Prove properties of a function satisfying a given functional equation

Assume $f(x)$ is a function defined for all $x \in \mathbb{R}$ such that

$f(x+y) = f(x)f(y).$

For parts (b)-(d) assume also that $f(x)$ is differentiable for all real $x$ (i.e., assume $f'(x)$ exists everywhere).

Using the given equation above, prove $f(0) = 0$ or $f(0) = 1$ . Prove that if $f(0) \neq 0$ then $f(x) \neq 0$ for all $x$ .
Prove that $f'(x)f(y) = f'(y)f(x)$ for all $x,y$ .
Prove that $f'(x) = cf(x)$ for some constant $c$ .
Prove that if $f(0) \neq 0$ then $f(x) = e^{cx}$ .

Proof. Assume $f(0) \neq 0$ . Then if $a \in \mathbb{R}$ by the functional equation we have
$f(a) f(0) = f(a+0) = f(a) \quad \implies \quad f(0) = 1.$

Therefore, if $f(0) \neq 0$ then we must have $f(0) = 1$ . Hence, we have either $f(0) = 0$ or $f(0) = 1$ .

Furthermore, if $f(0) \neq 0$ , then for any $x \in \mathbb{R}$ by the functional equation we have

$f(x) f(-x) = f(x+(-x)) = f(0) \neq 0.$

Hence, $f(x) \neq 0$ for any $x \in \mathbb{R}. \qquad \blacksquare$
Proof. Since the derivative of $f$ exists everywhere by hypothesis, we use the limit definition of the derivative and the functional equation $f(x+y)=f(x)f(y)$ to write,
$\begin{align*} f'(x)f(y) &= \left( \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \right) f(y) \\ &= \lim_{h \to 0} \frac{f(y)f(x)f(h) - f(x)f(y)}{h} \\ &= \lim_{h \to 0} \frac{f(y+h)f(x) - f(y)f(x)}{h} \\ &= \left( \lim_{h \to 0} \frac{f(y+h) - f(y)}{h} \right) f(x) \\ &= f'(y) f(x). \qquad \blacksquare \end{align*}$
Proof. First, if $f(0) = 0$ then by the functional equation we know for any $x \in \mathbb{R}$ we have
$f(x) = f(x+0) = f(x)f(0) = 0.$

So, if $f(0) = 0$ then $f(x)$ is the constant function which is 0 everywhere. Hence, $f'(x) = 0 = cf(x)$ for all $x$ .
Next, we consider the case that $f(0) = 1$ . Since $f'(x)$ exists for all $x$ we use the definition of derivative and the functional equation again to write,

$\begin{align*} f'(x) &= \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \\ &= \lim_{h \to 0} \frac{f(x)f(h) - f(x)}{h} \\ &= \lim_{h \to 0} \frac{f(x)(f(h) - 1)}{h} \\ &= f(x) \lim_{h \to 0} \frac{f(h) - 1}{h} \\ &= f(x) \lim_{h \to 0} \frac{f(0+h) - f(0)}{h} \\ &= f(x) f'(0) \\ &= cf(x). \end{align*}$

The final line follows since $f'(0)$ exists by assumption so we can write $c = f'(0)$ for some $c. \qquad \blacksquare$
Proof. Since $f'(x) = cf(x)$ (by part (c)) we know, by the previous exercise (Section 6.17, Exercise #39) that $f(x) = Ke^{cx}$ for some constant $K$ . Since $f(0) \neq 0$ by assumption, we know by part (a) that $f(0) = 1$ . Hence,
$f(0) = Ke^{c\cdot 0} = K = 1 \quad \implies \quad f(x) = e^{cx}. \qquad \blacksquare$