Proving that $(-infty,0]subseteq sigma(Delta)$

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Consider $w,gin L^2(mathbbR^n)$ such that $lambda w-Delta w=g$. Show that for all $lambda>0, lambdain rho(Delta)$. That is, $sigma(Delta)=(-infty,0]$. Where $sigma(Delta)$ is the spectrum of $Delta$.



In my answer, I have used Fourier transform by as in the following:



$hatlambda w-hatDelta w=hatg$ which led me to $hatw=hatg/(lambda+left|xiright|^2)$. Thus for $g in L^2$, $(lambda I - Delta)^-1 g = left((lambda +|xi|^2)^-1 hatgright)^vee$



I followed the solution on this link:



[What is spectrum for Laplacian in $mathbbR^n$?



From which I was able to prove that $sigma(Delta)subseteq(-infty,0]$.



However, I have found it hard to elementarily show that $(-infty,0]subseteq sigma(Delta)$, in order to establish the equality. Even the proof on the link is complicated.




functional-analysis pde fourier-analysis




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  • It is solved here: math.stackexchange.com/questions/790401/…
    – Hugocito
    Jul 16 at 20:58














up vote
1
down vote

favorite












Consider $w,gin L^2(mathbbR^n)$ such that $lambda w-Delta w=g$. Show that for all $lambda>0, lambdain rho(Delta)$. That is, $sigma(Delta)=(-infty,0]$. Where $sigma(Delta)$ is the spectrum of $Delta$.



In my answer, I have used Fourier transform by as in the following:



$hatlambda w-hatDelta w=hatg$ which led me to $hatw=hatg/(lambda+left|xiright|^2)$. Thus for $g in L^2$, $(lambda I - Delta)^-1 g = left((lambda +|xi|^2)^-1 hatgright)^vee$



I followed the solution on this link:



[What is spectrum for Laplacian in $mathbbR^n$?



From which I was able to prove that $sigma(Delta)subseteq(-infty,0]$.



However, I have found it hard to elementarily show that $(-infty,0]subseteq sigma(Delta)$, in order to establish the equality. Even the proof on the link is complicated.




functional-analysis pde fourier-analysis




share|cite|improve this question



















  • It is solved here: math.stackexchange.com/questions/790401/…
    – Hugocito
    Jul 16 at 20:58












up vote
1
down vote

favorite









up vote
1
down vote

favorite











Consider $w,gin L^2(mathbbR^n)$ such that $lambda w-Delta w=g$. Show that for all $lambda>0, lambdain rho(Delta)$. That is, $sigma(Delta)=(-infty,0]$. Where $sigma(Delta)$ is the spectrum of $Delta$.



In my answer, I have used Fourier transform by as in the following:



$hatlambda w-hatDelta w=hatg$ which led me to $hatw=hatg/(lambda+left|xiright|^2)$. Thus for $g in L^2$, $(lambda I - Delta)^-1 g = left((lambda +|xi|^2)^-1 hatgright)^vee$



I followed the solution on this link:



[What is spectrum for Laplacian in $mathbbR^n$?



From which I was able to prove that $sigma(Delta)subseteq(-infty,0]$.



However, I have found it hard to elementarily show that $(-infty,0]subseteq sigma(Delta)$, in order to establish the equality. Even the proof on the link is complicated.




functional-analysis pde fourier-analysis




share|cite|improve this question











Consider $w,gin L^2(mathbbR^n)$ such that $lambda w-Delta w=g$. Show that for all $lambda>0, lambdain rho(Delta)$. That is, $sigma(Delta)=(-infty,0]$. Where $sigma(Delta)$ is the spectrum of $Delta$.



In my answer, I have used Fourier transform by as in the following:



$hatlambda w-hatDelta w=hatg$ which led me to $hatw=hatg/(lambda+left|xiright|^2)$. Thus for $g in L^2$, $(lambda I - Delta)^-1 g = left((lambda +|xi|^2)^-1 hatgright)^vee$



I followed the solution on this link:



[What is spectrum for Laplacian in $mathbbR^n$?



From which I was able to prove that $sigma(Delta)subseteq(-infty,0]$.



