compute the norm of a linear operator












0












$begingroup$


good morning everyone,
I am preparing my functional analysis exam and I can't resolve this exercise:



Let $$ H = L^2([0, 3])$$ and let T : H → H be defined by



$$Tu(t) = (1 + t^2)u(t)$$



i) Prove that T is linear and continuous and compute its norm.



I find out that the norm of T is lower than 10 but I cannot show that It is greater or equal than 10 to conclude that it is actually equal to 10.
can anyone help me?



thank you in advance






functional-analysis hilbert-spaces norm







share|cite|improve this question









$endgroup$












  • $begingroup$
    HINT: Consider a sequence $f_nin L^2(0,3)$ of continuous functions which has unit $L^2$ norm and that concentrates on $t=3$, where $1+t^2$ is maximal, as $nto infty$.
    $endgroup$
    – Giuseppe Negro
    Jan 15 at 10:10










  • $begingroup$
    How did you get that $10$?
    $endgroup$
    – José Carlos Santos
    Jan 15 at 10:10










  • $begingroup$
    @JoséCarlosSantos using Holder's inequality. 10 is the maximum of $$1+t^2$$ in [0,3]
    $endgroup$
    – whowho
    Jan 15 at 10:19










  • $begingroup$
    @GiuseppeNegro thank you!
    $endgroup$
    – whowho
    Jan 15 at 10:22










  • $begingroup$
    @whowho You don't need Holder. Just write down norm square of $Tu$ and pull out the maximum of $(1+t^{2})^{2}$.
    $endgroup$
    – Kavi Rama Murthy
    Jan 15 at 10:23
















0












$begingroup$


good morning everyone,
I am preparing my functional analysis exam and I can't resolve this exercise:



Let $$ H = L^2([0, 3])$$ and let T : H → H be defined by



$$Tu(t) = (1 + t^2)u(t)$$



i) Prove that T is linear and continuous and compute its norm.



I find out that the norm of T is lower than 10 but I cannot show that It is greater or equal than 10 to conclude that it is actually equal to 10.
can anyone help me?



thank you in advance






functional-analysis hilbert-spaces norm







share|cite|improve this question









$endgroup$












  • $begingroup$
    HINT: Consider a sequence $f_nin L^2(0,3)$ of continuous functions which has unit $L^2$ norm and that concentrates on $t=3$, where $1+t^2$ is maximal, as $nto infty$.
    $endgroup$
    – Giuseppe Negro
    Jan 15 at 10:10










  • $begingroup$
    How did you get that $10$?
    $endgroup$
    – José Carlos Santos
    Jan 15 at 10:10










  • $begingroup$
    @JoséCarlosSantos using Holder's inequality. 10 is the maximum of $$1+t^2$$ in [0,3]
    $endgroup$
    – whowho
    Jan 15 at 10:19










  • $begingroup$
    @GiuseppeNegro thank you!
    $endgroup$
    – whowho
    Jan 15 at 10:22










  • $begingroup$
    @whowho You don't need Holder. Just write down norm square of $Tu$ and pull out the maximum of $(1+t^{2})^{2}$.
    $endgroup$
    – Kavi Rama Murthy
    Jan 15 at 10:23














0












0








0





$begingroup$


good morning everyone,
I am preparing my functional analysis exam and I can't resolve this exercise:



Let $$ H = L^2([0, 3])$$ and let T : H → H be defined by



$$Tu(t) = (1 + t^2)u(t)$$



i) Prove that T is linear and continuous and compute its norm.



I find out that the norm of T is lower than 10 but I cannot show that It is greater or equal than 10 to conclude that it is actually equal to 10.
can anyone help me?



thank you in advance






functional-analysis hilbert-spaces norm







share|cite|improve this question









$endgroup$




good morning everyone,
I am preparing my functional analysis exam and I can't resolve this exercise:



Let $$ H = L^2([0, 3])$$ and let T : H → H be defined by



$$Tu(t) = (1 + t^2)u(t)$$



i) Prove that T is linear and continuous and compute its norm.



