Find the resolvent set of a differential operator












2












$begingroup$


Let $X=C[0,1]$ and define:
$$Af=f'' text{for } f in D(A)={fin C^{2}[0,1]: f(0)=f(1)=0 }.$$
I want to find $rho(A)$ that is the resolvent set of operator $A.$



I'm not sure how to proceed with this. From the definition $rho(A)$ is the set of all $zin mathbb{C}$ such that $(A-zT)^{-1}$ is $1-1$ and onto.



My take on this problem is following. First I find the solutions to $Af-zf=f''-zf=0$ to get that $$f(t) = c_{1}e^{sqrt{z}t}+c_{2}e^{-sqrt{z}t}$$ solves the equation. Now by applying the boundary conditions I get that there is no $z in mathbb{C}$ such that the BC are satisfied and thus $sigma(A)$ is empty and since $rho(A) =mathbb{C}setminussigma(A) = mathbb{C}.$ Is this correct?






functional-analysis differential-equations eigenvalues-eigenvectors operator-theory







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$endgroup$












  • $begingroup$
    $A$ acts from subspace $C^{2}[0,1] subset C[0,1]$ into $C[0,1].$ If $f in C[0,1] setminus C^{2}[0,1]$ operator $A$ does not make sense and thus the definition. I follow the usual definition of the resolvent set and my approach was developed based on math.stackexchange.com/questions/2008302/….
    $endgroup$
    – data
    Dec 29 '18 at 20:06












  • $begingroup$
    @BenW Just to be clear I used definition which can be found here people.math.gatech.edu/~loss/14SPRINGTEA/spectraltheory.pdf
    $endgroup$
    – data
    Dec 29 '18 at 20:14






  • 2




    $begingroup$
    You made a mistake when solving the differential equation. Your solution is valid only when $z >0$, but what about the case $z<0$? And more generally $z in mathbb{C}$?
    $endgroup$
    – Michh
    Dec 29 '18 at 21:37










  • $begingroup$
    @Michh Yes - my bad - thanks!
    $endgroup$
    – data
    Dec 29 '18 at 21:55
















2












$begingroup$


Let $X=C[0,1]$ and define:
$$Af=f'' text{for } f in D(A)={fin C^{2}[0,1]: f(0)=f(1)=0 }.$$
I want to find $rho(A)$ that is the resolvent set of operator $A.$



I'm not sure how to proceed with this. From the definition $rho(A)$ is the set of all $zin mathbb{C}$ such that $(A-zT)^{-1}$ is $1-1$ and onto.



My take on this problem is following. First I find the solutions to $Af-zf=f''-zf=0$ to get that $$f(t) = c_{1}e^{sqrt{z}t}+c_{2}e^{-sqrt{z}t}$$ solves the equation. Now by applying the boundary conditions I get that there is no $z in mathbb{C}$ such that the BC are satisfied and thus $sigma(A)$ is empty and since $rho(A) =mathbb{C}setminussigma(A) = mathbb{C}.$ Is this correct?






functional-analysis differential-equations eigenvalues-eigenvectors operator-theory







share|cite|improve this question











$endgroup$












  • $begingroup$
    $A$ acts from subspace $C^{2}[0,1] subset C[0,1]$ into $C[0,1].$ If $f in C[0,1] setminus C^{2}[0,1]$ operator $A$ does not make sense and thus the definition. I follow the usual definition of the resolvent set and my approach was developed based on math.stackexchange.com/questions/2008302/….
    $endgroup$
    – data
    Dec 29 '18 at 20:06












  • $begingroup$
    @BenW Just to be clear I used definition which can be found here people.math.gatech.edu/~loss/14SPRINGTEA/spectraltheory.pdf
    $endgroup$
    – data
    Dec 29 '18 at 20:14






  • 2




    $begingroup$
    You made a mistake when solving the differential equation. Your solution is valid only when $z >0$, but what about the case $z<0$? And more generally $z in mathbb{C}$?
    $endgroup$
    – Michh
    Dec 29 '18 at 21:37










