I am having trouble finding easily $h$ defined and continuous on $\mathbf{R}$ verifying for all $x$ in $\mathbf{R}$, $$ h(x) + h(2x) + h(4x) = {x^n} $$ where $n$ is a fixed natural number.
I have a solution using induction, but it is far too long in my opinion. Can anyone help me finding a simple solution ? Thanks much.
Let $\displaystyle h(x) = \frac{(2^n-1)x^n}{8^n-1}.$ It is easy to check that this function satisfies the given functional equation.
Now suppose that we have two solutions $h_1(x)$ and $h_2(x)$ to the given equation (for a fixed $n$). Consider the function $g(x) = h_1(x) - h_2(x)$. We have that $g$ is continuous and that $g(x) + g(2x) + g(4x) = 0$ for all $x$. In particular, this implies that $g(0) = 0$. Also, $g(x/2) + g(x) + g(2x) = 0$ by substituting $x/2$ for $x$. Hence $g(4x) = g(x/2)$ for all $x$; equivalently, $g(x) = g(x/8)$ for all $x$.
So for all $x$ we have, for all $k$, $g(x) = g(x/8^k)$, and hence $$g(x) = \lim_{k\rightarrow \infty} g(x/8^k) = \lim_{z\rightarrow 0} g(z) = g(0) = 0.$$
This means that any two solutions to the functional equation are equal everywhere; that is, that the $h(x)$ given above is the only solution.