Properties of Exponential Functions from the Functional Equation

Exponential Functions

An exponential function is a continuous nonconstant function $E\colon\mathbb{R}\to\mathbb{R}$ satisfying the law of exponents $E(x+y)=E(x)E(y)$ for all $x,y\in\mathbb{R}$.

Exponentials are Nonzero

If $E$ is an exponential function then $E(x)\neq 0$ for all $x$.

Let $E$ be an exponential function and assume there is some $z\in\RR$ such that $E(z)=0$. Then for any $x\in\RR$ we may write $x=x-z+z=(x-z)+z=y+z$ for $y=x-z\in\RR$. Evaluating $E(x)$ using the law of exponents, $E(x)=E(y+z)=E(y)E(z)=E(y)\cdot 0 =0$ Thus $E$ is constantly zero, which is a contradiction as we assumed $E$ was a nonconstant solution to the Law of Exponents.

E(0)=1

If $E$ is any exponential function, then $E(0)=1$.

Since $0=0+0$ apply the law of exponents: $E(0)=E(0)E(0)$ Since we know $E(0)$ to be nonzero we can divide through by it, yielding $1=E(0)$

Exponential functions are positive.

Exponential on Natural Numbers

Prove that if $E$ is an exponential function, $x\in\RR$ and $n\in\mathbb{N}$ then $E\left(nx\right)=E(x)^{n}$

Since $nx=x+x+\cdots+x$, we may inductively apply the law of exponents: $E\left(\underbrace{x+x+\cdots+x}_n\right)=\underbrace{E(x)E(x)\cdots E(x)}_n=E(x)^n$

Exponential on Integers

Prove that if $E$ is an exponential function, $x\in\RR$ and $j\in\mathbb{Z}$ then $E\left(jx\right)=E(x)^{j}$

If $j$ is positive, the result is already proven for natural numbers above. If $j=0$ then the claim reduces to $E(0)=1$, also already proven. So, it remains to consider only the case $j=-n$ for $n\in\mathbb{N}$. Here, we apply the law of exponents to $n-n=0$ to get $1=E(0)=E(nx-nx)=E(nx)E(-nx)$ Thus $E(-nx)=1/E(nx)$, and we know $E(nx)=E(x)^n$ from the natural number case. Putting this together with the definition of negative exponents, $E(jx):=E(-nx)=\frac{1}{E(x)^n}=E(x)^{-n}=E(x)^j$

Exponential on Rationals

Prove that if $E$ is an exponential function, $x\in\RR$ and $r\in\mathbb{Q}$ then $E\left(xr\right)=E(x)^{r}$

Let $r=p/q$ for integers $p,q$. We can assume without loss of generality that $p\neq 0$ (as we’ve covered this case) and $q > 1$ ($q=0$ is not allowed, and we can always collect negative signs into the numerator, and $q=1$ is the integer case, also already covered.)

We proceed in two steps: first we look at fractions of the form $1/n$, so we must evaluate $E(\frac{1}{n}x)$. Here we apply the law of exponents to the identity $1=\frac{1}{n}+\frac{1}{n}+\cdots+\frac{1}{n}$:

Since $E(x)$, $E(x/n)$ are known to be positive, we can uniquely take the $n^{th}$ root of both sides to get

$E(x)^{\frac{1}{n}}=E\left(\tfrac{1}{n}x\right)$

Now let $r=p/q$ be an arbitrary fraction, and consider $E(\tfrac{p}{q}x)$. We compute as

Thus $E(rx)=E(x)^r$ as claimed.

Exponentials Agreeing at a Point

If $E,F$ are exponentials which take the same value at any nonzero point $c$, then they agree on the entire real line.

Assume $E(c)=F(c)=a$, and consider the set $S$ of rational multiples of $c$: $S=\{ c r\mid r\in\mathbb{Q}\}$ This is dense in $\RR$ and for every $s\in S$ we can calculate $E(s)$ and $F(s)$ using the previous proposition $E(s)=E(rc)=E(c)^r=a^r$ $F(s)=F(rc)=F(c)^r=a^r$ Thus $E(s)=F(s)$ on all of $S$, so the functions agree on a dense set, and by continuity, agree on all of $\RR$.

Base of an Exponential

If $E$ is an exponential function, its base is defined as the value $a=E(1)$.

Uniqueness with Given Base

If two exponentials have the same base, they are equal on the entire real line.

By defintion they agree at $1$, so apply the previous proposition.

If $E$ is an exponential, then restricted to the positive reals $E(x)$ is either (i) always greater than 1, or (ii) always less than 1.

Assume there are two positive numbers $a,b$ with $E(a) > 1$ and $E(b) < 1$. Then by the intermediate value theorem there must be a point $c\in (a,b)$ where $E(c)=1$.
Now consider the set of rational multiples of $c$: $S=\{ c r\mid r\in\mathbb{Q}\}$ This is dense in $\RR$ and for every element $s\in S$ we see $E(s)=E(rc)=E(c)^r=1^r=1$ Thus $E$ is constantly equal to 1 on the dense set $S$, and so by continuity is constantly equal to $1$ on all of $\RR$. This contradicts the definition of exponentials, but even more immediately contradicts the premise of this proposition, that $E(a) >1$ and $E(b) <1$.

Exponentials are Monotone

If $E$ is an exponential function then $E$ is strictly monotone: it is monotone increasing if $E(x) > 1$ on the positive reals, and decreasing if $E(x) < 1$ there.

Without loss of generality we work with the case that $E > 1$ on the positives. Let $x < y$ be arbitrary real numbers, we wish to show that $E(x) < E(y)$. Since $y-x > 0$ we know $E(y-x) > 1$ and by the law of exponents $E(y)= E(y-x+x)=E(y-x)E(x) > 1 E(x)$ So, $E(y) > E(x)$ as required.

Convex Function

A function $f$ is convex on an interval if the secant line between any two points in the interval lies strictly above the graph of $f$.

Exponentials are Convex

Let $E$ be an exponential. Then $E$ is convex on all of $\mathbb{R}$.