Finding a unique continuous $g$ on $[0,1]$ satisfying $g(x) = \frac{1}{2}g\left(\frac{x+1}{2}\right) + f(x)$ for given $f$ Let $f$ be a given continuous function on $[0,1]$. How do you prove that there is a unique continuous function $g$ on $[0,1]$ satisfying $$g(x) = \frac{1}{2}g\left(\frac{x+1}{2}\right) + f(x)$$ for all $x\in[0,1]$?
 A: Let $\Phi$ the map $$\Phi:C[0,1]\to C[0,1]\\ (\Phi(g))(x)=\frac12 g\left(\frac{x+1}{2}\right)+f(x)$$
Now, $\left\lVert \Phi(g)-\Phi(h)\right\rVert_\infty\le \frac12\left\lVert g-h\right\rVert_\infty$.
Therefore, $\Phi$ is a contraction. By contraction mapping theorem, there exists exactly one $g\in C[0,1]$ such that $\Phi(g)=g$, which is exactly the solution to this problem. $\ \,\ \square$
A: Without contraction mapping: let $a\in [0,1]$. Set $a_0 = a$ and $a_n = (a_{n-1} + 1)/2$. We have $$g(a_{n-1}) = \frac{1}{2} g(a_n) + f(a_{n-1})$$ so by induction, $$g(a_0) = \frac{g(a_{n+1})}{2^{n+1}} + \sum_{k=0}^{n} \frac{f(a_k)}{2^k}$$ for all $n\ge 0$. Taking $n\to\infty$, we see $$g(a) = \sum_{k=0}^{\infty} \frac{f(a_k)}{2^k}$$ which converges since $f$, being continuous on the compact interval $[0,1]$, is bounded. This proves both existence and uniqueness of $g$.
A: The functions $\hat g(t):=g(1-t)$ and $\hat f(t):=f(1-t)$ together satisfy the functional equation
$$\hat g(t)=\hat f(t)+{1\over2}\>\hat g\!\left({t\over2}\right)\ .\tag{1}$$
In particular one has $\hat g(0)=2\hat f(0)$. Iterating $(1)$ we obtain the condition
$$\hat g(t)=\sum_{k=0}^n {\hat f(2^{-k}t)\over 2^k}+{1\over2^{n+1}}\hat g(2^{-(n+1)}t)\qquad (n\geq0)\ .$$
for $\hat g$.  Letting $n\to\infty$ we can conclude that necessarily
$$\hat g(t)=\sum_{k=0}^\infty {\hat f(2^{-k}t)\over 2^k}\ .$$
It is easy to check that this $\hat g$ is continuous and indeed satisfies $(1)$. It follows that the originally given functional equation has the unique solution
$$g(x)=\hat g(1-x)=\sum_{k=0}^\infty{f\bigl(1-2^{-k}(1-x)\bigr)\over 2^k}\ .$$
