Finding Value, Related To Functional Equation $f(x)$ is continuous for $\forall x \in R$ and $f(2x)-f(x)=x^{3}$    
(1) $f(x)+f(-x)$ is constant ?
(2) $f(0)=0$ ?    
I don't know how to use the continuity.
especially for $f(0)=0$ ?
 A: Hint:
Simple definition as my guide says: If you can draw graph for the given function $f(x)$ without lifting your hand at all points, that's a continuous function. :)
Therefore, the given function has a graph which is continuous at every real point.
$f(2x)-f(x)=x^3 \implies f(x)=f(2x)-x^3$
$f(-2x)-f(-x)=-x^3 \implies f(-x)=x^3+f(-2x)$  
A: Instruction
The key of the answer below is in $\color{red}{\text{red}}$. Think carefully what does it mean. After your finish reading my answer, look back at Inceptio's answer again and think
how the same approach will allow you to prove $f(x) + f(-x)$ is a constant without solving $f(x)$ explicitly.
Answer
Let $f(x)$ be any continuous solution for the functional equation:
$$f(2x) - f(x) = x^3$$
Let $g(x) = f(x) - \frac{x^3}{7}$, we have:
$$g(2x) - g(x) = f(2x) - f(x) - \frac{2^3-1}{7}x^3 = x^3 - x^3 = 0$$
Since $f(x)$ is continuous over $\mathbb{R}$, so does $g(x)$. For any $x \ne 0$,
we thus have:
$$\color{red}{g(x) = g(\frac{x}{2}) = g(\frac{x}{2^2}) = \ldots} = g(\frac{x}{2^n})
\implies \color{red}{g(x) = \lim_{n\to\infty} g(\frac{x}{2^n})} = g(0)$$
From this, we can conclude $g(x)$ is a constant and hence $f(x)$ has the form:
$$f(x) = K + \frac{x^3}{7}$$ where $K$ is some constant. As a result, we have:


*

*$f(x) + f(-x) = 2K$, a constant!

*$f(0)  = K$, need not be zero!

