Is this linear mapping discontinuous?











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Let $C[0,1]$ denote the vector space of all continuous functions on the interval $[0,1]$, endowed with the supremum norm, and let $D$ denote the subspace



$D:={xin C[0,1]: x'' text{ is continuous and } x(0)=x'(0)=0 }subset C[0,1]$.



Is it true that the linear mapping $T: Dlongrightarrow C[0,1]$, given by $Tx(t):=x''(t)+x(t)$, is continuous? And does there exist the inverse $T^{-1}$? If so, what is $T^{-1}$?



I think/guess that the mapping is not continuous. Any hint/comment will be appreciated.






functional-analysis operator-theory







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  • For the first part consider the functions $frac {x^{n}} n$.
    – Kavi Rama Murthy
    Nov 29 at 5:25















up vote
1
down vote

favorite












Let $C[0,1]$ denote the vector space of all continuous functions on the interval $[0,1]$, endowed with the supremum norm, and let $D$ denote the subspace



$D:={xin C[0,1]: x'' text{ is continuous and } x(0)=x'(0)=0 }subset C[0,1]$.



Is it true that the linear mapping $T: Dlongrightarrow C[0,1]$, given by $Tx(t):=x''(t)+x(t)$, is continuous? And does there exist the inverse $T^{-1}$? If so, what is $T^{-1}$?



I think/guess that the mapping is not continuous. Any hint/comment will be appreciated.






functional-analysis operator-theory







share|cite|improve this question
























  • For the first part consider the functions $frac {x^{n}} n$.
    – Kavi Rama Murthy
    Nov 29 at 5:25













up vote
1
down vote

favorite









up vote
1
down vote

favorite











Let $C[0,1]$ denote the vector space of all continuous functions on the interval $[0,1]$, endowed with the supremum norm, and let $D$ denote the subspace



$D:={xin C[0,1]: x'' text{ is continuous and } x(0)=x'(0)=0 }subset C[0,1]$.



Is it true that the linear mapping $T: Dlongrightarrow C[0,1]$, given by $Tx(t):=x''(t)+x(t)$, is continuous? And does there exist the inverse $T^{-1}$? If so, what is $T^{-1}$?



I think/guess that the mapping is not continuous. Any hint/comment will be appreciated.






functional-analysis operator-theory







share|cite|improve this question















Let $C[0,1]$ denote the vector space of all continuous functions on the interval $[0,1]$, endowed with the supremum norm, and let $D$ denote the subspace



$D:={xin C[0,1]: x'' text{ is continuous and } x(0)=x'(0)=0 }subset C[0,1]$.



Is it true that the linear mapping $T: Dlongrightarrow C[0,1]$, given by $Tx(t):=x''(t)+x(t)$, is continuous? And does there exist the inverse $T^{-1}$? If so, what is $T^{-1}$?



I think/guess that the mapping is not continuous. Any hint/comment will be appreciated.






functional-analysis operator-theory




functional-analysis operator-theory






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share|cite|improve this question













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edited Nov 29 at 7:52

























asked Nov 29 at 4:59









mathmax

1896




1896












  • For the first part consider the functions $frac {x^{n}} n$.
    – Kavi Rama Murthy
    Nov 29 at 5:25


















  • For the first part consider the functions $frac {x^{n}} n$.
    – Kavi Rama Murthy
    Nov 29 at 5:25
















For the first part consider the functions $frac {x^{n}} n$.
– Kavi Rama Murthy
Nov 29 at 5:25




For the first part consider the functions $frac {x^{n}} n$.
– Kavi Rama Murthy
Nov 29 at 5:25










1 Answer
1






active

oldest

votes

















up vote
1
down vote



accepted










For the first question, let $x_n(t)=t^n$. Then $x_nin D$, and $Tx=n(n-1)t^{n-2}+t^n$. So $|x_n|=1$ for all $n$, while $|Tx_n|=n(n-1)+1$ (since $Tx_n$ is increasing). So
$$
frac{|Tx_n|}{|x_n|}=n(n-1)+1,
$$
and so $T$ is not bounded.



An inverse (as an unbounded operator) does exist: the initial value problem $$tag1x''+x=f, x(0)=x'(0)=0$$ always has a unique solution if $f$ is continuous. So $T$ is onto. It is also one-to-one, because the homogeneous problem only has the solution $x=0$.



The linear map $T^{-1}$ maps $f$ to the solution $x$ of the IVP $(1)$. Since a fundamental set of solutions for the homogeneous problem in $(1)$ is given by $cos t$, $sin t$, one gets that
$$
(T^{-1}f)(x)=int_0^x(cos tsin x-cos xsin t),f(t),dt.
$$
The second part of your question was asked earlier today.






