Some functional Analysis

Functional Analysis
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Tsakanikas Nickos
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Posts: 314
Joined: Tue Nov 10, 2015 8:25 pm

Some functional Analysis

#1

Post by Tsakanikas Nickos »

Let \( \displaystyle X , Y \) be normed spaces and let \( \displaystyle B(X,Y) \) be the space of all bounded linear operators from \( \displaystyle X \) to \( \displaystyle Y \). Suppose that \( \displaystyle Y \) is a Banach space, let \( \displaystyle X_{0} \) be a dense subspace of \( \displaystyle X \) and let \( \displaystyle T_{0} \in B(X_{0},Y) \). Show that there exists a unique \( \displaystyle T \in B(X,Y) \) such that \( \displaystyle T|_{X_{0}} = T_{0} \) and \( \displaystyle ||T|| = ||T_{0}|| \).
Papapetros Vaggelis
Community Team
Posts: 426
Joined: Mon Nov 09, 2015 1:52 pm

Re: Some functional Analysis

#2

Post by Papapetros Vaggelis »

Hi Nickos.

If \(\displaystyle{T:X\longrightarrow Y}\) is a bounded linear operator such that \(\displaystyle{T|_{X_{0}}=T_{0}\,(I)}\) and \(\displaystyle{||T||=||T_{0}||}\), then

it is the only one with those properties. Here is a proof.

Let \(\displaystyle{f:X\longrightarrow Y}\) be another bounded linear operator such that \(\displaystyle{f|_{X_{0}}=T_{0}\,(II)}\) and \(\displaystyle{||f||=||T_{0}||}\) .

If \(\displaystyle{x\in X_{0}}\), then : \(\displaystyle{T(x)=T|_{X_{0}}(x)=T_{0}(x)=f|_{X_{0}}(x)=f(x)}\).

Let now \(\displaystyle{x\in X-X_{0}}\). We have that \(\displaystyle{x\in \overline{X_{0}}}\) and there exists a sequence \(\displaystyle{\left(x_{n}\right)_{n\in\mathbb{N}}}\)

such that \(\displaystyle{x_{n}\in X_{0}\,,\forall\,n\in\mathbb{N}}\) and \(\displaystyle{x_{n}\to x}\), when \(\displaystyle{n\to \infty}\) . So, using the

fact that \(\displaystyle{f\,,T}\) are continuous and the relations \(\displaystyle{(I)\,,(II)}\), we get :

\(\displaystyle{\begin{aligned} f(x)&=f\,\left(\lim_{n\to \infty}x_{n}\right)\\&=\lim_{n\to \infty}f(x_{n})\\&=\lim_{n\to \infty}T_{0}(x_{n})\\&=\lim_{n\to \infty}T(x_{n})\\&=T\,\left(\lim_{n\to \infty}x_{n}\right)\\&=T(x)\end{aligned}}\)

so :

\(\displaystyle{\forall\,x\in X: T(x)=f(x)\implies T=f}\) .

If \(\displaystyle{T_{0}=\mathbb{O}}\), then \(\displaystyle{T=\mathbb{O}}\) and everything is ok. Let \(\displaystyle{T_{0}\neq \mathbb{O}}\) .

Let \(\displaystyle{\left(z_{n}\right)_{n\in\mathbb{N}}}\) be a Cauchy-sequence of \(\displaystyle{\left(Z,||\cdot||\right)}\) and \(\displaystyle{\epsilon>0}\) .

Then, there is \(\displaystyle{n_{0}\in\mathbb{N}}\) such that \(\displaystyle{||z_{n}-z_{m}||<\dfrac{\epsilon}{||T_{0}||}\,,\forall\,n\,,m\geq n_0}\) .

For every \(\displaystyle{n\,,m\geq n_0}\) we have that :

\(\displaystyle{||T_{0}(z_{n})-T_{0}(z_{m})||=||T_{0}\,(z_{n}-z_{m})||\leq ||T_{0}||\,||z_{n}-z_{m}||<\epsilon}\).

