Spivak Derivation, Ch. 18: Let $f(x+y)=f(x)\cdot f(y)$. How does the condition $f(1)=10$ imply $f(x)=10^x$ for rational $x$?

Question

I am on Chapter 18 "The Logarithm and Exponential Functions" of Spivak's Calculus. The first few pages of this chapter seem a bit less well written than previous chapters and I am having a difficult time following the flow of ideas.

Let me try to write out here these ideas.

Consider the function $f(x)=10^x$.

This function is assumed to be defined for all $x$ and to have an inverse function defined for positive $x$, which is the "logarithm to the base 10"

$$f^{-1}(x)=\log_{10}x$$

At this point Spivak tries to review the definition of $10^x$ for rational $x$.

The symbol $10^n$ is first defined for natural numbers $n$.

This notation turns out to be extremely convenient, especially for multiplying very large numbers, because

$$10^n\cdot 10^m=10^{n+m}\tag{1}$$

He doesn't actually say what the definition is.

Next, since we want the equation $10^0\cdot 10^n=10^{0+n}=10^n$ to be true, we have to define $10^0=1$

Since we want the equation $10^{-n}\cdot 10^n= 10^{-n+n}=10^0=1$ to be true, we define $10^{-n}=1/10^n$.

Since we want the equation $10^{1/n}\cdot10^{1/n}\cdot10^{1/n}\cdot ... \cdot 10^{1/n}=10^{1/n+1/n+...+1/n}=10^1=10$ to be true, where the product is of $n$ terms $1/n$, we must define $10^{1/n}=\sqrt[n]{10}$.

Finally, since we want the equation $10^{1/n}\cdot10^{1/n}\cdot10^{1/n}\cdot ... \cdot 10^{1/n}=10^{1/n+1/n+...+1/n}=10^{m/n}$ to be true, where the product is of $m$ terms $1/n$, we must define $10^{m/n}=(\sqrt[n]{10})^m$

Okay, at this point there is the following snippet

Unfortunately, at this point the program comes to a dead halt. We have been guided by the principle that $10^x$ should be defined so as to ensure that $10^{x+y}=10^x10^y$; but this principle does not suggest any simple algebraic way of defining $10^x$ for irrational $x$. For this reason we will try some more sophisticated ways of finding a function $f$ such that

$$f(x+y)=f(x)\cdot f(y), \text{ for all } x \text{ and } y\tag{2}$$

Of course, we are interested in a function which is not always zero, so we might add the condition $f(1)\neq 0$. If we add the more specific condition $f(1)=10$, then $(2)$ will imply $f(x)=10^x$ for rational $x$, and $10^x$ could be defined as $f(x)$ for other $x$; in general $f(x)$ will equal $[f(1)]^x$ for rational $x$.

How does the condition $f(1)=10$ imply $f(x)=10^x$ for rational $x$?

Answer 1

The first half of the text tells you that there is a unique way of defining the symbol $10^x$ for $x\in\Bbb{Q}$, such that $10^1=10$ and for all $x,y\in\Bbb{Q}$, $10^{x+y}=10^x\cdot 10^y$. So, there is (i.e existence) a definition given in the first half of the text, and if you follow the argument carefully, it also gives you uniqueness. Here's the same thing repeated:

Theorem (Definition and Existence).

Given any $a>0$, one can define the symbol $a^x$ for $x\in\Bbb{Q}$ in such a manner that $a^1=1$, and for all $x,y\in\Bbb{Q}$, we have $a^{x+y}=a^x\cdot a^y$.

The first half of Spivak's text, as motivation, gives you such a definition in the special case $a=10$. But clearly, nothing special was used about the number $10$, other than it being positive (so the roots were all defined). Now, if you go through the existence proof carefully, you'll see that it also establishes uniqueness:

Theorem (Uniqueness).

For any $a>0$, if $f:\Bbb{Q}\to\Bbb{R}$ is a function satisfying $f(1)=a$ and for all $x,y\in\Bbb{Q}$, $f(x+y)=f(x)f(y)$, then for all $x\in\Bbb{Q}$, we have $f(x)=a^x$.

The proof is literally obtained by reading over the existence proof again:

• For $n\in\Bbb{N}$, $f(n)= f(\underbrace{1+\dots +1}_{\text{$n$ times}})= \underbrace{f(1)\cdots f(1)}_{\text{$n$ times}}= a^n$.
• $a=f(1)=f(1+0)=f(1)\cdot f(0)=a\cdot f(0)$, so $f(0)=1=a^0$.
• $a=f(1)=f\left(\underbrace{\frac{1}{n}+\dots +\frac{1}{n}}_{\text{$n$ times}}\right)=\underbrace{f\left(\frac{1}{n}\right)\cdots f\left(\frac{1}{n}\right)}_{\text{$n$ times}}=\left[f\left(\frac{1}{n}\right)\right]^n$. Thus, $f(1/n)=\sqrt[n]{a}=a^{1/n}$.
• For any $m,n\in\Bbb{N}$, we have $f\left(\frac{m}{n}\right)=f\left(\underbrace{\frac{1}{n}+\dots+\frac{1}{n}}_{\text{$m$ times}}\right)=\underbrace{f\left(\frac{1}{n}\right)\cdots f\left(\frac{1}{n}\right)}_{\text{$m$ times}}= [a^{1/n}]^m=a^{m/n}$.

So, we have shown that for all rational $x\geq 0$, $f(x)=a^x$.

• FInally, for all $x\in\Bbb{Q}^+$, $1=f(0)=f(x-x)=f(x)\cdot f(-x)$, so $f(-x)=\frac{1}{f(x)}=\frac{1}{a^x}=a^{-x}$.

So, for all rational $x$, we have $f(x)=a^x$. Note that in each bullet point, the final equality is always by definition of the symbol $a^x$. Also, to be fully rigorous, each step requires a proof by induction.

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I suggest you read over the argument carefully again, because there's three things: definition, existence, and uniqueness and all three are intimately related and really just contained in the first half of the text. The motivation for the definition is "we want to preserve this functional equation". The proof for existence is "let's see how far we can push the functional equation", and the proof of uniqueness is "reword the proof of existence".

This existence and uniqueness argument heavily relies on the structure of $\Bbb{Q}$. Spivak uses this discussion as motivation for the question "are there functions $f:\Bbb{R}\to\Bbb{R}$ such that $f(1)>0$ and which satisfy for all $x,y\in\Bbb{R}$, $f(x+y)=f(x)\cdot f(y)$?" Also, an implicit question is whether $f$ is unique.

The answer is that if $f$ is continuous, then it is unique (why)? Proving existence of continuous $f$ satisfying the functional equation is the difficult part, and for that, read the rest of the chapter. See also this related answer.