Determine $f$ (continuous function) Determine a continuous function $ f:(0,\infty )\rightarrow \mathbb{R} $, knowing that $ F(2x)-F(x)=x(\ln x+a), \forall x\in (0,\infty ) $, where $ a\in \mathbb{R} $ and $ F:(0,\infty )\rightarrow \mathbb{R}, F'(x)=f(x) $.
I've managed to find that $ 2^{n}f(2^{n}x)-f(x)=(2^{n}-1)\cdot (\ln x+a+1)+ \ln 2\cdot (2+2^{2}\cdot 2+2^{3}\cdot 3+...+2^{n-1}\cdot (n-1)) $, but I've got stuck.
 A: Consider $G(x)=F(x)-ax$; then $G'(x)=f(x)-a$ and
$$
G(2x)-G(x)=F(2x)-2ax-F(x)+ax=x\ln x
$$
Substituting $x/2$ for $x$ we get successively
\begin{align}
G(x)-G(x/2)&=\frac{x}{2}\ln\frac{x}{2}\\[4px]
G(x/2)-G(x/4)&=\frac{x}{4}\ln\frac{x}{4}\\
&\;\,\vdots\\
G(x/2^{n-1})-G(x/2^n)&=\frac{x}{2^n}\ln\frac{x}{2^n}
\end{align}
Summing up,
$$
G(x)-G(x/2^n)=\sum_{k=1}^n \frac{x}{2^k}\ln\frac{x}{2^k}=
x\sum_{k=1}^n\frac{\ln x-k\ln2}{2^k}
$$
and
$$
\lim_{n\to\infty}G(x/2^n)=
G(x)-x\sum_{k=1}^{\infty}\frac{\ln x-k\ln2}{2^k}=
G(x)-x\ln x+2x\ln2
$$
If we set $c=\lim_{t\to0}G(t)$, we have
$$
G(x)=x\ln x-2x\ln2+c
$$
so
$$
F(x)=x\ln x+x(a-2\ln2)+c
$$
and finally
$$
f(x)=\ln x+1+a-2\ln2
$$
A: Differentiating the functional equation
$$F(2x) - F(x) = x(\ln x + a),$$
we find that
$$2f(2x) - f(x) = \ln x + a + 1.$$
The constant part $a + 1$ is easily taken care of by simply adding $a + 1$ to a function $g(x)$ which satisfies
$$2g(2x) - g(x) = \ln x.$$
We might guess a solution of the form $g(x) = \ln(cx)$ for some $c > 0$. Indeed,
$$2\ln(2cx) - \ln(cx) = \ln(4c^2x^2) - \ln(cx) = \ln(4cx),$$
so taking $c = \tfrac{1}{4}$ gives us the solution
$$f(x) = \ln(\tfrac{x}{4}) + a + 1.$$
The primitives of $f(x)$ are
$$F(x) = x\ln(\tfrac{x}{4}) + ax + C.$$
You can check that
$$F(2x) - F(x) = x(\ln x + a).$$