  • $begingroup$
    @Michh Yes - my bad - thanks!
    $endgroup$
    – data
    Dec 29 '18 at 21:55














2












2








2





$begingroup$


Let $X=C[0,1]$ and define:
$$Af=f'' text{for } f in D(A)={fin C^{2}[0,1]: f(0)=f(1)=0 }.$$
I want to find $rho(A)$ that is the resolvent set of operator $A.$



I'm not sure how to proceed with this. From the definition $rho(A)$ is the set of all $zin mathbb{C}$ such that $(A-zT)^{-1}$ is $1-1$ and onto.



My take on this problem is following. First I find the solutions to $Af-zf=f''-zf=0$ to get that $$f(t) = c_{1}e^{sqrt{z}t}+c_{2}e^{-sqrt{z}t}$$ solves the equation. Now by applying the boundary conditions I get that there is no $z in mathbb{C}$ such that the BC are satisfied and thus $sigma(A)$ is empty and since $rho(A) =mathbb{C}setminussigma(A) = mathbb{C}.$ Is this correct?






functional-analysis differential-equations eigenvalues-eigenvectors operator-theory







share|cite|improve this question











$endgroup$




Let $X=C[0,1]$ and define:
$$Af=f'' text{for } f in D(A)={fin C^{2}[0,1]: f(0)=f(1)=0 }.$$
I want to find $rho(A)$ that is the resolvent set of operator $A.$



I'm not sure how to proceed with this. From the definition $rho(A)$ is the set of all $zin mathbb{C}$ such that $(A-zT)^{-1}$ is $1-1$ and onto.



My take on this problem is following. First I find the solutions to $Af-zf=f''-zf=0$ to get that $$f(t) = c_{1}e^{sqrt{z}t}+c_{2}e^{-sqrt{z}t}$$ solves the equation. Now by applying the boundary conditions I get that there is no $z in mathbb{C}$ such that the BC are satisfied and thus $sigma(A)$ is empty and since $rho(A) =mathbb{C}setminussigma(A) = mathbb{C}.$ Is this correct?






functional-analysis differential-equations eigenvalues-eigenvectors operator-theory




functional-analysis differential-equations eigenvalues-eigenvectors operator-theory






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share|cite|improve this question













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edited Dec 29 '18 at 20:30







data

















asked Dec 29 '18 at 19:29









datadata

757510




757510












  • $begingroup$
    $A$ acts from subspace $C^{2}[0,1] subset C[0,1]$ into $C[0,1].$ If $f in C[0,1] setminus C^{2}[0,1]$ operator $A$ does not make sense and thus the definition. I follow the usual definition of the resolvent set and my approach was developed based on math.stackexchange.com/questions/2008302/….
    $endgroup$
    – data
    Dec 29 '18 at 20:06












  • $begingroup$
    @BenW Just to be clear I used definition which can be found here people.math.gatech.edu/~loss/14SPRINGTEA/spectraltheory.pdf
    $endgroup$
    – data
    Dec 29 '18 at 20:14






  • 2




    $begingroup$
    You made a mistake when solving the differential equation. Your solution is valid only when $z >0$, but what about the case $z<0$? And more generally $z in mathbb{C}$?
    $endgroup$
    – Michh
    Dec 29 '18 at 21:37










  • $begingroup$
    @Michh Yes - my bad - thanks!
    $endgroup$
    – data
    Dec 29 '18 at 21:55


















  • $begingroup$
    $A$ acts from subspace $C^{2}[0,1] subset C[0,1]$ into $C[0,1].$ If $f in C[0,1] setminus C^{2}[0,1]$ operator $A$ does not make sense and thus the definition. I follow the usual definition of the resolvent set and my approach was developed based on math.stackexchange.com/questions/2008302/….
    $endgroup$
    – data
    Dec 29 '18 at 20:06