Therefore, the sequence \(\displaystyle{\left(T(z_{n})\right)_{n\in\mathbb{N}}}\) is a Cauchy-sequence of elements of \(\displaystyle{\left(Y,||\cdot||\right)}\)

and since \(\displaystyle{\left(Y,||\cdot||\right)}\) is a \(\displaystyle{\rm{Banach}}\) space, the \(\displaystyle{\left(T(z_{n}\right))_{n\in\mathbb{N}}}\)

converges in \(\displaystyle{\left(Y,||\cdot||\right)}\). We define \(\displaystyle{T:X\longrightarrow Y}\) by

\(\displaystyle{T(x)=T_{0}(x)\,,x\in X_{0}}\) and if \(\displaystyle{x\in X-X_{0}}\), then

\(\displaystyle{T(x)=\lim_{n\to \infty}T_{0}(x_{n})\in Y}\), where

\(\displaystyle{\left(x_{n}\right)_{n\in\mathbb{N}}\subseteq X_{0}\,,x_{n}\to x\,,n\to \infty}\). If

\(\displaystyle{\left(y_{n}\right)_{n\in\mathbb{N}}\subseteq X_{0}}\)

such that \(\displaystyle{y_{n}\to x\,,n\to \infty}\), then the sequence

\(\displaystyle{z_{n}=\begin{cases}
x_{n}\,,n=2\,k-1\\
y_{n}\,\,\,,n=2\,k
\end{cases}}\)

of \(\displaystyle{X_{0}}\) converges to \(\displaystyle{x}\) and then \(\displaystyle{T_{0}(z_{n})\to k\in Y}\). The sequences

\(\displaystyle{\left(T_{0}(x_{n})\right)_{n\in\mathbb{N}}\,,\left(T_{0}(y_{n})\right)_{n\in\mathbb{N}}}\) are subsequences of \(\displaystyle{T_{0}(z_{n})}\)

and thus:\(\displaystyle{\lim_{n\to \infty}T_{0}(x_{n})=\lim_{n\to \infty}T_{0}(y_{n})}\) .

In conclusion, the function \(\displaystyle{T}\) is well defined and such that \(\displaystyle{T|_{X_{0}}=T}\).

It is easy to see that \(\displaystyle{T}\) is linear (algebra of limits) . Let \(\displaystyle{x\in B_{X}\,\left[0,1\right]}\). Then \(\displaystyle{||x||\leq 1}\) and

\(\displaystyle{x\in \overline{X_{0}}\implies \exists\,\left(x_{n}\right)_{n\in\mathbb{N}}\in X_{0}: x_{n}\to x\,,n\to \infty}\).

\(\displaystyle{x_{n}\to x\implies ||x_{n}||\to ||x||\leq 1\stackrel{(\ast)}{\implies}||x_{n}||\leq 1\implies x_{n}\in B_{X_{0}}\,\left[0,1\right]}\) .

\(\displaystyle{(\ast)}\) : Left as an exercise.

Then,

\(\displaystyle{T(x)=\lim_{n\to \infty}T_{0}(x_{n})\implies ||T(x)||=\lim_{n\to \infty}||T_{0}(x_{n})||\leq ||T_{0}||}\).

So, \(\displaystyle{||T(x)||\leq ||T_{0}||\,,x\in B_{X}\,\left[0,1\right]}\), which means that \(\displaystyle{T}\) is bounded and

\(\displaystyle{||T||\leq ||T_{0}||}\) by definition. Also, \(\displaystyle{||T_{0}||\leq ||T||}\) and thus :

\(\displaystyle{||T||=||T_{0}||}\) .
Papapetros Vaggelis
Community Team
Posts: 426
Joined: Mon Nov 09, 2015 1:52 pm

Re: Some functional Analysis

#3

Post by Papapetros Vaggelis »

Some thoughts:

Let \(\displaystyle{\left(X,||\cdot||_{X}\right)=\left(Y,||\cdot||_{Y}\right)=\left(\mathbb{R},\left|\cdot\right|\right)}\).

