Functional equation involving three different functions: $ f ( x + y ) = g ( x ) + h ( y ) $ 
If $ f , g , h : \mathbb R \to \mathbb R $ all are continuous functions such that
$$ f ( x + y ) = g ( x ) + h ( y ) \text , \quad \forall x , y \in \mathbb R \text , $$
find $ f $, $ g $ and $ h $.

I have literally no idea where to begin. I mean, what can I even say or claim and prove? How should I go about this?
Also, a quick doubt I had was if a function is continuous over $\mathbb{R}$, can we safely say that it must be a linear function? If not, Why?
 A: Note that
$$g(x) + h(y) = f(x + y) = f(y + x) = g(y) + h(x) \implies h(x) - g(x) = h(y) - g(y)$$
for all $x, y \in \Bbb{R}$. That is, $h - g$ is a constant function, i.e. there exists some $k \in \Bbb{R}$ such hat $h(x) = g(x) + k$ for all $x \in \Bbb{R}$.
This gives us the equivalent functional equation
$$f(x + y) = g(x) + g(y) + k.$$
Note that, when $y = 0$, we simply see that
$$f(x) = g(x) + g(0) + k,$$
hence
$$g(x + y) + g(0) + k = g(x) + g(y) + k \implies g(x + y) + g(0) = g(x) + g(y).$$
Let $L(x) = g(x) - g(0)$. Then, the above equation simplifies to
$$L(x + y) = L(x) + L(y),$$
which is Cauchy's functional equation. Since $g$ is continuous, so is $L$, and hence $L$ is linear. On $\Bbb{R}$, this means $L(x) = ax$ for some $a \in \Bbb{R}$.
So, rebuilding, we have
\begin{align*}
g(x) &= ax + c \\
h(x) &= ax + c + k \\
f(x) &= ax + 2c + k,
\end{align*}
where $a, c, k \in \Bbb{R}$ are parameters. Checking this family of possible solutions, we get
$$f(x + y) = a(x + y) + 2c + k = ax + c + ay + c + k = g(x) + h(y),$$
verifying that all functions of the above form are indeed solutions, yielding a complete family of solutions.