  • $begingroup$
    @BenW Just to be clear I used definition which can be found here people.math.gatech.edu/~loss/14SPRINGTEA/spectraltheory.pdf
    $endgroup$
    – data
    Dec 29 '18 at 20:14






  • 2




    $begingroup$
    You made a mistake when solving the differential equation. Your solution is valid only when $z >0$, but what about the case $z<0$? And more generally $z in mathbb{C}$?
    $endgroup$
    – Michh
    Dec 29 '18 at 21:37










  • $begingroup$
    @Michh Yes - my bad - thanks!
    $endgroup$
    – data
    Dec 29 '18 at 21:55
















$begingroup$
$A$ acts from subspace $C^{2}[0,1] subset C[0,1]$ into $C[0,1].$ If $f in C[0,1] setminus C^{2}[0,1]$ operator $A$ does not make sense and thus the definition. I follow the usual definition of the resolvent set and my approach was developed based on math.stackexchange.com/questions/2008302/….
$endgroup$
– data
Dec 29 '18 at 20:06






$begingroup$
$A$ acts from subspace $C^{2}[0,1] subset C[0,1]$ into $C[0,1].$ If $f in C[0,1] setminus C^{2}[0,1]$ operator $A$ does not make sense and thus the definition. I follow the usual definition of the resolvent set and my approach was developed based on math.stackexchange.com/questions/2008302/….
$endgroup$
– data
Dec 29 '18 at 20:06














$begingroup$
@BenW Just to be clear I used definition which can be found here people.math.gatech.edu/~loss/14SPRINGTEA/spectraltheory.pdf
$endgroup$
– data
Dec 29 '18 at 20:14




$begingroup$
@BenW Just to be clear I used definition which can be found here people.math.gatech.edu/~loss/14SPRINGTEA/spectraltheory.pdf
$endgroup$
– data
Dec 29 '18 at 20:14




2




2




$begingroup$
You made a mistake when solving the differential equation. Your solution is valid only when $z >0$, but what about the case $z<0$? And more generally $z in mathbb{C}$?
$endgroup$
– Michh
Dec 29 '18 at 21:37




$begingroup$
You made a mistake when solving the differential equation. Your solution is valid only when $z >0$, but what about the case $z<0$? And more generally $z in mathbb{C}$?
$endgroup$
– Michh
Dec 29 '18 at 21:37












$begingroup$
@Michh Yes - my bad - thanks!
$endgroup$
– data
Dec 29 '18 at 21:55




$begingroup$
@Michh Yes - my bad - thanks!
$endgroup$
– data
Dec 29 '18 at 21:55










2 Answers
2






active

oldest

votes


















1












$begingroup$

The resolvent $R(z)=(A-zI)^{-1}$ does not exist if $z=-n^2pi^2$ where $n=1,2,3,cdots$. This is because $sin(npi x)$ vanishes at $0,1$ and
$$
(A+n^2I)sin(npi x) = 0,;; n=1,2,3,cdots.
$$



For $z ne -n^2pi^2$, the resolvent is determined by solving
$$
f''+lambda f=g,;;; f(0)=f(1)=0.
$$
Variation of parameters can be used to solve this equation. The solution is
begin{align}
f(x)&=frac{sin(sqrt{lambda}(x-1))}{sqrt{lambda}sin(sqrt{lambda})}int_{0}^{x}sin(sqrt{lambda}y)g(y)dy\&+frac{sin(sqrt{lambda}x)}{sqrt{lambda}sin(sqrt{lambda})}int_{x}^{1}sin(sqrt{lambda}(y-1))g(y)dy.
end{align}






share|cite|improve this answer











$endgroup$













  • $begingroup$
    So $rho(A)=mathbb{C} setminus {-n^{2}pi^{2}: ninmathbb{N}}$ right?
    $endgroup$
    – data
    Dec 30 '18 at 14:37










