Fixed point

Functional Analysis
Post Reply
Papapetros Vaggelis
Community Team
Posts: 426
Joined: Mon Nov 09, 2015 1:52 pm

Fixed point

#1

Post by Papapetros Vaggelis »

Let \(\displaystyle{\left(X,||\cdot||\right)}\) be a real \(\displaystyle{\rm{Banach}}\) space and \(\displaystyle{T\in\mathbb{B}(X,X)}\)

such that \(\displaystyle{\sum_{n=0}^{\infty}||T^{n}||<\infty}\) . If \(\displaystyle{y\in X}\), then we define

\(\displaystyle{S_{y}:X\longrightarrow X}\) by \(\displaystyle{S_{y}(x)=y+T(x)}\) . Prove that the map \(\displaystyle{S_{y}}\)

has a fixed piont (\(\displaystyle{\exists\,x_0\in X: S_{y}(x_0)=x_0}\)) .
Papapetros Vaggelis
Community Team
Posts: 426
Joined: Mon Nov 09, 2015 1:52 pm

Re: Fixed point

#2

Post by Papapetros Vaggelis »

Since the normed space \(\displaystyle{\left(\mathbb{B}(X,X),||\cdot||\right)}\) is a \(\displaystyle{\rm{Banach}}\)

space and the series \(\displaystyle{\sum_{n=0}^{\infty}T^n}\) converges absolutely, we have that

the series \(\displaystyle{\sum_{n=0}^{\infty}T^{n}}\) converges. Now,

\(\displaystyle{\sum_{n=0}^{0}T^n\circ (I-T)=I-T=(I-T)\circ \sum_{n=0}^{0}T^n}\)

\(\displaystyle{\begin{aligned} \sum_{n=0}^{1}T^{n}\circ (I-T)&=(I+T)\circ (I-T)\\&=I+T-T-T^2\\&=I-T^2\\&=(I-T)\circ \sum_{n=0}^{1}T^n\end{aligned}}\)

Let \(\displaystyle{\sum_{n=0}^{k}T^n\circ (I-T)=I-T^{k+1}=(I-T)\circ \sum_{n=0}^{k}T^{n}\,\,(\ast)}\)

Then,

\(\displaystyle{\begin{aligned}\sum_{n=0}^{k+1}T^n\circ (I-T)&=\left(\sum_{n=0}^{k}T^n+T^{k+1}\right)\circ (I-T)\\&=\sum_{n=0}^{k}T^{n}\circ (I-T)+T^{k+1}-T^{k+2}\\&=I-T^{k+1}+T^{k+1}-T^{k+2}\\&=I-T^{k+2}=(I-T)\circ \sum_{n=0}^{k+1}T^{n}\end{aligned}}\)



So, by induction,

\(\displaystyle{\sum_{n=0}^{k}T^n\circ (I-T)=I-T^{k+2}=(I-T)\circ \sum_{n=0}^{k}T^{n}\,,\forall\,k\in\mathbb{N}}\)

and by taking limits, we have :

\(\displaystyle{\sum_{n=0}^{\infty}T^{n}\circ (I-T)=I=(I-T)\circ \sum_{n=0}^{\infty}T^{n}}\) .

So, the operator \(\displaystyle{I-T}\) is invertible, that is, one to one and onto \(\displaystyle{X}\)

with \(\displaystyle{(I-T)^{-1}=\sum_{n=0}^{\infty}T^n}\) .

So, \(\displaystyle{y\in X\implies \exists\,x_0\in X: (I-T)(x_0)=y\implies S_{y}(x_0)=x_0}\) .
admin
Administrator
Administrator
Posts: 40
Joined: Mon Oct 26, 2015 12:27 pm

Re: Fixed point

#3

Post by admin »

Solution by Gigaster

Hello Vaggelis!

We will show that \( S_y\) is continuous and that \(\exists x_0\in X\) such that \(\lim_{n\to\infty} S_{y}^{n}(z)=x_0\) for some \(z\in X\).

From the above we can easily deduce that \( S_y\) has a fixed point since :

\(\lim_{n\to\infty} S_y(S_{y}^{n}(z))=\lim_{n\to\infty} S_{y}^{n+1}(z)=S_y(x_0)=x_0\) (from continuity and uniqueness of limit)

So,considering a sequence \( (x_n)\subset X\) such that \( x_n\rightarrow x\in X\) we have that:

\(S_y(x_n)=y+T(x_n)\rightarrow y+T(x)=S_y(x)\) as \(n\rightarrow\infty\) (since \(T\) is bounded) which proves that \( S_y\) is continuous.