I find out that the norm of T is lower than 10 but I cannot show that It is greater or equal than 10 to conclude that it is actually equal to 10.
can anyone help me?



thank you in advance






functional-analysis hilbert-spaces norm




functional-analysis hilbert-spaces norm






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 15 at 10:05









whowhowhowho

202




202












  • $begingroup$
    HINT: Consider a sequence $f_nin L^2(0,3)$ of continuous functions which has unit $L^2$ norm and that concentrates on $t=3$, where $1+t^2$ is maximal, as $nto infty$.
    $endgroup$
    – Giuseppe Negro
    Jan 15 at 10:10










  • $begingroup$
    How did you get that $10$?
    $endgroup$
    – José Carlos Santos
    Jan 15 at 10:10










  • $begingroup$
    @JoséCarlosSantos using Holder's inequality. 10 is the maximum of $$1+t^2$$ in [0,3]
    $endgroup$
    – whowho
    Jan 15 at 10:19










  • $begingroup$
    @GiuseppeNegro thank you!
    $endgroup$
    – whowho
    Jan 15 at 10:22










  • $begingroup$
    @whowho You don't need Holder. Just write down norm square of $Tu$ and pull out the maximum of $(1+t^{2})^{2}$.
    $endgroup$
    – Kavi Rama Murthy
    Jan 15 at 10:23


















  • $begingroup$
    HINT: Consider a sequence $f_nin L^2(0,3)$ of continuous functions which has unit $L^2$ norm and that concentrates on $t=3$, where $1+t^2$ is maximal, as $nto infty$.
    $endgroup$
    – Giuseppe Negro
    Jan 15 at 10:10










  • $begingroup$
    How did you get that $10$?
    $endgroup$
    – José Carlos Santos
    Jan 15 at 10:10










  • $begingroup$
    @JoséCarlosSantos using Holder's inequality. 10 is the maximum of $$1+t^2$$ in [0,3]
    $endgroup$
    – whowho
    Jan 15 at 10:19










  • $begingroup$
    @GiuseppeNegro thank you!
    $endgroup$
    – whowho
    Jan 15 at 10:22










  • $begingroup$
    @whowho You don't need Holder. Just write down norm square of $Tu$ and pull out the maximum of $(1+t^{2})^{2}$.
    $endgroup$
    – Kavi Rama Murthy
    Jan 15 at 10:23
















$begingroup$
HINT: Consider a sequence $f_nin L^2(0,3)$ of continuous functions which has unit $L^2$ norm and that concentrates on $t=3$, where $1+t^2$ is maximal, as $nto infty$.
$endgroup$
– Giuseppe Negro
Jan 15 at 10:10




$begingroup$
HINT: Consider a sequence $f_nin L^2(0,3)$ of continuous functions which has unit $L^2$ norm and that concentrates on $t=3$, where $1+t^2$ is maximal, as $nto infty$.
$endgroup$
– Giuseppe Negro
Jan 15 at 10:10












$begingroup$
How did you get that $10$?
$endgroup$
– José Carlos Santos
Jan 15 at 10:10




$begingroup$
How did you get that $10$?
$endgroup$
– José Carlos Santos
Jan 15 at 10:10












$begingroup$
@JoséCarlosSantos using Holder's inequality. 10 is the maximum of $$1+t^2$$ in [0,3]
$endgroup$
– whowho
Jan 15 at 10:19




$begingroup$
@JoséCarlosSantos using Holder's inequality. 10 is the maximum of $$1+t^2$$ in [0,3]
$endgroup$
– whowho
Jan 15 at 10:19












$begingroup$
@GiuseppeNegro thank you!
$endgroup$
– whowho
Jan 15 at 10:22




$begingroup$
@GiuseppeNegro thank you!
$endgroup$
– whowho
Jan 15 at 10:22












$begingroup$
@whowho You don't need Holder. Just write down norm square of $Tu$ and pull out the maximum of $(1+t^{2})^{2}$.
$endgroup$
– Kavi Rama Murthy
Jan 15 at 10:23