  • $begingroup$
    @data : NO. $rho(A) = mathbb{C}setminus { -n^2pi^2 : n=1,2,3,cdots }$. $0$ is in the resolvent set. The resolvent expression I gave you has a removable singularity at $lambda=0$, and is equal to $(x-1)int_0^x y g(y)dy + xint_x^1 (y-1)g(y)dy$.
    $endgroup$
    – DisintegratingByParts
    Dec 30 '18 at 15:00










  • $begingroup$
    Yep - I use convention to exclude $0$ from $mathbb{N}.$ Anyway thanks for your answer!
    $endgroup$
    – data
    Dec 30 '18 at 16:09










  • $begingroup$
    @data : oops, my blunder. Been rather I'll lately.
    $endgroup$
    – DisintegratingByParts
    Dec 30 '18 at 18:41



















1












$begingroup$

It seems not correct.E.g.:The spectrum is not empty. Consider the function
$$f(t)=sin(pi t).$$
Then $$f^{''}(t)=-pi^2sin(pi t)$$
thus $Af+pi^2f=0$ and $f(0)=f(1)=0$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks for that comment! Can you please comment on my general approach to this problem? Is there an alternative way to look at this problem?
    $endgroup$
    – data
    Dec 29 '18 at 21:57










  • $begingroup$
    I´m not yet sure. I´mthinking about that. By slightly generalizing the above example to $f_k(t)=sin(kpi t)$ for $kinmathbb{Z}$ You can see that the spectrum has at least countably many elements.
    $endgroup$
    – Peter Melech
    Dec 29 '18 at 22:05











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

The resolvent $R(z)=(A-zI)^{-1}$ does not exist if $z=-n^2pi^2$ where $n=1,2,3,cdots$. This is because $sin(npi x)$ vanishes at $0,1$ and
$$
(A+n^2I)sin(npi x) = 0,;; n=1,2,3,cdots.
$$



For $z ne -n^2pi^2$, the resolvent is determined by solving
$$
f''+lambda f=g,;;; f(0)=f(1)=0.
$$
Variation of parameters can be used to solve this equation. The solution is
begin{align}
f(x)&=frac{sin(sqrt{lambda}(x-1))}{sqrt{lambda}sin(sqrt{lambda})}int_{0}^{x}sin(sqrt{lambda}y)g(y)dy\&+frac{sin(sqrt{lambda}x)}{sqrt{lambda}sin(sqrt{lambda})}int_{x}^{1}sin(sqrt{lambda}(y-1))g(y)dy.
end{align}






share|cite|improve this answer











$endgroup$













  • $begingroup$
    So $rho(A)=mathbb{C} setminus {-n^{2}pi^{2}: ninmathbb{N}}$ right?
    $endgroup$
    – data
    Dec 30 '18 at 14:37










  • $begingroup$
    @data : NO. $rho(A) = mathbb{C}setminus { -n^2pi^2 : n=1,2,3,cdots }$. $0$ is in the resolvent set. The resolvent expression I gave you has a removable singularity at $lambda=0$, and is equal to $(x-1)int_0^x y g(y)dy + xint_x^1 (y-1)g(y)dy$.
    $endgroup$
    – DisintegratingByParts
    Dec 30 '18 at 15:00










  • $begingroup$
    Yep - I use convention to exclude $0$ from $mathbb{N}.$ Anyway thanks for your answer!
    $endgroup$
    – data
    Dec 30 '18 at 16:09










  • $begingroup$
    @data : oops, my blunder. Been rather I'll lately.
    $endgroup$
    – DisintegratingByParts
    Dec 30 '18 at 18:41
















1












$begingroup$

The resolvent $R(z)=(A-zI)^{-1}$ does not exist if $z=-n^2pi^2$ where $n=1,2,3,cdots$. This is because $sin(npi x)$ vanishes at $0,1$ and
$$
(A+n^2I)sin(npi x) = 0,;; n=1,2,3,cdots.
$$



