Understanding and applying Banach's contraction principle.

Question

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.

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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

$$||K(w)||_{C([0,1])}\leq(||w||_{C([0,1])})^2+A\leq1/4+A. \tag7$$

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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 $$||K(x)-K(y)||_X\leq \lambda ||x-y||_X.$$

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QUESTIONS:

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

2. 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 $z$?

3. 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.

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

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 $1/2$?

6. 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!

Answer 1

Answers:

1. That is what they did.
2. 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.
3. They wrote absolute values because it makes it easier to go from eq(4) to eq(5). Normal parentheses would have preserved equality aswell.
4. 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).
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.
6. Since $K$ is a contraction on $B$, the Banach fixed point Theorem (or contraction principle) implies that $K$ must have a fixed point in $B$.