$begingroup$
@whowho You don't need Holder. Just write down norm square of $Tu$ and pull out the maximum of $(1+t^{2})^{2}$.
$endgroup$
– Kavi Rama Murthy
Jan 15 at 10:23










1 Answer
1






active

oldest

votes


















3












$begingroup$

Let $u_n(t)=sqrt nI_{(3-frac 1 n,3)}$. Then $|u_n|=1$ for all $n$. Now $|Tu_n|^{2}=nint _{3-1/n} ^{3} (1+t^{2})^{2} , dt to 100$ so $|T| geq 10$. As you have already noted the bound $(1+t^{2})^{2} leq 100$ gives the opposite inequality.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you very much. The exercise also ask to show that T is compact. To do so, I am trying to show that it is weak-strong continuous, since L2 is reflexive, but I don't think it will work because I can't manage to demonstrate the strong convergence in L2. Do you suggest other ways?
    $endgroup$
    – whowho
    Jan 15 at 11:43










  • $begingroup$
    @whowho $T$ is not compact. If ${e_n}$ is an orthonormal basis for $L^{2}[0,3]$ then it is easy to see (using the fact that $1+t^{2} geq 1$) that no subsequence of ${Te_n}$ is Cauchy in the norm.
    $endgroup$
    – Kavi Rama Murthy
    Jan 15 at 11:49












  • $begingroup$
    again, thank you very much!
    $endgroup$
    – whowho
    Jan 15 at 11:50












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1 Answer
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active

oldest

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









3












$begingroup$

Let $u_n(t)=sqrt nI_{(3-frac 1 n,3)}$. Then $|u_n|=1$ for all $n$. Now $|Tu_n|^{2}=nint _{3-1/n} ^{3} (1+t^{2})^{2} , dt to 100$ so $|T| geq 10$. As you have already noted the bound $(1+t^{2})^{2} leq 100$ gives the opposite inequality.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you very much. The exercise also ask to show that T is compact. To do so, I am trying to show that it is weak-strong continuous, since L2 is reflexive, but I don't think it will work because I can't manage to demonstrate the strong convergence in L2. Do you suggest other ways?
    $endgroup$
    – whowho
    Jan 15 at 11:43










  • $begingroup$
    @whowho $T$ is not compact. If ${e_n}$ is an orthonormal basis for $L^{2}[0,3]$ then it is easy to see (using the fact that $1+t^{2} geq 1$) that no subsequence of ${Te_n}$ is Cauchy in the norm.
    $endgroup$
    – Kavi Rama Murthy
    Jan 15 at 11:49












  • $begingroup$
    again, thank you very much!
    $endgroup$
    – whowho
    Jan 15 at 11:50
















3












$begingroup$

Let $u_n(t)=sqrt nI_{(3-frac 1 n,3)}$. Then $|u_n|=1$ for all $n$. Now $|Tu_n|^{2}=nint _{3-1/n} ^{3} (1+t^{2})^{2} , dt to 100$ so $|T| geq 10$. As you have already noted the bound $(1+t^{2})^{2} leq 100$ gives the opposite inequality.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you very much. The exercise also ask to show that T is compact. To do so, I am trying to show that it is weak-strong continuous, since L2 is reflexive, but I don't think it will work because I can't manage to demonstrate the strong convergence in L2. Do you suggest other ways?
    $endgroup$
    – whowho
    Jan 15 at 11:43










  • $begingroup$
    @whowho $T$ is not compact. If ${e_n}$ is an orthonormal basis for $L^{2}[0,3]$ then it is easy to see (using the fact that $1+t^{2} geq 1$) that no subsequence of ${Te_n}$ is Cauchy in the norm.
    $endgroup$
    – Kavi Rama Murthy
    Jan 15 at 11:49












  • $begingroup$
    again, thank you very much!
    $endgroup$
    – whowho
    Jan 15 at 11:50