For $z ne -n^2pi^2$, the resolvent is determined by solving
$$
f''+lambda f=g,;;; f(0)=f(1)=0.
$$
Variation of parameters can be used to solve this equation. The solution is
begin{align}
f(x)&=frac{sin(sqrt{lambda}(x-1))}{sqrt{lambda}sin(sqrt{lambda})}int_{0}^{x}sin(sqrt{lambda}y)g(y)dy\&+frac{sin(sqrt{lambda}x)}{sqrt{lambda}sin(sqrt{lambda})}int_{x}^{1}sin(sqrt{lambda}(y-1))g(y)dy.
end{align}






share|cite|improve this answer











$endgroup$













  • $begingroup$
    So $rho(A)=mathbb{C} setminus {-n^{2}pi^{2}: ninmathbb{N}}$ right?
    $endgroup$
    – data
    Dec 30 '18 at 14:37










  • $begingroup$
    @data : NO. $rho(A) = mathbb{C}setminus { -n^2pi^2 : n=1,2,3,cdots }$. $0$ is in the resolvent set. The resolvent expression I gave you has a removable singularity at $lambda=0$, and is equal to $(x-1)int_0^x y g(y)dy + xint_x^1 (y-1)g(y)dy$.
    $endgroup$
    – DisintegratingByParts
    Dec 30 '18 at 15:00










  • $begingroup$
    Yep - I use convention to exclude $0$ from $mathbb{N}.$ Anyway thanks for your answer!
    $endgroup$
    – data
    Dec 30 '18 at 16:09










  • $begingroup$
    @data : oops, my blunder. Been rather I'll lately.
    $endgroup$
    – DisintegratingByParts
    Dec 30 '18 at 18:41














1












1








1





$begingroup$

The resolvent $R(z)=(A-zI)^{-1}$ does not exist if $z=-n^2pi^2$ where $n=1,2,3,cdots$. This is because $sin(npi x)$ vanishes at $0,1$ and
$$
(A+n^2I)sin(npi x) = 0,;; n=1,2,3,cdots.
$$



For $z ne -n^2pi^2$, the resolvent is determined by solving
$$
f''+lambda f=g,;;; f(0)=f(1)=0.
$$
Variation of parameters can be used to solve this equation. The solution is
begin{align}
f(x)&=frac{sin(sqrt{lambda}(x-1))}{sqrt{lambda}sin(sqrt{lambda})}int_{0}^{x}sin(sqrt{lambda}y)g(y)dy\&+frac{sin(sqrt{lambda}x)}{sqrt{lambda}sin(sqrt{lambda})}int_{x}^{1}sin(sqrt{lambda}(y-1))g(y)dy.
end{align}






share|cite|improve this answer











$endgroup$



The resolvent $R(z)=(A-zI)^{-1}$ does not exist if $z=-n^2pi^2$ where $n=1,2,3,cdots$. This is because $sin(npi x)$ vanishes at $0,1$ and
$$
(A+n^2I)sin(npi x) = 0,;; n=1,2,3,cdots.
$$



For $z ne -n^2pi^2$, the resolvent is determined by solving
$$
f''+lambda f=g,;;; f(0)=f(1)=0.
$$
Variation of parameters can be used to solve this equation. The solution is
begin{align}
f(x)&=frac{sin(sqrt{lambda}(x-1))}{sqrt{lambda}sin(sqrt{lambda})}int_{0}^{x}sin(sqrt{lambda}y)g(y)dy\&+frac{sin(sqrt{lambda}x)}{sqrt{lambda}sin(sqrt{lambda})}int_{x}^{1}sin(sqrt{lambda}(y-1))g(y)dy.
end{align}







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 30 '18 at 7:28

























answered Dec 30 '18 at 6:57









DisintegratingByPartsDisintegratingByParts

58.8k42579




58.8k42579












  • $begingroup$
    So $rho(A)=mathbb{C} setminus {-n^{2}pi^{2}: ninmathbb{N}}$ right?
    $endgroup$
    – data
    Dec 30 '18 at 14:37










