Integral operator

Functional Analysis
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Papapetros Vaggelis
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Posts: 426
Joined: Mon Nov 09, 2015 1:52 pm

Integral operator

#1

Post by Papapetros Vaggelis »

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 \(\displaystyle{||T||=\sup\,\left\{||T(f)||_{\infty}\geq 0: f\in C\,(\left[0,1\right])\,,||f||_{\infty}\leq 1\right\}}\) .
Papapetros Vaggelis
Community Team
Posts: 426
Joined: Mon Nov 09, 2015 1:52 pm

Re: Integral operator

#2

Post by Papapetros Vaggelis »

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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