Cauchy equation, then

$\displaystyle f\left(\dfrac{k}{n}t\right)=\dfrac{k}{n}f(t),\indent t\in\mathbb{R}^{>0}, k,n\in \mathbb{Z}^{\geq 1}$

Proof. Let $n \in \mathbb{Z}^{\geq 1}$ and $t \in \mathbb{R}^{>0}$. Then

$\displaystyle f(t)=\underbrace{f\left(\dfrac{1}{n}t\right)+\cdots+f\left(\dfrac{1}{n}t\right)}_{n\text{ times}}=nf\left(\dfrac{1}{n}t\right),$

so dividing both sides by $n$ yields the case $k = 1$. Replacing $t$ by $kt$ and applying additivity again yields the formula. $\Box$

Observe that if $f$ is continuous, then by the density of $\mathbb{Q}$ in $\mathbb{R}$, we see that $f(x) = xf(1)$ for all $x \in \mathbb{R}^{> 0}$.

Lemma 2. If $f: \mathbb{R}^{> 0}$ is additive, then in an interval $(t_{0}-\delta,t_{0}+\delta)$, where $0 < \delta, $f$ is bounded from above if and only if $f$ is bounded from below.

Proof. Suppose $f$ is bounded from above by $M_{1}$. For $t \in (t_{0}-\delta, t_{0}+\delta)$, set $t' := 2t_{0} - t$. I claim that $t' \in (t_{0} - \delta, t_{0}+\delta)$. Indeed,

$\displaystyle t_{0}-\delta=2t_{0}-(t_{0}+\delta)<{2t_{0}-t}<2t_{0}-(t_{0}-\delta)=t_{0}+\delta$

$\displaystyle f(t')+f(t)=f(2t_{0})=2f(t_{0})\Rightarrow f(t)=2f(t_{0})-f(t')>2f(t_{0})-M_{1}$

Similarly, if $f$ is bounded from below by $M_{2}$, then

$\displaystyle f(t)=2f(t_{0})-f(t')<2f(t_{0})-M_{2}$

$\Box$

Lemma 3. Suppose $f: \mathbb{R}^{> 0}\rightarrow \mathbb{R}$ is additive and is bounded by $M$ in an interval $(t_{0}-\delta,t_{0}+\delta)$. Then

$\displaystyle\left|f(t) - f(t_{0})\right|\leq\frac{2M}{\delta}\left|t-t_{0}\right|, \indent t \in (t_{0}-\delta, t_{0}+\delta)$

In particular, $f$ is locally Lipschitz at $t_{0}$ and therefore continuous at $t_{0}$.

Proof. For $t \in (t_{0}-\delta, t_{0}+\delta)$, let $t'$ be as above. Since $f(t)+f(t')=2f(t_{0})$ and $t+t'=2t_{0}$, we see that $\left|f(t) - f(t_{0})\right|=\left|f(t')-f(t_{0})\right|$ and $\left|t-t_{0}\right|=\left|t'-t_{0}\right|$. Hence, it suffices to consider $t \in (t_{0}-\delta, t_{0})$. I claim that we may choose a sequence $(t_{n})_{n=1}^{\infty}$ such that $t_{n} \downarrow t_{0}-\delta$, $t_{1}, and

$\displaystyle\alpha_{n}:=\frac{t_{0} - t}{t_{0}-t_{n}}\in\mathbb{Q}$

Indeed, the function $\alpha(s)=\frac{t_{0} - t}{t_{0}-s}$ is continuous, so the claim follows from by the intermediate value theorem and the density of the rationals. Rewriting the above expression for $\alpha_{n}$, we see that

$\displaystyle t=t_{0}-\alpha_{n}(t_{0}-t_{n})=(1-\alpha_{n})t_{0}+\alpha_{n}t_{n}$

so that

$\begin{array}{lcl}&\displaystyle f(t)&\displaystyle =f\left((1-\alpha_{n})t_{0}+\alpha_{n}t_{n}\right)=(1-\alpha_{n})f(t_{0})+\alpha_{n}f(t_{n})\\&\Longrightarrow&\displaystyle f(t)-f(t_{0})=\alpha_{n}\left(f(t_{n})-f(t_{0})\right)\leq 2M\frac{t_{0}-t}{t_{0}-t_{n}}\rightarrow\frac{2M}{\delta}(t_{0}-t)\end{array}$

as $n\rightarrow\infty$.

