existence of a unique continues function Prove that there is exactly one continuous function  $f\colon [0,\infty)\rightarrow \mathbb{R}$ that for every $x\geq0$
$f (x)=3f (2x)+e^x$
I've tried a lot, but All I know is:
$f (0)=\frac{-1}{2}$ & $f (x)<0$ for every  $x$.
Is there any hint?
Thanks a lot 
 A: Suppose you have that $f$ and $g$ are continuous and solve your equation for all $x\ge 0$. Then
$$f(x) = 3f(2x)+e^x \quad\text{and}\quad g(x) = 3g(2x)+e^x$$
Subtracting these two equations gives
$$f(x)-g(x) = 3f(2x)+e^x - 3g(2x)-e^x = 3f(2x)-3g(2x).$$
Defining $h=f-g$ gives us
$$h(x) = 3h(2x).$$
Setting $x=0$, we have that $h(0) = 3h(0)$, i.e. $h(0)=0$. Our recursion relation can be written instead as $h\left(\frac{x}{2}\right) = 3h(x)$. Suppose $h(a)\neq 0$ for some $a> 0$, then $h\left(\frac{a}{2}\right)=3h(a)$ and hence $h\left(\frac{a}{4}\right) = 9h(a)$. Repeating, we have that $h\left(\frac{a}{2^n}\right) = 3^nh(a)$.
Since $f,g$ are continuous, so is $h$. You can use continuity of $h$ (at $0$) to get a contradiction from here. Can you see how to proceed?
A: Existence: For $a>0$ let $C_a$ denote the set of all continuous functions on $[0,a]$ equipped with the sup-norm $||\cdot||_a$. Then $(C_a,||\cdot||_a)$ is a Banach-Space. Define $T_a:C_a\to C_a$ by
\begin{align*}
(T_af)(x):=\frac{1}{3}\left(f(x/2)-e^{x/2}\right).
\end{align*}
Then
\begin{align*}
||T_af-T_ag||_a=\frac{1}{3}\sup_{x\in[0,a]}|f(x/2)-g(x/2)|\leq\frac{1}{3}||f-g||_a.
\end{align*}
The Banach Fixed Point Theorem ensures that $T_a$ has a unique fixed point $f_a$. In particular,
\begin{align*}
\frac{1}{3}\left(f_a(x/2)-e^{x/2}\right)=f_a(x),\qquad x\in[0,a].
\end{align*}
Now, note that for $a<b$ the uniqueness of this fixed point guarantees that $f_b(x)=f_a(x)$ for $x\in[0,a]$. Therefore, $f_b$ is the unique continuous continuation of $f_a$ to $[0,b]$. Then $f(x):=f_x(x)$ is well-defined, continuous on $[0,\infty)$ and satisfies $f(x)=f_a(x)$ for $x\in[0,a]$ and 
\begin{align*}
\frac{1}{3}\left(f(x/2)-e^{x/2}\right)=f(x),\qquad x\in[0,\infty).\qquad(1)
\end{align*}
Rearranging this equality finally gives
\begin{align*}
f(x)=3f(2x)+e^x,\qquad x\in[0,\infty).\qquad(2)
\end{align*}
Uniqueness: Suppose $g$ is another continuous function on $[0,\infty)$ satisfying (2). Then rearranging again yields that $g$ satisfies (1). In particular, $g|_{[0,a]}$ is a fixed point of $T_a$ for each $a>0$. The uniqueness of the fixed points then says that $g(x)=f_a(x)$ for each $a>0$ and $x\in[0,a]$; in particular, $g=f$.
A: Let $$f(x)=\displaystyle -\sum_{k=1}^\infty  \frac{1}{3^k}{e^\left(\tfrac{x}{2^k}\right)}=-\frac{e^\frac{x}{2}}{3}- \frac{e^\frac{x}{4}}{9}- \frac{e^\frac{x}{8}}{27}-\cdots.$$ Then
$$e^x+3\cdot f(2x)= e^x -e^x- \frac{e^\frac{x}{2}}{3}- \frac{e^\frac{x}{4}}{9}-\cdots=f(x).$$
The series is easily shown to converge for $x\ge0$, so $f(x)$ solves the original equation.