However, I have found it hard to elementarily show that $(-infty,0]subseteq sigma(Delta)$, in order to establish the equality. Even the proof on the link is complicated.





functional-analysis pde fourier-analysis





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asked Jul 16 at 20:39









Sulayman

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  • It is solved here: math.stackexchange.com/questions/790401/…
    – Hugocito
    Jul 16 at 20:58
















  • It is solved here: math.stackexchange.com/questions/790401/…
    – Hugocito
    Jul 16 at 20:58















It is solved here: math.stackexchange.com/questions/790401/…
– Hugocito
Jul 16 at 20:58




It is solved here: math.stackexchange.com/questions/790401/…
– Hugocito
Jul 16 at 20:58










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The resolvent of $Delta$ is $(Delta-lambda I)^-1$, which is defined at least for $lambdanotinmathbbR$, and can be found with the Fourier transform to be
$$
(Delta-lambda I)^-1f = frac1(2pi)^n/2intfrac1-hatf(xi)dxi,;;; lambdainmathbbCsetminusmathbbR.
$$
From this you can also see that resolvent set includes $mathbbCsetminus(-infty,0]$. So $sigma(Delta)subseteq(-infty,0]$.



To show that $(-infty,0]subseteqsigma(Delta)$, let $r > 0$. For any $0 < epsilon < r$, let $f_r,epsilon$ be defined so that
$$
hatf_r,epsilon(xi) = chi_[r-epsilon,r+epsilon](|xi|).
$$
Then $|f_r,epsilon|=|hatf_r,epsilon|=|chi_[r-epsilon,r+epsilon]|$, and
$$
|(Delta+rI)f_r,epsilon|=|(|xi|-r)chi_[r-epsilon,r+epsilon](|xi|)| le epsilon|chi_[r-epsilon,r+epsilon]|=epsilon|f_r,epsilon|.
$$
So $(Delta +rI)$ cannot have a bounded inverse for any $r > 0$. So $(-infty,0]=sigma(Delta)$.






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  • @Sulyman : I'm curious why you unchecked my answer.
    – DisintegratingByParts
    Jul 23 at 23:27










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The resolvent of $Delta$ is $(Delta-lambda I)^-1$, which is defined at least for $lambdanotinmathbbR$, and can be found with the Fourier transform to be
$$
(Delta-lambda I)^-1f = frac1(2pi)^n/2intfrac1-hatf(xi)dxi,;;; lambdainmathbbCsetminusmathbbR.
$$
From this you can also see that resolvent set includes $mathbbCsetminus(-infty,0]$. So $sigma(Delta)subseteq(-infty,0]$.



To show that $(-infty,0]subseteqsigma(Delta)$, let $r > 0$. For any $0 < epsilon < r$, let $f_r,epsilon$ be defined so that
$$
hatf_r,epsilon(xi) = chi_[r-epsilon,r+epsilon](|xi|).
$$
Then $|f_r,epsilon|=|hatf_r,epsilon|=|chi_[r-epsilon,r+epsilon]|$, and
$$
|(Delta+rI)f_r,epsilon|=|(|xi|-r)chi_[r-epsilon,r+epsilon](|xi|)| le epsilon|chi_[r-epsilon,r+epsilon]|=epsilon|f_r,epsilon|.
$$
So $(Delta +rI)$ cannot have a bounded inverse for any $r > 0$. So $(-infty,0]=sigma(Delta)$.






share|cite|improve this answer























  • @Sulyman : I'm curious why you unchecked my answer.
    – DisintegratingByParts
    Jul 23 at 23:27














up vote
0
down vote













The resolvent of $Delta$ is $(Delta-lambda I)^-1$, which is defined at least for $lambdanotinmathbbR$, and can be found with the Fourier transform to be
$$
(Delta-lambda I)^-1f = frac1(2pi)^n/2intfrac1-hatf(xi)dxi,;;; lambdainmathbbCsetminusmathbbR.
$$
From this you can also see that resolvent set includes $mathbbCsetminus(-infty,0]$. So $sigma(Delta)subseteq(-infty,0]$.