Analogously, we can choose a sequence $(t'_{n})_{n=1}^{\infty}$ such that $t'_{n} \uparrow t_{0} + \delta$, $t'_{1} > t_{0}$, and $\alpha'_{n} = \frac{t_{0} - t}{t'_{n}-t} \in \mathbb{Q}$. Then

$\displaystyle t_{0}=\alpha'_{n}t'_{n}+(1-\alpha'_{n})t$

so that

$\displaystyle f(t_{0})-f(t)=\alpha'_{n}f(t'_{n})-\alpha'_{n}f(t)\leq 2M\alpha'_{n}\rightarrow\frac{2M}{t_{0}+\delta-t}(t_{0}-t)\leq\frac{M}{\delta}(t_{0}-t)$

as $n \rightarrow \infty$. $\Box$

Lemma 4. If $f: \mathbb{R}^{> 0} \rightarrow \mathbb{R}$ is additive and is bounded from above in an interval $(t_{0}-\delta, t_{0}+\delta)$, where $t_{0} > \delta > 0$, then it is also bounded from above in the interval $(t_{1}-\delta_{1},t_{1}+\delta_{1})$, where $t_{1} \in \mathbb{R}$ and $\delta_{1} := \min\left\{t_{1},\delta\right\}$.

Proof. Fix $t_{1} > 0$, and without loss of generality, we may assume that $0 <\delta. For $t \in (t_{1}-\delta,t_{1}+\delta)$,

$\displaystyle t':=t+t_{0}-t_{1}\in (t_{0}-\delta,t_{0}+\delta)$

$\begin{array}{lcl} \displaystyle f(t)=f(t')+f(t_{1}-t_{0})&=&\displaystyle f(t')+\text{sgn}(t_{1}-t_{0})f(\left|t_{1}-t_{0}\right|)\\&\leq&\displaystyle\sup_{t'\in (t_{0}-\delta,t_{0}+\delta)}f(t')+\text{sgn}(t_{1}-t_{0})f(\left|t_{1}-t_{0}\right|)\end{array}$

where we know that the supremum is finite by hypothesis that $f$ is bounded on the interval $(t_{0}-\delta,t_{0}+\delta)$. $\Box$

Theorem 5. If $f:\mathbb{R}^{> 0}\rightarrow \mathbb{R}$ satisfies the Cauchy functional equation and is measurable, then it is of the form

$\displaystyle f(x)=xf(1),\indent\forall x\in\mathbb{R}^{ > 0}$

Proof. By a previous observation, it suffices to show that $f$ is continuous. Since $\mathbb{R}^{> 0}=\bigcup_{n=1}^{\infty}\left\{t \in \mathbb{R}^{> 0} : f(t) \leq n\right\}$, we see that $\left|\left\{t : f(t) \leq n\right\}\right|>0$ for some $n\in\mathbb{Z}^{\geq 1}$. Set

$\displaystyle A:=\left\{t:f(t)\leq n\right\}+\left\{t: f(t)\leq n\right\}$

so that for all $t \in A$, $f(t)\leq 2n$. By Steinhaus’s theorem, $A$ contains an open interval $(t_{0}-\delta,t_{0}+\delta)$, where $\delta > 0$. In particular, $f$ is bounded on the interval $(t_{0}-\delta,t_{0}+\delta)$, so by the preceding lemma, $f$ is bounded from above on the interval $(t_{1}-\delta_{1},t_{1}+\delta_{1})$, where $t_{1}>0$ and $\delta_{1}=\min\left\{t_{1},\delta\right\}$. By Lemma 3, $f$ is continuous at $t_{1}$. $\Box$