To complete the proof,we will show that \( (S_{y}^{n}(z))_n\subset\ X\) is a Cauchy sequence:

We have that :
\( S_{y}^{n}(z)=y+\sum_{k=1}^{n-1}T^{k}(y) + T^{n}(z)\) and also:\( \sum_{n=1}^{\infty}\|T^{n}\|<\infty \Rightarrow \|T^{n}\|\rightarrow 0\).

Finally,choosing \(m,n\in\mathbb N \) such that \(n>m\) it follows that :

\(\|S_{y}^{n}(z)-S_{y}^{m}(z)\|=\|\sum_{k=m}^{n-1}T^{k}(y) +T^{n}(z)-T^{m}(z)\|\leq\sum_{k=m}^{n-1}\|T^{k}\|\|y\|+\|T^{n}\|\|z\|+\|T^{m}\|\|z\|=\)

\(=(\sum_{k=1}^{n-1}\|T^{k}\|-\sum_{k=1}^{m-1}\|T^{k}\|)\|y\|+(\|T^{n}\|+\|T^{m}\|)\|x\|\rightarrow 0\) as \(n,m\rightarrow\infty\).

Since \(X\) is Banach and \((S_{y}^{n}(z))_n\) is Cauchy,there is a \(x_0\in X\) such that \(\lim_{n\to\infty} S_{y}^{n}(z)=x_0\).

As shown above,\(x_0\) is a fixed point of \(S_y\).
admin
Papapetros Vaggelis
Community Team
Posts: 426
Joined: Mon Nov 09, 2015 1:52 pm

Re: Fixed point

#4

Post by Papapetros Vaggelis »

Thank you Gigaster.

According to your solution, we can choose $\displaystyle{x_0=y+f(y)\in X}$ where $\displaystyle{f=\sum_{n=1}^{\infty}T^{n}}$.

Indeed:

$\displaystyle{\begin{aligned} S_{y}(y+f(y))&=y+T(y+f(y))\\&=y+T(y)+T(f(y))\\&=y+T(y)+T\,\left(\lim_{n\to +\infty}\sum_{k=1}^{n}T^{k}(y)\right)\\&=y+T(y)+\lim_{n\to +\infty}T\,\left(\sum_{k=1}^{n}T^{k}(y)\right)\\&=y+T(y)+\lim_{n\to +\infty}\,\sum_{k=1}^{n}T(T^{k}(y))\\&=y+T(y)+\sum_{n=1}^{\infty}T^{n+1}(y)\\&=y+\sum_{n=0}^{\infty}T^{n+1}(y)\\&=y+\sum_{n=1}^{\infty}T^{n}(y)\\&=y+f(y)\end{aligned}}$
User avatar
Tolaso J Kos
Administrator
Administrator
Posts: 867
Joined: Sat Nov 07, 2015 6:12 pm
Location: Larisa
Contact:

Re: Fixed point

#5

Post by Tolaso J Kos »

Additional question - Application :

Let $g:\left[0,1\right]\rightarrow \mathbb{R}$ and $K:\left[0,1\right]\times \left[0,1\right]\rightarrow \mathbb{R}$ be two continuous functions. Prove that there exists a continuous function $f:\left[0,1\right]\rightarrow \mathbb{R}$ such that

$$f(t)=g(t)+\int_{0}^{t}K(t,s)\,f(s)\,\mathrm{d}s\,\,,\forall\,t\in\left[0,1\right]$$

(Volterra's equation)
Imagination is much more important than knowledge.
Papapetros Vaggelis
Community Team
Posts: 426
Joined: Mon Nov 09, 2015 1:52 pm

Re: Fixed point

#6

Post by Papapetros Vaggelis »

We give a solution :

Consider the real \(\displaystyle{\rm{Banach}}\) space \(\displaystyle{\left(C(\left[0,1 \right ]),||\cdot||_{\infty} \right )}\)

and the function \(\displaystyle{T:C(\left[0,1 \right ])\longrightarrow C(\left[0,1 \right ])} \) given by

\(\displaystyle{T(f)(t):=\int_{0}^{t}K(t,s)\,f(s)\,\mathrm{d}s\,\,,\forall\,f\in C(\left[0,1 \right ])\,\,,\forall\,t\in\left[0,1 \right ] }\)

Also, let \(\displaystyle{M=\max\,\left\{\left|K(t,s)\right|\in\mathbb{R}: \left(t,s\right)\in\left[0,1\right]^2\right\}\in\mathbb{R}}\)

since the function \(\displaystyle{K}\) is continuous at the compact set \(\displaystyle{\left[0,1\right]^2}\) .

