Proving there exists a continuous function Let $f: [0,1] \rightarrow \mathbb{R}$ be a continuous function and let $T_f: [0,1] \rightarrow \mathbb{R}$ and $S_f: [0, 1] \rightarrow \mathbb{R}$ be defined as:
$$T_f(x) = 1+\int_{0}^{x} f(s) ds\\ S_f(x) = \begin{cases} f(x+1/2) \ \text{if } x < 1/2 \\ f(1) \ \text{if } x \ge 1/2  \end{cases}$$
Furthermore, define $W_f = \alpha T_f + \beta S_f$ for some real numbers $\alpha$ and $\beta$. 

Prove that if $|\alpha| + |\beta|<1$, then there exists a continuous
  function $f: [0,1] \rightarrow \mathbb{R}$ such that $W_f = f$.

 A: Use Banach's fixed point theorem link in the space $C([0,1])$. You have
$$|W_f(x)-W_g(x)|\le |\alpha| \int_0^x|f(s)-g(s)|\,ds+|\beta||S_f(x)-S_g(x)|
\le (|\alpha|+|\beta|)\Vert f-g\Vert_\infty$$ and so
$$\Vert W_f-W_g\Vert_\infty\le  (|\alpha|+|\beta|)\Vert f-g\Vert_\infty$$
A: Let's first try to work out what $f$ is. We'll expand the equation $f = W_f$:
$$f(x) = \alpha\left(1+\int_0^xf(s)\mathrm ds\right) + \beta S_f(x)$$
And now we'll differentiate it:
$$f'(x) - \alpha f(x) = \beta S_f'(x).\tag{1}$$
Now we need to work out what $S_f'(x)$ is. So we think very hard and we write down
$$S_f'(x) = \left\{ \begin{matrix}f'\left(x+\frac12\right)&x\le\frac12\\0&x\ge\frac12\end{matrix}\right.$$
It would probably be nice if both cases held at $x=\frac12$ but we only need $f$ to be continuous and not differentiable (everywhere).
It looks like we can solve equation $(1)$ for $x>\frac12$ as $f'=\alpha f$ so $f=A\mathrm e^{\alpha x}$. Now we can write down what $f'(x+\frac12)$ is for $x\in\left[0,\frac12\right]:$
$$f'(x)-\alpha f(x)=\alpha\beta A\mathrm e^{\alpha x}\mathrm e^{\frac\alpha2}.\tag{2}$$
We know then that the solution to this will look like $f=(A+Bx)\mathrm e^{\alpha x}$ so we solve for $B$ to get $B=\alpha\beta A \mathrm e^{\frac\alpha2}.$ We can now write down the solution so far:
$$f(x)=\left\{\begin{matrix}
A\left(1+\alpha\beta x \mathrm e^{\frac\alpha2}\right)\mathrm e^{\alpha x} & x<\frac12
\\
A'\mathrm e^{\alpha x} & x>\frac12
\end{matrix}\right.$$
We now put in the continuity condition to get $A'=A\left(1+\frac12\alpha\beta\mathrm e^{\frac\alpha2}\right)$
Finally the boundary conditions were lost when we differentiated $f=W_f$ so let's put those in to work out what $A$ is.
\begin{align*}
f(0)&=\alpha+\beta f(1)\\
A &=\alpha + \beta A\left(1+\frac12\alpha\beta\mathrm e^{\frac\alpha2}\right)\mathrm e^\alpha\\
A &= \frac{-2\alpha}{2-2\beta\mathrm e^\alpha - \alpha\beta^2\mathrm e^{\frac32\alpha}}
\end{align*}
This seems to be what the solution looks like. There may well be mistakes in the working. The next steps are:


*

*To simplify the above

*To show that the function is continuous

*To show that the function satisfies the desired equation

*To show that the function is well defined

*To show that these are all true so long as $|\alpha|+|\beta|<1$

A: This can be easily shown using the Banach fixed-point theorem. All we need to do is show that $$W: C([0,1]) \rightarrow C([0,1]): f\mapsto W_f$$ is a contraction, i.e.:
$$\lVert W_f - W_g\rVert \leq c\lVert f- g\rVert$$
where $0<c<1$ (in this case $c=|\alpha|+|\beta|$ works) and with a suitable norm (in this case the supremum norm works).
Then the banach fixed-point theorem tells you there is a unique fixed-point, so a unique function $f \in C([0,1])$ such that $W_f = f$