Consider \(\displaystyle{W\subseteq \mathbb{R}}\) a subspace of \(\displaystyle{\left(\mathbb{R},+,\cdot\right)}\) over \(\displaystyle{\mathbb{R}}\).

\(\displaystyle{\dim_{\mathbb{R}}\,W\leq 1\implies \dim_{\mathbb{R}}\,W=0\,\lor \dim_{\mathbb{R}}\,W=1\implies W=\left\{0\right\}\,\lor W=\mathbb{R}}\)

and thus, \(\displaystyle{\mathbb{R}}\) is the only dense subspace of \(\displaystyle{\mathbb{R}}\).

Let \(\displaystyle{f\in\,\mathbb{B}\,\left(\mathbb{R},\mathbb{R}\right)}\). Then, the real function \(\displaystyle{f}\) is continuous at \(\displaystyle{\mathbb{R}}\)

and satisfies the relations :

\(\displaystyle{f(x+y)=f(x)+f(y)\,,x\,,y\in\mathbb{R}\,\,,f(a\,x)=a\,f(x)\,,a\in\mathbb{R}\,,x\in\mathbb{R}}\). So,

\(\displaystyle{a\in\mathbb{R}\implies f(a)=f\left(a\cdot 1\right)=a\,f(1)}\).

Now, the functions \(\displaystyle{f_{c}:\mathbb{R}\longrightarrow \mathbb{R}\,,f_{c}(x)=c\,x\,,c\in\mathbb{R}}\) are continuous and linear and then :

\(\displaystyle{f_{c}\in \mathbb{B}\,\left(\mathbb{R},\mathbb{R}\right)\,,\forall\,c\in\mathbb{R}}\).

So, \(\displaystyle{\mathbb{B}\,\left(\mathbb{R},\mathbb{R}\right)=\left\{f_{c}\,,c\in\mathbb{R}\right\}}\)

If \(\displaystyle{f_{c}\in\,\mathbb{B}\,\left(\mathbb{R},\mathbb{R}\right)}\), then :

\(\displaystyle{||f||=\sup\,\left\{\left|f_{c}(x)\right|\,,x\in\mathbb{R}\,\left|x\right|\leq 1\right\}=\sup\,\left\{\left|c\right|\,\left|x\right|\,x\in\mathbb{R}\,\left|x\right|\leq 1\right\}=\left|c\right|}\) .

Let \(\displaystyle{\left(f_{n}\right)_{n\in\mathbb{N}}}\) be a sequence of \(\displaystyle{\left(\mathbb{B}\,\left(\mathbb{R},\mathbb{R}\right),||\cdot||\right)}\) .

Then, there exists real sequence \(\displaystyle{\left(c_{n}\right)_{n\in\mathbb{N}}}\) such that

\(\displaystyle{f_{n}(x)=c_{n}\,x\,,\forall\,n\in\mathbb{N}\,,\forall\,x\in\mathbb{R}}\).

Consider \(\displaystyle{\epsilon>0}\) . There is \(\displaystyle{n_{0}\in\mathbb{N}}\) such that \(\displaystyle{||f_{n}-f_{m}||<\epsilon\,,\forall\,n\,m\geq n_{0}}\).

Let \(\displaystyle{n\,,m\geq n_{0}}\). The function \(\displaystyle{f_{n}-f_{m}}\) is given by

\(\displaystyle{(f_{n}-f_{m})(x)=f_{n}(x)-f_{m}(x)=\left(c_{n}-c_{m}\right)\,x\,,x\in\mathbb{R}}\), so :

\(\displaystyle{||f_{n}-f_{m}||=\left|c_{n}-c_{m}\right|<\epsilon}\).

