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?
Best Answer
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.