  • $begingroup$
    @data : NO. $rho(A) = mathbb{C}setminus { -n^2pi^2 : n=1,2,3,cdots }$. $0$ is in the resolvent set. The resolvent expression I gave you has a removable singularity at $lambda=0$, and is equal to $(x-1)int_0^x y g(y)dy + xint_x^1 (y-1)g(y)dy$.
    $endgroup$
    – DisintegratingByParts
    Dec 30 '18 at 15:00










  • $begingroup$
    Yep - I use convention to exclude $0$ from $mathbb{N}.$ Anyway thanks for your answer!
    $endgroup$
    – data
    Dec 30 '18 at 16:09










  • $begingroup$
    @data : oops, my blunder. Been rather I'll lately.
    $endgroup$
    – DisintegratingByParts
    Dec 30 '18 at 18:41


















  • $begingroup$
    So $rho(A)=mathbb{C} setminus {-n^{2}pi^{2}: ninmathbb{N}}$ right?
    $endgroup$
    – data
    Dec 30 '18 at 14:37










  • $begingroup$
    @data : NO. $rho(A) = mathbb{C}setminus { -n^2pi^2 : n=1,2,3,cdots }$. $0$ is in the resolvent set. The resolvent expression I gave you has a removable singularity at $lambda=0$, and is equal to $(x-1)int_0^x y g(y)dy + xint_x^1 (y-1)g(y)dy$.
    $endgroup$
    – DisintegratingByParts
    Dec 30 '18 at 15:00










  • $begingroup$
    Yep - I use convention to exclude $0$ from $mathbb{N}.$ Anyway thanks for your answer!
    $endgroup$
    – data
    Dec 30 '18 at 16:09










  • $begingroup$
    @data : oops, my blunder. Been rather I'll lately.
    $endgroup$
    – DisintegratingByParts
    Dec 30 '18 at 18:41
















$begingroup$
So $rho(A)=mathbb{C} setminus {-n^{2}pi^{2}: ninmathbb{N}}$ right?
$endgroup$
– data
Dec 30 '18 at 14:37




$begingroup$
So $rho(A)=mathbb{C} setminus {-n^{2}pi^{2}: ninmathbb{N}}$ right?
$endgroup$
– data
Dec 30 '18 at 14:37












$begingroup$
@data : NO. $rho(A) = mathbb{C}setminus { -n^2pi^2 : n=1,2,3,cdots }$. $0$ is in the resolvent set. The resolvent expression I gave you has a removable singularity at $lambda=0$, and is equal to $(x-1)int_0^x y g(y)dy + xint_x^1 (y-1)g(y)dy$.
$endgroup$
– DisintegratingByParts
Dec 30 '18 at 15:00




$begingroup$
@data : NO. $rho(A) = mathbb{C}setminus { -n^2pi^2 : n=1,2,3,cdots }$. $0$ is in the resolvent set. The resolvent expression I gave you has a removable singularity at $lambda=0$, and is equal to $(x-1)int_0^x y g(y)dy + xint_x^1 (y-1)g(y)dy$.
$endgroup$
– DisintegratingByParts
Dec 30 '18 at 15:00












$begingroup$
Yep - I use convention to exclude $0$ from $mathbb{N}.$ Anyway thanks for your answer!
$endgroup$
– data
Dec 30 '18 at 16:09




$begingroup$
Yep - I use convention to exclude $0$ from $mathbb{N}.$ Anyway thanks for your answer!
$endgroup$
– data
Dec 30 '18 at 16:09












$begingroup$
@data : oops, my blunder. Been rather I'll lately.
$endgroup$
– DisintegratingByParts
Dec 30 '18 at 18:41




$begingroup$
@data : oops, my blunder. Been rather I'll lately.
$endgroup$
– DisintegratingByParts
Dec 30 '18 at 18:41











1












$begingroup$

It seems not correct.E.g.:The spectrum is not empty. Consider the function
$$f(t)=sin(pi t).$$
Then $$f^{''}(t)=-pi^2sin(pi t)$$
thus $Af+pi^2f=0$ and $f(0)=f(1)=0$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks for that comment! Can you please comment on my general approach to this problem? Is there an alternative way to look at this problem?
    $endgroup$
    – data
    Dec 29 '18 at 21:57