The function \(\displaystyle{T}\) is well defined and \(\displaystyle{\mathbb{R}}\) - linear. Indeed,

let \(\displaystyle{f\,,g\in C(\left[0,1 \right ]\,\,,a\,,b\in\mathbb{R})}\) and \(\displaystyle{t\in\left[0,1\right]}\) .

Since the \(\displaystyle{\rm{Riemann}}\) integral is linear, we get :

\(\displaystyle{\begin{aligned}T(a\,f+b\,g)(t)&=\int_{0}^{t}K(t,s)\left(a\,f+b\,g \right )(s)\,\mathrm{d}s\\&=\int_{0}^{t}K(t,s)\left(a\,f(s)+b\,g(s) \right )\,\mathrm{d}s\\&=a\,\int_{0}^{t}K(t,s)\,f(s)\,\mathrm{d}s+b\,\int_{0}^{t}K(t,s)\,g(s)\,\mathrm{d}s\\&=a\,T(f)(t)+b\,T(g)(t)\\&=\left(a\,T(f)+b\,T(g) \right )(t) \end{aligned}}\)

so: \(\displaystyle{T(a\,f+b\,g)=a\,T(f)+b\,T(g)}\).

Also, the function \(\displaystyle{T}\) is bounded since :

\(\displaystyle{\begin{aligned}\left|T(f)(t)\right|&=\left|\int_{0}^{t}K(t,s)\,f(s)\,\mathrm{d}s\right|\\&\leq \int_{0}^{t}\left|K(t,s)\right|\,\left|f(s)\right|\,\mathrm{d}s\\&\leq M\,\int_{0}^{t}\left|f(s)\right|\,\mathrm{d}s\\&\leq M\,\int_{0}^{t}\left|f(s)\right|\,\mathrm{d}s\\&\leq M\,\int_{0}^{t}\sup\,\left\{\left|f(s)\right|:s\in\left[0,1 \right ]\right\}\,\mathrm{d}s\\&=M\,t\,||f||_{\infty}\end{aligned}}\)

So, \(\displaystyle{T\in\mathbb{B}(C(\left[0,1\right]),C(\left[0,1\right]))}\) and thus :

\(\displaystyle{||T||\leq M=\dfrac{M^{1}}{1!}}\) .

Also,

\(\displaystyle{\begin{aligned}\left|T^2(f)(t)\right|&=\left|T(T(f))(t)\right|\\&=\left|\int_{0}^{t}K(t,s)\,T(f)(s)\,\mathrm{d}s\right|\\&\leq \int_{0}^{t}\left|K(t,s)\right|\,\left|T(f)(s)\right|\,\mathrm{d}s\\&\leq M\,\int_{0}^{t}\left|T(f)(s)\right|\,\mathrm{d}s\\&\leq M\,\int_{0}^{t}M\,s\,||f||_{\infty}\mathrm{d}s\\&=M^2\,||f||_{\infty}\,\int_{0}^{t}s\,\mathrm{d}s\\&=\left[\dfrac{M^2\,s^2\,||f||_{\infty}}{2} \right ]_{0}^{t}\\&=\dfrac{M^2\,t^2\,||f||_{\infty}}{2!}\end{aligned}}\)

which means that \(\displaystyle{||T^2||\leq \dfrac{M^2}{2!}}\) .

and we can easily prove that \(\displaystyle{||T^{n}||\leq \dfrac{M^{n}}{n!}\,\,,\forall\,,n\in\mathbb{N}}\) .

Since, \(\displaystyle{\sum_{n=1}^{\infty}\dfrac{M^{n}}{n!}=\sum_{n=0}^{\infty}\dfrac{M^{n}}{n!}-1=e^{M}-1<\infty}\)

we have that the series \(\displaystyle{\sum_{n=1}^{\infty}||T^{n}||}\) converges.

According to the exercise, for \(\displaystyle{y=g}\) , we have that there exists \(\displaystyle{f\in C(\left[0,1\right])}\)

such that \(\displaystyle{S_{g}(f)=g+T(f)=f}\) and then :

\(\displaystyle{f(t)=g(t)+\int_{0}^{t}K(t,s)\,f(s)\,\mathrm{d}s\,\,,\forall\,t\in\left[0,1\right]}\) .
Post Reply

Create an account or sign in to join the discussion

You need to be a member in order to post a reply

Create an account

Not a member? register to join our community
Members can start their own topics & subscribe to topics
It’s free and only takes a minute

Register

Sign in

Who is online

Users browsing this forum: No registered users and 8 guests