# Proof

• Assume that $$f:\mathbb R\to \mathbb R$$ obeys the functional equation $$f(x+y)=f(x)\cdot f(y)$$ for all $$x,y\in\mathbb R$$.
• Because of the functional equation, we must have $f(1)=f\left(\frac 12\right)^2.$
• Therefore, in any case $$f(1)\ge 0$$.
• Suppose $$f(1)=0$$.
• Then, $$f$$ is a constant function and equals $$0$$, since:

$\begin{array}{rcll} f(x)&=&f(x+0)&\text{existence of real zero}\\ f(x)&=&f(x+(-1)+1)&\text{existence of inverse real numbers with respect to addition}\\ &=&f((x-1)+1)&\text{associativity of addition}\\ &=&f(x-1)\cdot f(1)&\text{by assumed functional equation}\\ &=&f(x-1)\cdot 0&\text{by assumption }f(1)=0\\ &=&0&\text{existence of real zero}\\ \end{array}$

Above, we have used the existence of real zero, the existence of inverse real numbers with respect to addition, and the associativity law of adding real numbers. Now, suppose $$f(1) > 0$$. Set $$a:=f(1)$$. Then, it follows from the uniqueness of 1 and from $$a=f(1 + 0)=f(1)\cdot f(0)=a\cdot f(0)$$ that $$f(0)=1$$.

Next, we will prove that

$$(i)$$ $$f(n)=a^n$$ for all natural numbers $$n\in \mathbb N$$

$$(ii)$$ $$f(n)=a^n$$ for all integers $$n\in \mathbb Z$$

$$(iii)$$ $$f\left(\frac pq\right)=\sqrt[q]a^p$$ for all rational numbers $$\frac pq\in \mathbb Q$$

$$(iv)$$ $$f(x)=a^x$$ for all real numbers $$x\in \mathbb R$$, if $$f$$ in addition is a continuous real function.

### ad $$(i)$$

By induction, for the base case $$n=1$$, $$f(1)=a$$, and for the induction step $$f(n+1)=f(n)\cdot f(1)=a^n\cdot a=a^{n+1}.$$

### ad $$(ii)$$

From the functional equation, it follows $1=0=f(n+(-n))=f(n)\cdot f(-n)$ and therefore $f(-n)=\frac 1{f(n)}=\frac 1{a^n}=a^{-n}.$

### ad $$(iii)$$

From $$a^p=f(p)=f\left(q\cdot \frac pq\right)=\left(f\left(\frac pq\right)\right)^q$$, it follows $$\sqrt[q]a^p=f\left(\frac pq\right)$$.

### ad $$(iv)$$

Let $$x$$ be a real number, i.e. $$x$$ is a class of all rational Cauchy sequences, which equal each other except a difference, which is a rational Cauchy sequence converging to zero, formally $x\in\mathbb R\Longleftrightarrow x=(x_n)_{n\in\mathbb N}+I$ where $$(x_n)_{n\in\mathbb N}$$ is a rational Cauchy sequence representing the real number $$x$$, and

$I:=\{(i_n)_{n\in\mathbb N}~|~i_n\in\mathbb Q,\lim i_n=0\}$ is the set of all rational sequences, which converge to $$0$$. Now, we have

$\begin{array}{rcll} a^x&=&\lim_{n\to\infty}a^{x_n+i_n}&\text{due to the definition of real numbers}\\ &=&\lim_{n\to\infty}f(x_n+i_n)&\text{follows from }( i i i )\\ &=&\lim_{n\to\infty}f(x_n)\cdot f(i_n)&\text{by the assumed functional equation}\\ &=&f(x)\cdot f(0)&\text{continuity of }f\\ &=&f(x)\cdot 1&\text{according to }f(0)=1\\ &=&f(x)&\text{multiplication by }1. \end{array}$

Above, we have used, in addition, the existence of the real number 1. Finally, it remains to be shown that $$f(x)=\exp_a(x)$$ for all $$x\in\mathbb R$$ and all positive real numbers $$a > 0$$. This follows immediately from $$(iv)$$ and the proposition about general powers of positive numbers.

Github: ### References

#### Bibliography

1. Forster Otto: "Analysis 1, Differential- und Integralrechnung einer Veränderlichen", Vieweg Studium, 1983