How to show that the eigenvalues of this Hermitian operator are non-negative?












1












$begingroup$


$V$ is the space of infinitely differentiable functions $f:mathbb R to mathbb C$ which are periodic of $2pi$, with the inner product $<f|g> = int_0^{2pi} overline f(t)g(t)dt$. Let $L:Vto V$ be defined as $L=-d^2/dt^2$. It is known that $L$ is self-adjoint, $L=L^*$. How to show that the eigenvalues of $L$ are non-negative? And how to write $L=A^*A$ for an operator $A:Vto V$?






linear-algebra functional-analysis







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$endgroup$












  • $begingroup$
    It is straightforward to show that if $A = ifrac{d}{dt}$, then $A^*A = L$.
    $endgroup$
    – whpowell96
    Dec 13 '18 at 4:46












  • $begingroup$
    @whpowell96 How did you come up with that?
    $endgroup$
    – user398843
    Dec 13 '18 at 4:53










  • $begingroup$
    I know that $frac{d}{dt}frac{d}{dt} = frac{d^2}{dt^2}$, so to make it negative I needed to multiply by a constant $alpha$ such that $alpha^2=-1$
    $endgroup$
    – whpowell96
    Dec 13 '18 at 4:54










  • $begingroup$
    @whpowell96 Is there a general method to write a self-adjoint operator like that, i.e., $L=A^*A$?
    $endgroup$
    – user398843
    Dec 13 '18 at 4:57






  • 1




    $begingroup$
    The spectral theorem guarantees the existence of such an operator if $L$ is bounded and positive semidefinite. I cannot remember off the top of my head if the proof is constructive or not
    $endgroup$
    – whpowell96
    Dec 13 '18 at 5:02
















1












$begingroup$


$V$ is the space of infinitely differentiable functions $f:mathbb R to mathbb C$ which are periodic of $2pi$, with the inner product $<f|g> = int_0^{2pi} overline f(t)g(t)dt$. Let $L:Vto V$ be defined as $L=-d^2/dt^2$. It is known that $L$ is self-adjoint, $L=L^*$. How to show that the eigenvalues of $L$ are non-negative? And how to write $L=A^*A$ for an operator $A:Vto V$?






linear-algebra functional-analysis







share|cite|improve this question









$endgroup$












  • $begingroup$
    It is straightforward to show that if $A = ifrac{d}{dt}$, then $A^*A = L$.
    $endgroup$
    – whpowell96
    Dec 13 '18 at 4:46












  • $begingroup$
    @whpowell96 How did you come up with that?
    $endgroup$
    – user398843
    Dec 13 '18 at 4:53










  • $begingroup$
    I know that $frac{d}{dt}frac{d}{dt} = frac{d^2}{dt^2}$, so to make it negative I needed to multiply by a constant $alpha$ such that $alpha^2=-1$
    $endgroup$
    – whpowell96
    Dec 13 '18 at 4:54










  • $begingroup$
    @whpowell96 Is there a general method to write a self-adjoint operator like that, i.e., $L=A^*A$?
    $endgroup$
    – user398843
    Dec 13 '18 at 4:57






  • 1




    $begingroup$
    The spectral theorem guarantees the existence of such an operator if $L$ is bounded and positive semidefinite. I cannot remember off the top of my head if the proof is constructive or not
    $endgroup$
    – whpowell96
    Dec 13 '18 at 5:02














1












1








1





$begingroup$


$V$ is the space of infinitely differentiable functions $f:mathbb R to mathbb C$ which are periodic of $2pi$, with the inner product $<f|g> = int_0^{2pi} overline f(t)g(t)dt$. Let $L:Vto V$ be defined as $L=-d^2/dt^2$. It is known that $L$ is self-adjoint, $L=L^*$. How to show that the eigenvalues of $L$ are non-negative? And how to write $L=A^*A$ for an operator $A:Vto V$?






