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 Post subject: Functional analysisPosted: Thu Jul 14, 2016 6:17 am
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$\displaystyle{1)}$ We define $\displaystyle{T:\left(C\left(\left[0,1\right]\right),||\cdot||_{\infty}\right)\longrightarrow \left(\mathbb{R},|\cdot|\right)}$ by

$\displaystyle{T(f)=\int_{0}^{1}f(t)\,\mathrm{d}t}$ .

Prove that the function $\displaystyle{T}$ is a bounded linear operator and find its norm.

$\displaystyle{2)}$ Let $\displaystyle{T:\left(\mathbb{l_{2}}\,(\mathbb{N}),||\cdot||_{2}\right)\longrightarrow \left(\mathbb{l_{2}}\,(\mathbb{N}),||\cdot||_{2}\right)}$

$\displaystyle{T\,(\left(a_{n}\right)_{n\in\mathbb{N}})=\left(0,a_1,a_2,...,a_{n},a_{n+1},...\right)}$.

Prove that $\displaystyle{T}$ is a linear isometry.

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 Post subject: Re: Functional analysisPosted: Thu Jul 14, 2016 6:18 am
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Joined: Mon Nov 09, 2015 1:52 pm
Posts: 426
$\displaystyle{1)}$ Let $\displaystyle{f\in C\,\left(\left[0,1\right]\right)}$. The function $\displaystyle{f}$ is Riemann integrable at $\displaystyle{\left[0,1\right]}$

as continuous at this interval, so $\displaystyle{\int_{0}^{1}f(t)\,\mathrm{d}t\in\mathbb{R}}$.

Therefore, the function $\displaystyle{T}$ is well defined.

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

Using the properties of $\displaystyle{\rm{Riemann}}$ integral, we get :

\displaystyle{\begin{aligned} T(f+g)&=\int_{0}^{1}\left(f+g\right)\,(t)\,\mathrm{d}t\\&=\int_{0}^{1}\left(f(t)+g(t)\right)\,\mathrm{d}t\\&=\int_{0}^{1}f(t)\,\mathrm{d}t+\int_{0}^{1}g(t)\,\mathrm{d}t\\&=T(f)+T(g)\end{aligned}}

and

$\displaystyle{T(a\,f)=\int_{0}^{1}(a\,f)(t)\,\mathrm{d}t=\int_{0}^{1}a\,f(t)\,\mathrm{d}t=a\,\int_{0}^{1}f(t)\,\mathrm{d}t=a\,T(f)}$

and thus the function $\displaystyle{T}$ is $\displaystyle{\mathbb{R}}$ - linear.

Let $\displaystyle{f\in C\,\left(\left[0,1\right]\right)}$. Then :

$\displaystyle{\forall\,t\in\left[0,1\right]: \left|f(t)\right|\leq \sup\,\left\{\left|f(t)\right|: 0\leq t\leq 1\right\}=||f||_{\infty}}$

so :

$\displaystyle{\int_{0}^{1}\left|f(t)\right|\,\mathrm{d}t\leq \int_{0}^{1}||f||_{\infty}\,\mathrm{d}t=||f||_{\infty}}$ and :

$\displaystyle{\left|T(f)\right|=\left|\int_{0}^{1}f(t)\,\mathrm{d}t\right|\leq \int_{0}^{1}\left|f(t)\right|\,\mathrm{d}t\leq ||f||_{\infty}}$.

In conclusion, $\displaystyle{\left|T(f)\right|\leq ||f||_{\infty}\,,\forall\,f\in C\,\left(\left[0,1\right]\right)}$, which means that $\displaystyle{T}$

is bounded, and since :

$\displaystyle{||T||=\sup\,\left\{\left|T(f)\right|: f\in C\,\left(\left[0,1\right]\right)\,,||f||_{\infty}\leq 1\right\}}$, we have that

$\displaystyle{||T||\leq 1}$. Now, the function $\displaystyle{f:\left[0,1\right]\longrightarrow \mathbb{R}\,,f(t)=1}$ is continuous, so :

$\displaystyle{f\in C\,\left(\left[0,1\right]\right)}$ and also :

$\displaystyle{||f||_{\infty}=\sup\,\left\{\left|f(t)\right|: 0\leq t\leq 1\right\}=\sup\,\left\{1\right\}=1}$

$\displaystyle{\left|T(f)\right|=\left|\int_{0}^{1}f(t)\,\mathrm{d}t\right|=\left|\int_{0}^{1}\mathrm{d}t\right|=\left|1\right|=1}$ .

