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 Post subject: Integral operator Posted: Fri Nov 13, 2015 5:53 pm
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Joined: Mon Nov 09, 2015 1:52 pm
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Consider the real normed space $\displaystyle{\left(C\,(\left[0,1\right]),||\cdot||_{\infty}\right)}$ and the map

$\displaystyle{T:C\,(\left[0,1\right])\longrightarrow C\,(\left[0,1\right])\,,f\mapsto T(f):\left[0,1\right]\longrightarrow \mathbb{R}\,,T(f)(t)=\int_{0}^{t}f(s)\,\mathrm{d}s}$ .

A. Prove that the map $\displaystyle{T}$ is $\displaystyle{\mathbb{R}}$ - linear, bounded and one to one.

B. Find $\displaystyle{T\,\left(C\,(\left[0,1\right]\right)\subseteq C(\left[0,1\right])}$ .

C. Is the operator $\displaystyle{T^{-1}}$ bounded ?

D. Find $\displaystyle{||T||=\sup\,\left\{||T(f)||_{\infty}\geq 0: f\in C\,(\left[0,1\right])\,,||f||_{\infty}\leq 1\right\}}$ .

Top   Post subject: Re: Integral operator Posted: Wed Jan 20, 2016 7:59 pm
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Joined: Mon Nov 09, 2015 1:52 pm
Posts: 426
We give a solution.

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

then for each $\displaystyle{t\in\left[0,1\right]}$ holds :

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

so, $\displaystyle{T(a\,f+b\,g)=a\,T(f)+b\,T(g)}$, that is $\displaystyle{T}$ is $\displaystyle{\mathbb{R}}$ - linear .

Also, for each $\displaystyle{t\in\left[0,1\right]}$ holds :

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

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

and $\displaystyle{T}$ is bounded. Thus, $\displaystyle{||T||\leq 1}$ .

Let now $\displaystyle{f\,,g\in C(\left[0,1\right])}$ such that $\displaystyle{T(f)=T(g)}$. Then,

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

and by differentiating the above relation with respect to $\displaystyle{t}$, we get :

$\displaystyle{f(t)=g(t)\,,\forall\,t\in\left[0,1\right]\implies f=g}$ .

So, the operator $\displaystyle{T}$ is one to one.

B. If $\displaystyle{g\in T(C(\left[0,1\right]))}$, then $\displaystyle{g=T(f)}$ for some

$\displaystyle{f\in C(\left[0,1\right])}$. Now, $\displaystyle{g\in C^1(\left[0,1\right]}$ and

$\displaystyle{g(0)=T(f)(0)=0}$ .

Conversely, if $\displaystyle{g\in C^1(\left[0,1\right])}$ with $\displaystyle{g(0)=0}$, then

$\displaystyle{g^\prime\in C(\left[0,1\right]}$ and

$\displaystyle{T(g^\prime)(t)=\int_{0}^{t}g^\prime(s)\,\mathrm{d}s=g(t)-g(0)=g(t)\implies T(g^\prime)=g\implies g\in T(C(\left[0,1\right])}$.

Therefore,

$\displaystyle{T(C(\left[0,1\right]))=\left\{g\in C^1(\left[0,1\right]): g(0)=0\right\}\leq C(\left[0,1\right])}$ .

C. We have that

$\displaystyle{T^{-1}: T(C(\left[0,1\right])\longrightarrow C(\left[0,1\right])\,,T(g)=g^\prime}$.

Let $\displaystyle{g_{n}:\left[0,1\right]\longrightarrow \mathbb{R}\,,g_{n}(x)=x^n\,,n\in\mathbb{N}}$.

Then, $\displaystyle{g_{n}\in T(C(\left[0,1\right]))}$ and

obviously, $\displaystyle{||g_{n}||_{\infty}=1\,,\forall\,n\in\mathbb{N}}$ . But,

$\displaystyle{T^{-1}(g_{n})(t)=g_{n}^\prime(t)=n\,t^{n-1}\,,0\leq t\leq 1\,,n\in\mathbb{N}}$

and $\displaystyle{||T^{-1}(g_{n})||_{\infty}=n}$.

So, if $\displaystyle{T^{-1}}$ is bounded, then

$\displaystyle{\left(\exists\,c>0\right)\,\left(\forall\,g\in T(C(\left[0,1\right]))\right),||T^{-1}(g)||_{\infty}\leq c\,||g||_{\infty}}$

Especially,

$\displaystyle{\forall\,n\in\mathbb{N}: n\leq c}$, a contradiction.

D. According to A. , we have that $\displaystyle{||T||\leq 1}$ .

Let $\displaystyle{f(t)=1\,,t\in\left[0,1\right]}$ with $\displaystyle{||f||_{\infty}=1}$ . Then,

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

and therefore, $\displaystyle{||T||\geq ||T(f)||_{\infty}=1}$.

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

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