In conclusion, the real \(\displaystyle{\left(c_{n}\right)_{n\in\mathbb{N}}}\) is a Cauchy-sequence and since \(\displaystyle{\left(\mathbb{R},\left|\cdot\right|\right)}\)

we get : \(\displaystyle{c_{n}\to c\,,n\to \infty}\) for some \(\displaystyle{c\in\mathbb{R}}\). Let \(\displaystyle{\epsilon>0}\). There is \(\displaystyle{k\in\mathbb{N}}\)

such that \(\displaystyle{\left|c_{n}-c\right|<\epsilon\,,\forall\,n\geq n_{0}}\). We define \(\displaystyle{f:\mathbb{R}\longrightarrow \mathbb{R}\,,f(x)=c\,x}\)

and then \(\displaystyle{f\in \mathbb{B}\,(\mathbb{R},\mathbb{R})}\). Also, for every \(\displaystyle{n\geq n_{0}}\) :

\(\displaystyle{||f_{n}-f||=\left|c_{n}-c\right|<\epsilon}\), which means that \(\displaystyle{f_{n}\to f\,,n\to \infty}\) .

This result was expected because, if \(\displaystyle{\left(Y,||\cdot||\right)}\), is a \(\displaystyle{\rm{Banach}}\) space, then so is \(\displaystyle{\left(\mathbb{B}\,(X,Y),||\cdot||\right)}\).

Additional question :

If \(\displaystyle{f_{n}:\mathbb{R}\longrightarrow \mathbb{R}\,,f_{n}(x)=\dfrac{x}{2^{n}}\,,n\in\mathbb{N}}\), then :

calculate the series \(\displaystyle{\sum_{n=1}^{\infty}f_{n}}\).

Also, by defining \(\displaystyle{g:\mathbb{B}\,(\mathbb{R},\mathbb{R})\longrightarrow \mathbb{R}\equiv \mathbb{M}_{1}\,(\mathbb{R})\,,f_{c}\mapsto c}\)

we have that :

\(\displaystyle{g\,(f_{c_1}+f_{c_{2}})=g\,(f_{c_1+c_2})=c_1+c_2=g(f_{c_1})+g(f_{c_2})\,,\forall\,,c_1\,,c_2\in\mathbb{R}}\)

\(\displaystyle{g\,(a\,f_{c})=g\,(f_{a\,c})=a\,c=a\,g(f_{c})\,,\forall\,a\in\mathbb{R}\,,\forall\,c\in\mathbb{R}}\)

and obviously \(\displaystyle{g}\) is one-one, and onto \(\mathbb{R}\).

Futhermore, \(\displaystyle{||g\,(f_{c})||=\left|c\right|=||f_{c}||\,,\forall\,c\in\mathbb{R}}\) .

Therefore, the \(\displaystyle{\mathbb{R}}\) - spaces \(\displaystyle{\left(\mathbb{B}\,(\mathbb{R},\mathbb{R}),+,\cdot\right)\,,\left(\mathbb{R},+,\cdot\right)}\)

are isometrically isomorphic and then : \(\displaystyle{\mathbb{B}\,(\mathbb{R},\mathbb{R})=<g^{-1}(1)>=<Id_{\mathbb{R}}>}\). Indeed,

\(\displaystyle{Id_{\mathbb{R}}\neq \mathbb{O}}\) and if \(\displaystyle{f\in \mathbb{B}\,(\mathbb{R},\mathbb{R})}\), then \(\displaystyle{f=f_{c}}\)

for some \(\displaystyle{c\in\mathbb{R}}\) and

\(\displaystyle{f(x)=f_{c}(x)=c\,x=c\,Id_{\mathbb{R}}\,(x)\,\forall\,x\in\mathbb{R}\implies f=c\,Id_{\mathbb{R}}}\) .


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