Problem source

The function $f$ satisfies the identity
\begin{equation}
f(x) +f(y) \equiv f(x+y) 
\tag{$*$}
\end{equation}
for all $x$ and $y$. Show that $2\f(x)\equiv \f(2x)$ and deduce that 
$f''(0)=0$.
By considering the Maclaurin series for $\f(x)$, find the most general function that satisfies $(*)$.
 [{\it Do not
consider issues of existence or convergence of Maclaurin series in this question.}]
\begin{questionparts}
\item
By considering the function $\G$, defined by $\ln\big(\g(x)\big) =\G(x)$, find the most general function that, for all $x$ and $y$, satisfies the identity
\[
\g(x) \g(y) \equiv \g(x+y)\,.
\]
\item By considering the function $H$, defined by
$\h(\e^u) =H(u)$, find the most general function that satisfies, for all positive $x$ and $y$, the identity
\[
\h(x) +\h(y) \equiv  \h(xy)
\,.
\]
\item Find the most general function $t$ that, for all $x$ and $y$, satisfies the identity
\begin{equation*}
t(x) + t(y) \equiv t(z)\,,
\end{equation*}
where $z= \dfrac{x+y}{1-xy}\,$.
\end{questionparts}

Solution source

\begin{align*}
&&2f(x) &\equiv f(x) + f(x) \\
&&&\equiv f(x+x) \\
&&&\equiv f(2x) \\
\\
\Rightarrow && 2f(0) &= f(0) \\
\Rightarrow && f(0) &= 0
\\
&& f''(0) &= \lim_{h \to 0} \frac{f(2h)-2f(0)+f(-2h)}{h^2} \\
&&&= \lim_{h \to 0} \frac{f(2h)+f(-2h)}{h^2} \\
&&&=  \lim_{h \to 0} \frac{f(0)}{h^2} \\
&&&= 0 \\
\Rightarrow && f''(0) &= 0
\end{align*}

If $f(x)$ satisfies the equation, then $f'(x)$ satisfies the equation. In particular this means that $f^{(n)}(0) = 0$ for all $n \geq 2$. Therefore the only non-zero term in the Maclaurin series is $x^1$. Therefore $f(x) = cx$

\begin{questionparts}
\item Suppose $g(x)g(y) \equiv g(x+y)$, then if $G(x) = \ln g(x)$ we must have $G(x)+G(y) \equiv G(x+y)$, ie $G(x) = cx \Rightarrow g(x) = e^{cx}$
\item Suppose $h(x)+h(y) \equiv h(xy)$, then if $h(e^u) = H(u)$ we must have that $H(u)+H(v) \equiv h(e^u) + h(e^v) \equiv h(e^{u+v}) \equiv H(u+v)$.Therefore $H(u) = cu$, ie $h(e^u) = cu$ or $h(x) = h(e^{\ln x}) = c \ln x$.

\item Finally if $t(x) + t(y) \equiv t(z)$, the considering $T(w) = t(\tan w)$ then $T(x) + T(y) \equiv t(\tan x) + t(\tan y) \equiv t( \frac{\tan x + \tan y}{1- \tan x \tan y}) \equiv t (\tan (x+y)) \equiv T(x+y)$. Therefore $T(x) = cx$ Therefore $t(\tan w) = c w \Rightarrow t(x) = c \tan^{-1} x$

\end{questionparts}