linear-algebra functional-analysis







share|cite|improve this question









$endgroup$




$V$ is the space of infinitely differentiable functions $f:mathbb R to mathbb C$ which are periodic of $2pi$, with the inner product $<f|g> = int_0^{2pi} overline f(t)g(t)dt$. Let $L:Vto V$ be defined as $L=-d^2/dt^2$. It is known that $L$ is self-adjoint, $L=L^*$. How to show that the eigenvalues of $L$ are non-negative? And how to write $L=A^*A$ for an operator $A:Vto V$?






linear-algebra functional-analysis




linear-algebra functional-analysis






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 13 '18 at 4:38









user398843user398843

648216




648216












  • $begingroup$
    It is straightforward to show that if $A = ifrac{d}{dt}$, then $A^*A = L$.
    $endgroup$
    – whpowell96
    Dec 13 '18 at 4:46












  • $begingroup$
    @whpowell96 How did you come up with that?
    $endgroup$
    – user398843
    Dec 13 '18 at 4:53










  • $begingroup$
    I know that $frac{d}{dt}frac{d}{dt} = frac{d^2}{dt^2}$, so to make it negative I needed to multiply by a constant $alpha$ such that $alpha^2=-1$
    $endgroup$
    – whpowell96
    Dec 13 '18 at 4:54










  • $begingroup$
    @whpowell96 Is there a general method to write a self-adjoint operator like that, i.e., $L=A^*A$?
    $endgroup$
    – user398843
    Dec 13 '18 at 4:57






  • 1




    $begingroup$
    The spectral theorem guarantees the existence of such an operator if $L$ is bounded and positive semidefinite. I cannot remember off the top of my head if the proof is constructive or not
    $endgroup$
    – whpowell96
    Dec 13 '18 at 5:02


















  • $begingroup$
    It is straightforward to show that if $A = ifrac{d}{dt}$, then $A^*A = L$.
    $endgroup$
    – whpowell96
    Dec 13 '18 at 4:46












  • $begingroup$
    @whpowell96 How did you come up with that?
    $endgroup$
    – user398843
    Dec 13 '18 at 4:53










  • $begingroup$
    I know that $frac{d}{dt}frac{d}{dt} = frac{d^2}{dt^2}$, so to make it negative I needed to multiply by a constant $alpha$ such that $alpha^2=-1$
    $endgroup$
    – whpowell96
    Dec 13 '18 at 4:54










  • $begingroup$
    @whpowell96 Is there a general method to write a self-adjoint operator like that, i.e., $L=A^*A$?
    $endgroup$
    – user398843
    Dec 13 '18 at 4:57






  • 1




    $begingroup$
    The spectral theorem guarantees the existence of such an operator if $L$ is bounded and positive semidefinite. I cannot remember off the top of my head if the proof is constructive or not
    $endgroup$
    – whpowell96
    Dec 13 '18 at 5:02
















$begingroup$
It is straightforward to show that if $A = ifrac{d}{dt}$, then $A^*A = L$.
$endgroup$
– whpowell96
Dec 13 '18 at 4:46






$begingroup$
It is straightforward to show that if $A = ifrac{d}{dt}$, then $A^*A = L$.
$endgroup$
– whpowell96
Dec 13 '18 at 4:46














$begingroup$
@whpowell96 How did you come up with that?
$endgroup$
– user398843
Dec 13 '18 at 4:53




$begingroup$
@whpowell96 How did you come up with that?
$endgroup$
– user398843
Dec 13 '18 at 4:53












$begingroup$
I know that $frac{d}{dt}frac{d}{dt} = frac{d^2}{dt^2}$, so to make it negative I needed to multiply by a constant $alpha$ such that $alpha^2=-1$
$endgroup$
– whpowell96
Dec 13 '18 at 4:54




$begingroup$
I know that $frac{d}{dt}frac{d}{dt} = frac{d^2}{dt^2}$, so to make it negative I needed to multiply by a constant $alpha$ such that $alpha^2=-1$
$endgroup$
– whpowell96
Dec 13 '18 at 4:54












$begingroup$
@whpowell96 Is there a general method to write a self-adjoint operator like that, i.e., $L=A^*A$?
$endgroup$
– user398843
Dec 13 '18 at 4:57




