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 Post subject: Functional analysis
PostPosted: 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 analysis
PostPosted: Thu Jul 14, 2016 6:18 am 
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\(\displaystyle{1)}\) 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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