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  • Thanks for the answer
    – mathmax
    Nov 29 at 13:23











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1 Answer
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active

oldest

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1






active

oldest

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oldest

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oldest

votes








up vote
1
down vote



accepted










For the first question, let $x_n(t)=t^n$. Then $x_nin D$, and $Tx=n(n-1)t^{n-2}+t^n$. So $|x_n|=1$ for all $n$, while $|Tx_n|=n(n-1)+1$ (since $Tx_n$ is increasing). So
$$
frac{|Tx_n|}{|x_n|}=n(n-1)+1,
$$
and so $T$ is not bounded.



An inverse (as an unbounded operator) does exist: the initial value problem $$tag1x''+x=f, x(0)=x'(0)=0$$ always has a unique solution if $f$ is continuous. So $T$ is onto. It is also one-to-one, because the homogeneous problem only has the solution $x=0$.



The linear map $T^{-1}$ maps $f$ to the solution $x$ of the IVP $(1)$. Since a fundamental set of solutions for the homogeneous problem in $(1)$ is given by $cos t$, $sin t$, one gets that
$$
(T^{-1}f)(x)=int_0^x(cos tsin x-cos xsin t),f(t),dt.
$$
The second part of your question was asked earlier today.






share|cite|improve this answer





















  • Thanks for the answer
    – mathmax
    Nov 29 at 13:23















up vote
1
down vote



accepted










For the first question, let $x_n(t)=t^n$. Then $x_nin D$, and $Tx=n(n-1)t^{n-2}+t^n$. So $|x_n|=1$ for all $n$, while $|Tx_n|=n(n-1)+1$ (since $Tx_n$ is increasing). So
$$
frac{|Tx_n|}{|x_n|}=n(n-1)+1,
$$
and so $T$ is not bounded.



An inverse (as an unbounded operator) does exist: the initial value problem $$tag1x''+x=f, x(0)=x'(0)=0$$ always has a unique solution if $f$ is continuous. So $T$ is onto. It is also one-to-one, because the homogeneous problem only has the solution $x=0$.



The linear map $T^{-1}$ maps $f$ to the solution $x$ of the IVP $(1)$. Since a fundamental set of solutions for the homogeneous problem in $(1)$ is given by $cos t$, $sin t$, one gets that
$$
(T^{-1}f)(x)=int_0^x(cos tsin x-cos xsin t),f(t),dt.
$$
The second part of your question was asked earlier today.






share|cite|improve this answer





















  • Thanks for the answer
    – mathmax
    Nov 29 at 13:23













up vote
1
down vote



accepted







up vote
1
down vote



accepted






For the first question, let $x_n(t)=t^n$. Then $x_nin D$, and $Tx=n(n-1)t^{n-2}+t^n$. So $|x_n|=1$ for all $n$, while $|Tx_n|=n(n-1)+1$ (since $Tx_n$ is increasing). So
$$
frac{|Tx_n|}{|x_n|}=n(n-1)+1,
$$
and so $T$ is not bounded.



An inverse (as an unbounded operator) does exist: the initial value problem $$tag1x''+x=f, x(0)=x'(0)=0$$ always has a unique solution if $f$ is continuous. So $T$ is onto. It is also one-to-one, because the homogeneous problem only has the solution $x=0$.



The linear map $T^{-1}$ maps $f$ to the solution $x$ of the IVP $(1)$. Since a fundamental set of solutions for the homogeneous problem in $(1)$ is given by $cos t$, $sin t$, one gets that
$$
(T^{-1}f)(x)=int_0^x(cos tsin x-cos xsin t),f(t),dt.
$$
The second part of your question was asked earlier today.






share|cite|improve this answer












For the first question, let $x_n(t)=t^n$. Then $x_nin D$, and $Tx=n(n-1)t^{n-2}+t^n$. So $|x_n|=1$ for all $n$, while $|Tx_n|=n(n-1)+1$ (since $Tx_n$ is increasing). So
$$
frac{|Tx_n|}{|x_n|}=n(n-1)+1,
$$
and so $T$ is not bounded.



An inverse (as an unbounded operator) does exist: the initial value problem $$tag1x''+x=f, x(0)=x'(0)=0$$ always has a unique solution if $f$ is continuous. So $T$ is onto. It is also one-to-one, because the homogeneous problem only has the solution $x=0$.



The linear map $T^{-1}$ maps $f$ to the solution $x$ of the IVP $(1)$. Since a fundamental set of solutions for the homogeneous problem in $(1)$ is given by $cos t$, $sin t$, one gets that
$$
(T^{-1}f)(x)=int_0^x(cos tsin x-cos xsin t),f(t),dt.
$$
The second part of your question was asked earlier today.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 29 at 6:23









Martin Argerami

123k1176174




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  • Thanks for the answer
    – mathmax
    Nov 29 at 13:23


















  • Thanks for the answer
    – mathmax
    Nov 29 at 13:23
















Thanks for the answer
– mathmax
Nov 29 at 13:23




Thanks for the answer
– mathmax
Nov 29 at 13:23


















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