Therefore, $\displaystyle{||T||=1}$ .

$\displaystyle{2)}$ Let $\displaystyle{\left(a_{n}\right)_{n\in\mathbb{N}}\in \rm{{l}_{2}}\,(\mathbb{N})}$ . The map $\displaystyle{T}$

corresponds the sequence $\displaystyle{\left(a_{n}\right)_{n\in\mathbb{N}}}$ with the sequence $\displaystyle{\left(b_{n}\right)_{n\in\mathbb{N}}}$, where

$\displaystyle{b_{n}=\begin{cases} 0\,\,\,\,\,\,\,\,\,\,\,,n=1\\ a_{n-1}\,\,\,,n\geq 2 \end{cases}}$

It's known that $\displaystyle{||a_{n}||_{2}=\sqrt{\sum_{n=1}^{\infty}a_{n}^2}\in\mathbb{R}\cap\left[0,+\infty\right)}$.

and then:

\displaystyle{\begin{aligned} ||b_{n}||_{2}&=\sqrt{\sum_{n=1}^{\infty}b_{n}^2}\\&=\sqrt{\sum_{n=2}^{\infty}a_{n-1}^2}\\&=\sqrt{\sum_{n=1}^{\infty}a_{n}^2}=||a_{n}||_{2}\in\mathbb{R}\cap\left[0,+\infty\right)\end{aligned}} .

So, the map $\displaystyle{T}$ is well defined and we observe that

$\displaystyle{||T\,\left(\left(a_{n}\right)_{n\in\mathbb{N}}\right)||_{2}=||\left(a_{n}\right)_{n\in\mathbb{N}}||_{2}\,,\forall\,\left(a_{n}\right)_{n\in\mathbb{N}}\in \rm{l_{2}}\,(\mathbb{N})\,\,(I)}$ .

If $\displaystyle{\left(a_{n}\right)_{n\in\mathbb{N}}\,,\left(b_{n}\right)_{n\in\mathbb{N}}\in \rm{l_{2}}\,(\mathbb{N})}$ and $\displaystyle{c\in\mathbb{R}}$, then :

\displaystyle{\begin{aligned} T\,\left(\left(a_{n}\right)_{n\in\mathbb{N}}+\left(b_{n}\right)_{n\in\mathbb{N}}\right)&=T\,\left(\left(a_{n}+b_{n}\right)_{n\in\mathbb{N}}\right)\\&=\left(0,a_{1}+b_{1},a_{2}+b_{2},...,a_{n}+b_{n},a_{n+1}+b_{n+1},...\right)\\&=\left(0,a_{1},a_{2},...,a_{n},a_{n+1},...\right)+\left(0,b_{1},b_{2},...,b_{n},b_{n+1},...\right)\\&=T\,\left(\left(a_{n}\right)_{n\in\mathbb{N}}\right)+T\,\left(\left(b_{n}\right)_{n\in\mathbb{N}}\right)\end{aligned}}

\displaystyle{\begin{aligned} T\,\left(c\,\left(a_{n}\right)_{n\in\mathbb{N}}\right)&=T\,\left(\left(c\,a_{n}\right)_{n\in\mathbb{N}}\right)\\&=\left(0,c\,a_1,c\,a_2,...,c\,a_n,c\,a_{n+1},...\right)\\&=c\,\left(0,a_1,a_2,...,a_n,a_{n+1},...\right)\\&=c\,T\,\left(\left(a_{n}\right)_{n\in\mathbb{N}}\right)\end{aligned}}

so, the function $\displaystyle{T}$ is $\displaystyle{\mathbb{R}}$ - linear and according to $\displaystyle{(I)}$ is bounded and isometry.

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