  • $begingroup$
    I´m not yet sure. I´mthinking about that. By slightly generalizing the above example to $f_k(t)=sin(kpi t)$ for $kinmathbb{Z}$ You can see that the spectrum has at least countably many elements.
    $endgroup$
    – Peter Melech
    Dec 29 '18 at 22:05
















1












$begingroup$

It seems not correct.E.g.:The spectrum is not empty. Consider the function
$$f(t)=sin(pi t).$$
Then $$f^{''}(t)=-pi^2sin(pi t)$$
thus $Af+pi^2f=0$ and $f(0)=f(1)=0$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks for that comment! Can you please comment on my general approach to this problem? Is there an alternative way to look at this problem?
    $endgroup$
    – data
    Dec 29 '18 at 21:57










  • $begingroup$
    I´m not yet sure. I´mthinking about that. By slightly generalizing the above example to $f_k(t)=sin(kpi t)$ for $kinmathbb{Z}$ You can see that the spectrum has at least countably many elements.
    $endgroup$
    – Peter Melech
    Dec 29 '18 at 22:05














1












1








1





$begingroup$

It seems not correct.E.g.:The spectrum is not empty. Consider the function
$$f(t)=sin(pi t).$$
Then $$f^{''}(t)=-pi^2sin(pi t)$$
thus $Af+pi^2f=0$ and $f(0)=f(1)=0$.






share|cite|improve this answer









$endgroup$



It seems not correct.E.g.:The spectrum is not empty. Consider the function
$$f(t)=sin(pi t).$$
Then $$f^{''}(t)=-pi^2sin(pi t)$$
thus $Af+pi^2f=0$ and $f(0)=f(1)=0$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 29 '18 at 21:27









Peter MelechPeter Melech

2,562813




2,562813












  • $begingroup$
    Thanks for that comment! Can you please comment on my general approach to this problem? Is there an alternative way to look at this problem?
    $endgroup$
    – data
    Dec 29 '18 at 21:57










  • $begingroup$
    I´m not yet sure. I´mthinking about that. By slightly generalizing the above example to $f_k(t)=sin(kpi t)$ for $kinmathbb{Z}$ You can see that the spectrum has at least countably many elements.
    $endgroup$
    – Peter Melech
    Dec 29 '18 at 22:05


















  • $begingroup$
    Thanks for that comment! Can you please comment on my general approach to this problem? Is there an alternative way to look at this problem?
    $endgroup$
    – data
    Dec 29 '18 at 21:57










  • $begingroup$
    I´m not yet sure. I´mthinking about that. By slightly generalizing the above example to $f_k(t)=sin(kpi t)$ for $kinmathbb{Z}$ You can see that the spectrum has at least countably many elements.
    $endgroup$
    – Peter Melech
    Dec 29 '18 at 22:05
















$begingroup$
Thanks for that comment! Can you please comment on my general approach to this problem? Is there an alternative way to look at this problem?
$endgroup$
– data
Dec 29 '18 at 21:57




$begingroup$
Thanks for that comment! Can you please comment on my general approach to this problem? Is there an alternative way to look at this problem?
$endgroup$
– data
Dec 29 '18 at 21:57












$begingroup$
I´m not yet sure. I´mthinking about that. By slightly generalizing the above example to $f_k(t)=sin(kpi t)$ for $kinmathbb{Z}$ You can see that the spectrum has at least countably many elements.
$endgroup$
– Peter Melech
Dec 29 '18 at 22:05




$begingroup$
I´m not yet sure. I´mthinking about that. By slightly generalizing the above example to $f_k(t)=sin(kpi t)$ for $kinmathbb{Z}$ You can see that the spectrum has at least countably many elements.
$endgroup$
– Peter Melech
Dec 29 '18 at 22:05


















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