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<t_{0}, 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

By additivity,

\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}<t_{0}, 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<t_{1}. For t \in (t_{1}-\delta,t_{1}+\delta),

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

By additivity,

\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

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