To show that $(-infty,0]subseteqsigma(Delta)$, let $r > 0$. For any $0 < epsilon < r$, let $f_r,epsilon$ be defined so that
$$
hatf_r,epsilon(xi) = chi_[r-epsilon,r+epsilon](|xi|).
$$
Then $|f_r,epsilon|=|hatf_r,epsilon|=|chi_[r-epsilon,r+epsilon]|$, and
$$
|(Delta+rI)f_r,epsilon|=|(|xi|-r)chi_[r-epsilon,r+epsilon](|xi|)| le epsilon|chi_[r-epsilon,r+epsilon]|=epsilon|f_r,epsilon|.
$$
So $(Delta +rI)$ cannot have a bounded inverse for any $r > 0$. So $(-infty,0]=sigma(Delta)$.






share|cite|improve this answer























  • @Sulyman : I'm curious why you unchecked my answer.
    – DisintegratingByParts
    Jul 23 at 23:27












up vote
0
down vote










up vote
0
down vote









The resolvent of $Delta$ is $(Delta-lambda I)^-1$, which is defined at least for $lambdanotinmathbbR$, and can be found with the Fourier transform to be
$$
(Delta-lambda I)^-1f = frac1(2pi)^n/2intfrac1-hatf(xi)dxi,;;; lambdainmathbbCsetminusmathbbR.
$$
From this you can also see that resolvent set includes $mathbbCsetminus(-infty,0]$. So $sigma(Delta)subseteq(-infty,0]$.



To show that $(-infty,0]subseteqsigma(Delta)$, let $r > 0$. For any $0 < epsilon < r$, let $f_r,epsilon$ be defined so that
$$
hatf_r,epsilon(xi) = chi_[r-epsilon,r+epsilon](|xi|).
$$
Then $|f_r,epsilon|=|hatf_r,epsilon|=|chi_[r-epsilon,r+epsilon]|$, and
$$
|(Delta+rI)f_r,epsilon|=|(|xi|-r)chi_[r-epsilon,r+epsilon](|xi|)| le epsilon|chi_[r-epsilon,r+epsilon]|=epsilon|f_r,epsilon|.
$$
So $(Delta +rI)$ cannot have a bounded inverse for any $r > 0$. So $(-infty,0]=sigma(Delta)$.






share|cite|improve this answer















The resolvent of $Delta$ is $(Delta-lambda I)^-1$, which is defined at least for $lambdanotinmathbbR$, and can be found with the Fourier transform to be
$$
(Delta-lambda I)^-1f = frac1(2pi)^n/2intfrac1-hatf(xi)dxi,;;; lambdainmathbbCsetminusmathbbR.
$$
From this you can also see that resolvent set includes $mathbbCsetminus(-infty,0]$. So $sigma(Delta)subseteq(-infty,0]$.



To show that $(-infty,0]subseteqsigma(Delta)$, let $r > 0$. For any $0 < epsilon < r$, let $f_r,epsilon$ be defined so that
$$
hatf_r,epsilon(xi) = chi_[r-epsilon,r+epsilon](|xi|).
$$
Then $|f_r,epsilon|=|hatf_r,epsilon|=|chi_[r-epsilon,r+epsilon]|$, and
$$
|(Delta+rI)f_r,epsilon|=|(|xi|-r)chi_[r-epsilon,r+epsilon](|xi|)| le epsilon|chi_[r-epsilon,r+epsilon]|=epsilon|f_r,epsilon|.
$$
So $(Delta +rI)$ cannot have a bounded inverse for any $r > 0$. So $(-infty,0]=sigma(Delta)$.







share|cite|improve this answer















share|cite|improve this answer



share|cite|improve this answer








edited Jul 23 at 23:25


























answered Jul 16 at 23:25









DisintegratingByParts

55.7k42273




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  • @Sulyman : I'm curious why you unchecked my answer.
    – DisintegratingByParts
    Jul 23 at 23:27
















  • @Sulyman : I'm curious why you unchecked my answer.
    – DisintegratingByParts
    Jul 23 at 23:27















@Sulyman : I'm curious why you unchecked my answer.
– DisintegratingByParts
Jul 23 at 23:27




@Sulyman : I'm curious why you unchecked my answer.
– DisintegratingByParts
Jul 23 at 23:27












 

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