Let $(X, ||\cdot||)$ be a Banach space and $T \in \mathbb{B} (X, X)$ with the property the series $\sum \limits_{n=1}^{\infty} ||T^n||$ converges. If $y \in X$ we define the map $S_y: X \rightarrow X$ such that:
$$S_y(x)= y+T(x)$$
Prove that $S_y$ has a fixed point.
Solution
It suffices to prove that the map $F= T- {\rm Id}$ is invertible. However, this is easy since we note that:
$$\left ( {\rm Id} - T \right )^{-1}= I+T+T^2 +\cdots $$
which is bounded from the hypothesis and of the triangular inequality. Hence the exercise comes to an end.
$$S_y(x)= y+T(x)$$
Prove that $S_y$ has a fixed point.
Solution
It suffices to prove that the map $F= T- {\rm Id}$ is invertible. However, this is easy since we note that:
$$\left ( {\rm Id} - T \right )^{-1}= I+T+T^2 +\cdots $$
which is bounded from the hypothesis and of the triangular inequality. Hence the exercise comes to an end.
More analytically :
ReplyDeleteSince the series \(\displaystyle{\sum_{n=1}^{\infty}T^{n}}\) converges absolutely and the normed space
\(\displaystyle{\left(\mathbb{B}(X,X),||\cdot||\right)}\) is a \(\displaystyle{\rm{Banach}}\) space, we have
that the series \(\displaystyle{\sum_{n=1}^{\infty}T^{n}}\) converges to \(\displaystyle{\mathbb{B}(X,X)}\) .
So, the series \(\displaystyle{\sum_{n=0}^{\infty}T^{n}}\) converges.
\(\displaystyle{\left(I-T\right)\circ I=I-T=I\circ \left(I-T\right)}\)
\(\displaystyle{\left(\left(I-T\right)\circ \left(I+T\right)\right)(x)=(I-T)(x+T(x))=x+T(x)-T(x+T(x))=x-T^2(x)\,,\forall\,x\in X}\)
Similarly, \(\displaystyle{\left(I+T\right)\circ (I-T)=I-T^2}\) .
Let \(\displaystyle{n\in\mathbb{N}}\) such that
\(\displaystyle{(I-T)\circ \sum_{k=0}^{n}T^{k}=I-T^{k+1}=\sum_{k=0}^{n}T^{k}\circ (I-T)\,\,(I)}\)
\(\displaystyle{n\to n+1}\) :
\(\displaystyle{\begin{aligned} (I-T)\circ \sum_{k=0}^{n+1}T^{k}&=(I-T)\circ \left(\sum_{k=0}^{n}T^{k}+T^{n+1}\right)\\&=(I-T)\circ \sum_{k=0}^{n}T^{k}+(I-T)\circ T^{n+1}\\&\stackrel{(I)}{=}I-T^{n+1}+T^{n+1}-T^{n+2}\\&=I-T^{n+2}\end{aligned}}\)
Similarly, \(\displaystyle{\sum_{k=0}^{n+1}T^{k}\circ (I-T)=I-T^{n+2}}\)
So, by induction,
\(\displaystyle{(I-T)\circ \sum_{k=0}^{n}T^{k}=I-T^{n+1}=\sum_{k=0}^{n}T^{k}\circ (I-T)\,\,(II)\,\,,\forall\,n\in\mathbb{N}}\) .
But, \(\displaystyle{I-T^{n+1}\to I}\) because :
\(\displaystyle{||(I-T^{n+1})-I||=||T^{n+1}||\leq ||T||\cdot ||T^{n}||\to 0\,\,,n\to +\infty}\)
and by taking limits, the relation \(\displaystyle{(II)}\) gives us :
\(\displaystyle{(I-T)\circ \sum_{n=0}^{\infty}T^{n}=I=\sum_{n=0}^{\infty}T^{n}\circ (I-T)}\), which means that
\(\displaystyle{\left(I-T\right)^{-1}=\sum_{n=0}^{\infty}T^{n}}\) .
We have to prove that there exists \(\displaystyle{x_0\in X}\) such that \(\displaystyle{S_{y}(x_0)=x_0}\)
or else \(\displaystyle{y+T(x_0)=x_0\iff \left(I-T\right)(x_0)=y}\) .
So, it is sufficient to prove that the map \(\displaystyle{I-T\in\mathbb{B}(X,X)}\) is onto \(\displaystyle{X}\)
which is true.
Additional question - Application :
ReplyDeleteLet \(\displaystyle{g:\left[0,1\right]\longrightarrow \mathbb{R}}\) and
\(\displaystyle{K:\left[0,1\right]\times \left[0,1\right]\longrightarrow \mathbb{R}}\) two continuous
functions. Prove that there exists continuous function \(\displaystyle{f:\left[0,1\right]\longrightarrow \mathbb{R}}\)
such that
\(\displaystyle{f(t)=g(t)+\int_{0}^{t}K(t,s)\,f(s)\,\mathrm{d}s\,\,,\forall\,t\in\left[0,1\right]}\) (\(\displaystyle{\rm{Volterra's}}\) equation)
ReplyDeleteWe 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 fumction \(\displaystyle{K}\) 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 \(\displaystyle{\rm{Riemann}}\) 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 :
\(\displaystyle{f(t)=g(t)+\int_{0}^{t}K(t,s)\,f(s)\,\mathrm{d}s\,\,,\forall\,t\in\left[0,1\right]}\) .
Hello Vaggelis, g-)
ReplyDelete