$begingroup$
@whpowell96 Is there a general method to write a self-adjoint operator like that, i.e., $L=A^*A$?
$endgroup$
– user398843
Dec 13 '18 at 4:57




1




1




$begingroup$
The spectral theorem guarantees the existence of such an operator if $L$ is bounded and positive semidefinite. I cannot remember off the top of my head if the proof is constructive or not
$endgroup$
– whpowell96
Dec 13 '18 at 5:02




$begingroup$
The spectral theorem guarantees the existence of such an operator if $L$ is bounded and positive semidefinite. I cannot remember off the top of my head if the proof is constructive or not
$endgroup$
– whpowell96
Dec 13 '18 at 5:02










1 Answer
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$begingroup$

The associated form for $L$ is
begin{align}
langle Lf,frangle &= int_{0}^{2pi} -f''(t)overline{f(t)}dt\
&= -f'overline{f}|_{0}^{2pi}+int_{0}^{2pi}f'(t)overline{f'(t)}dt \
&= int_{0}^{2pi}|f'(t)|^2dt.
end{align}
So, if $Lf=lambda f$, then $lambda|f|^2=|f'|^2$ forces $lambda$ to be real and non-negative. Furthermore, $lambda =0$ iff $f'=0$ or $f$ is constant.






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    $begingroup$

    The associated form for $L$ is
    begin{align}
    langle Lf,frangle &= int_{0}^{2pi} -f''(t)overline{f(t)}dt\
    &= -f'overline{f}|_{0}^{2pi}+int_{0}^{2pi}f'(t)overline{f'(t)}dt \
    &= int_{0}^{2pi}|f'(t)|^2dt.
    end{align}
    So, if $Lf=lambda f$, then $lambda|f|^2=|f'|^2$ forces $lambda$ to be real and non-negative. Furthermore, $lambda =0$ iff $f'=0$ or $f$ is constant.






    share|cite|improve this answer











    $endgroup$


















      2












      $begingroup$

      The associated form for $L$ is
      begin{align}
      langle Lf,frangle &= int_{0}^{2pi} -f''(t)overline{f(t)}dt\
      &= -f'overline{f}|_{0}^{2pi}+int_{0}^{2pi}f'(t)overline{f'(t)}dt \
      &= int_{0}^{2pi}|f'(t)|^2dt.
      end{align}
      So, if $Lf=lambda f$, then $lambda|f|^2=|f'|^2$ forces $lambda$ to be real and non-negative. Furthermore, $lambda =0$ iff $f'=0$ or $f$ is constant.






      share|cite|improve this answer











      $endgroup$
















        2












        2








        2





        $begingroup$

        The associated form for $L$ is
        begin{align}
        langle Lf,frangle &= int_{0}^{2pi} -f''(t)overline{f(t)}dt\
        &= -f'overline{f}|_{0}^{2pi}+int_{0}^{2pi}f'(t)overline{f'(t)}dt \
        &= int_{0}^{2pi}|f'(t)|^2dt.
        end{align}
        So, if $Lf=lambda f$, then $lambda|f|^2=|f'|^2$ forces $lambda$ to be real and non-negative. Furthermore, $lambda =0$ iff $f'=0$ or $f$ is constant.






        share|cite|improve this answer











        $endgroup$



        The associated form for $L$ is
        begin{align}
        langle Lf,frangle &= int_{0}^{2pi} -f''(t)overline{f(t)}dt\
        &= -f'overline{f}|_{0}^{2pi}+int_{0}^{2pi}f'(t)overline{f'(t)}dt \
        &= int_{0}^{2pi}|f'(t)|^2dt.
        end{align}
        So, if $Lf=lambda f$, then $lambda|f|^2=|f'|^2$ forces $lambda$ to be real and non-negative. Furthermore, $lambda =0$ iff $f'=0$ or $f$ is constant.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 13 '18 at 4:58

























        answered Dec 13 '18 at 4:45









        DisintegratingByPartsDisintegratingByParts

        59.4k42580




        59.4k42580






























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