3












3








3





$begingroup$

Let $u_n(t)=sqrt nI_{(3-frac 1 n,3)}$. Then $|u_n|=1$ for all $n$. Now $|Tu_n|^{2}=nint _{3-1/n} ^{3} (1+t^{2})^{2} , dt to 100$ so $|T| geq 10$. As you have already noted the bound $(1+t^{2})^{2} leq 100$ gives the opposite inequality.






share|cite|improve this answer









$endgroup$



Let $u_n(t)=sqrt nI_{(3-frac 1 n,3)}$. Then $|u_n|=1$ for all $n$. Now $|Tu_n|^{2}=nint _{3-1/n} ^{3} (1+t^{2})^{2} , dt to 100$ so $|T| geq 10$. As you have already noted the bound $(1+t^{2})^{2} leq 100$ gives the opposite inequality.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 15 at 10:14









Kavi Rama MurthyKavi Rama Murthy

76.9k53471




76.9k53471












  • $begingroup$
    Thank you very much. The exercise also ask to show that T is compact. To do so, I am trying to show that it is weak-strong continuous, since L2 is reflexive, but I don't think it will work because I can't manage to demonstrate the strong convergence in L2. Do you suggest other ways?
    $endgroup$
    – whowho
    Jan 15 at 11:43










  • $begingroup$
    @whowho $T$ is not compact. If ${e_n}$ is an orthonormal basis for $L^{2}[0,3]$ then it is easy to see (using the fact that $1+t^{2} geq 1$) that no subsequence of ${Te_n}$ is Cauchy in the norm.
    $endgroup$
    – Kavi Rama Murthy
    Jan 15 at 11:49












  • $begingroup$
    again, thank you very much!
    $endgroup$
    – whowho
    Jan 15 at 11:50


















  • $begingroup$
    Thank you very much. The exercise also ask to show that T is compact. To do so, I am trying to show that it is weak-strong continuous, since L2 is reflexive, but I don't think it will work because I can't manage to demonstrate the strong convergence in L2. Do you suggest other ways?
    $endgroup$
    – whowho
    Jan 15 at 11:43










  • $begingroup$
    @whowho $T$ is not compact. If ${e_n}$ is an orthonormal basis for $L^{2}[0,3]$ then it is easy to see (using the fact that $1+t^{2} geq 1$) that no subsequence of ${Te_n}$ is Cauchy in the norm.
    $endgroup$
    – Kavi Rama Murthy
    Jan 15 at 11:49












  • $begingroup$
    again, thank you very much!
    $endgroup$
    – whowho
    Jan 15 at 11:50
















$begingroup$
Thank you very much. The exercise also ask to show that T is compact. To do so, I am trying to show that it is weak-strong continuous, since L2 is reflexive, but I don't think it will work because I can't manage to demonstrate the strong convergence in L2. Do you suggest other ways?
$endgroup$
– whowho
Jan 15 at 11:43




$begingroup$
Thank you very much. The exercise also ask to show that T is compact. To do so, I am trying to show that it is weak-strong continuous, since L2 is reflexive, but I don't think it will work because I can't manage to demonstrate the strong convergence in L2. Do you suggest other ways?
$endgroup$
– whowho
Jan 15 at 11:43












$begingroup$
@whowho $T$ is not compact. If ${e_n}$ is an orthonormal basis for $L^{2}[0,3]$ then it is easy to see (using the fact that $1+t^{2} geq 1$) that no subsequence of ${Te_n}$ is Cauchy in the norm.
$endgroup$
– Kavi Rama Murthy
Jan 15 at 11:49






$begingroup$
@whowho $T$ is not compact. If ${e_n}$ is an orthonormal basis for $L^{2}[0,3]$ then it is easy to see (using the fact that $1+t^{2} geq 1$) that no subsequence of ${Te_n}$ is Cauchy in the norm.
$endgroup$
– Kavi Rama Murthy
Jan 15 at 11:49














$begingroup$
again, thank you very much!
$endgroup$
– whowho
Jan 15 at 11:50




$begingroup$
again, thank you very much!
$endgroup$
– whowho
Jan 15 at 11:50


















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