Understanding and applying Banach's contraction principle. 
Consider the following operator
$$K(x)(t)=\int_\limits{0}^{1}x^2(s) \ ds + At^2 \quad \forall \
 t\in[0,1], \tag1$$
acting on the Banach space $C([0,1])$ of continous functions
  $||x||=\sup_{t\in[0,1]}|x(t)|.$ Find, using Banachs contraction
  principle, conditions on the real constant $A$ such that the operator
  $K(x)(t)$ has a fixed point.


SOLUTION: 
Consider first the $C([0,1])$ norm of the operator $K(x)$, that is
$$||K(x)||_{C([0,1])}=\sup\limits_{t\in{[0,1]}}\left|\int\limits_0^1x^2(s) \ ds + At^2\right|\leq\left(||x||_{C([0,1])}\right)^2+A.\tag2$$
The operator $K$ will have a fixed point $y\in C([0,1])$ according to Banach's contraction principle if it would be a contraction in a subset $B\subset C([0,1]).$ The last means that for any continous functions $z(t)$ and $w(t)$ in $B$ the estimate $||K(z)-K(w)||_{C([0,1])}<\lambda||z-w||_{C([0,1])}$ must be satisfied with a number $0<\lambda <1.$
Consider
\begin{align}
||K(z)-K(w)||_{C([0,1])}&=||\int\limits_0^1z^2(s) \ ds - \int\limits_0^1w^2(s) \ ds||_{C([0,1])}\tag3\\
&=||\int\limits_0^1|z(s)-w(s)|\cdot|z(s)+w(s)| \ ds||_{C([0,1])}\tag4 \\
&\leq\left(\sup\limits_{t\in[0,1]}|z(t)|+\sup\limits_{t\in[0,1]}|w(t)|\right)\left(\sup\limits_{t\in[0,1]}|z(t)-w(t)|\right)\tag5\\
&=\left(||z||_{C([0,1])}+||w||_{C([0,1])}\right)||z-w||_{C([0,1])}.\tag6
\end{align}
It implies that for $||z||_{C([0,1])}$ and $||w||_{C([0,1])}$ strictly smaller than $1/2$ the norm $||K(z)-K(w)||_{C([0,1])}<1$. To check conditions of the Banach principle we need to find for which conditions the ball $B=\{w\in C([0,1]):||w||_{C([0,1])}<1/2\}$ in $C([0,1])$ is mapped by the operator $K$ onto itself, namely $||K(w)||_{C([0,1])}<1/2$. The estimate at the beginning of the consideration shows that it will be valid for $A<1/4$ because for $w\in B$ it implies that
$$||K(w)||_{C([0,1])}\leq(||w||_{C([0,1])})^2+A\leq1/4+A. \tag7$$

I have a lot of questions about this but let's begin by two handy definitions:
Definition 1 - A point $\bar{x}\in C$ is called the fixed point of the operator $K$ if $K(\bar{x})=\bar{x}$.
Definition 2 - The operator $K:A\rightarrow A$, where $A\subset X$ and $X$ is a Banach space is called a contraction on $A$ if there is a constant $0< \lambda < 1$ such that for any $x,y\in A$ we have that
$$||K(x)-K(y)||_X\leq \lambda ||x-y||_X.$$

QUESTIONS:


*

*In eq(1), when they obtained the estimate, did they simply set $t=1$, used the triangle inequality for integrals to get to
$$...\leq\sup\limits_{t\in[0,1]}\left|\int\limits_0^1x^2(s) \ ds\right| + A = \sup_{t\in[0,1]}|x^2(t)|+A=(||x||_{C([0,1])})^2+A \ ?$$

*I assume that $y$ is the same as $\bar{x}$ in definition 1. I don't see why they introduce $z$ and $w$. Why not just use $x$ and $\bar{x}$ instead? do we really need $y,\ w,$ and $z$?

*Going from eq(3) to eq(4) why do we need absolute values in the integrand? Why not normal parenthesis? I don't see which conditions force us to have a positive integrand.

*I don't understand at all what happens from eq(4) to eq(5).

*I don't understand the implication stated right after eq(6). I assume that we now have $$\left(||z||_{C([0,1])}+||w||_{C([0,1])}\right) = \lambda \in (0,1).$$
So if $||z||_{C([0,1])}=0.1$ and $||w||_{C([0,1])}=0.7$ then $\lambda=0.8$ which still is in the interval $(0,1)$. So why do both of these terms need to be strictly less than $1/2$?

*I don't understand where the conclusion $K(\bar{x})=\bar{x}$ has been made?
I appreciate all the help I can get to understand this wizardry here!
 A: Answers:


*

*That is what they did.

*The name $\bar x$ was used as a general definition (in the sense "for all $\bar x$: $\bar x$ is a fixed point if ...). They wanted to fix a $y$ instead. You definitely could have used the name $\bar x$ instead of $y$, but that might have confused some readers.
About the introduction of $w$ and $z$: Once you fix a point (in this case, the point is a continuous function) $\bar x$, it is extremely confusing (but still possible) to reuse that name in a universal statement. So it makes sense to use a new name here. However, they could have written something like $||K(\color{red}x)-K(w)||_{C([0,1])}<\lambda||\color{red}x-w||_{C([0,1])}$ without being less confusing. But this is not an important issue anyway.

*They wrote absolute values because it makes it easier to go from eq(4) to eq(5). Normal parentheses would have preserved equality aswell.

*Note that they shouldn't be using the norm around the integral in eq(3) and eq(4) (which is a number), but rather absolute value by definition of their norm. Then we have
\begin{split}
\left|\int_0^1|z(s)-w(s)|\cdot|z(s)+w(s)| \, \mathrm ds\right|
&\le \left|\int_0^1|z(s)-w(s)|\cdot\sup_{s\in[0,1]}\big(|z(s)+w(s)|\big)\, \mathrm ds\right| \\
&\le \sup_{s\in[0,1]}\big(|z(s)+w(s)|\big)\cdot\left|\int_0^1|z(s)-w(s)|\, \mathrm ds\right| \\
&\le \sup_{s\in[0,1]}\big(|z(s)+w(s)|\big)\cdot\sup_{s\in[0,1]}\big(|z(s)-w(s)|\big) \\
&\le \sup_{s\in[0,1]}\big(|z(s)|+|w(s)|\big)\cdot\sup_{s\in[0,1]}\big(|z(s)-w(s)|\big),
\end{split}
where the last expression is equal to the right-hand-side of eq(5).

*They said "$\|z\|_{C([0,1])}<\frac12$ and $\|w\|_{C([0,1])}<\frac12$ implies $\|K(z)-K(w)\|_{C([0,1])}<1$", which is clearly correct. They didn't claim the other (incorrect) direction.

*Since $K$ is a contraction on $B$, the Banach fixed point Theorem (or contraction principle) implies that $K$ must have a fixed point in $B$.

