Prove some properties of a differentiable function satisfying a given functional equation

Let $g$ be a function differentiable everywhere such that

$g'(0) = 2 \quad \text{and} \quad g(x+y) = e^y g(x) + e^x g(y) \qquad \text{for all } x,y \in \mathbb{R}.$

Prove that $g(2x) = 2e^x g(x)$ and conjecture and prove a similar formula for $g(3x)$ .
Conjecture and prove a formula for $g(nx)$ in terms of $g(x)$ for all positive integers $n$ .
Prove that $g(0) = 0$ and compute
$\lim_{h \to 0} \frac{g(h)}{h}.$
Prove that there exists a constant $C$ such that
$g'(x) = g(x) + Ce^x$

for all $x$ . Find the value of the constant $C$ .

Proof. We can compute this using the functional equation:
$g(2x) = g(x+x) = e^x g(x) + e^x g(x) = 2e^x g(x). \qquad \blacksquare$

Next, we conjecture

$g(3x) = 3e^{2x} g(x).$

Proof. Again, we compute using the functional equation, and the above formula for $g(2x)$ ,

$\begin{align*} g(3x) &= g(2x + x) \\ &= e^x g(2x) + e^{2x}g(x) \\ &= e^x (2e^x g(x)) + e^{2x} g(x) \\ &= 2e^{2x}g(x) + e^{2x}g(x) \\ &= 3e^{2x}g(x). \qquad \blacksquare \end{align*}$
We conjecture
$g(nx) = ne^{(n-1)x} g(x).$

Proof. The proof is by induction. We have already established the cases $n = 2$ and $n = 3$ (and the $n = 1$ case is the trivial $g(x) = g(x)$ ). Assume then that the formula holds for some integer $k \geq 1$ . Then we have

$\begin{align*} g((k+1)x) &= g(kx + x) \\ &= e^x g(kx) + e^{kx} g(x) &(\text{functional equation})\\ &= e^x (ke^{(k-1)x} g(x)) + e^{kx}g(x) &(\text{induction hypothesis}) \\ &= ke^{kx} g(x) + e^{kx} g(x) \\ &= (k+1)e^{kx} g(x). \end{align*}$

Thus, if the formula holds for $k$ , it also holds for $k + 1$ . Hence, by induction it holds for all integers $k. \qquad \blacksquare$
Proof. Using the functional equation
$g(0) = g(0+0) = e^0 g(0) + e^0 g(0) = 2g(0) \quad \implies \quad g(0) = 0.$

Then, since the derivative $g'(x)$ exists for all $x$ (by hypothesis) we know it must exist in particular at $x = 0$ . Using the limit definition of derivative, and the facts that $g(0) = 0$ and $g'(0) = 2$ we have

$g'(0) = \lim_{h \to 0} \frac{g(0+h) - g(0)}{h} = \lim_{h \to 0} \frac{g(h)}{h} = 2. \qquad \blacksquare$
Proof. Since the derivative $g'(x)$ must exist for all $x$ (by hypothesis) we know that the limit
$\lim_{h \to 0} \frac{g(x+h) - g(x)}{h}$

must exist for all $x$ . Using the functional equation for $g$ we have

$\begin{align*} \lim_{h \to 0} \frac{g(x+h) - g(x)}{h} &= \lim_{h \to 0} \frac{e^h g(x) + e^x g(h) - g(x)}{h} \\[9pt] &= \lim_{h \to 0} \frac{g(x)(e^h - 1) + e^x g(h)}{h}. \end{align*}$

But then, from part (c) we know $\lim_{h \to 0} \frac{g(h)}{h} = 2$ and from this exercise (Section 6.17, Exercise #38) we know

$\lim_{x \to 0} \frac{e^x - 1}{x} = 1.$

Therefore, we have

$\begin{align*} \lim_{h \to 0} \frac{g(x)(e^h - 1) + e^x g(h)}{h} &= \lim_{h \to 0} \left( \frac{e^x g(h)}{h} \right) + \lim_{h \to 0} \left( \frac{g(x)(e^h - 1)}{h} \right) \\ &= e^x \lim_{h \to 0} \left(\frac{g(h)}{h} \right) + g(x) \lim_{h \to 0} \left( \frac{e^h-1}{h} \right) \\ &= 2e^x + g(x). \qquad \blacksquare \end{align*}$