Mixing
Find all solutions $T$ of $x^{2006} T = 0$ in the space of tempered distributions, $mathcal{S}'(mathbb{R})$
$begingroup$
I'm modeling my solution after this answer to a similar question. This is as far as I've gotten:
Every $phi in mathcal{S}(mathbb{R})$ that vanishes at $0$ can be expressed as $phi(x) = x psi (x)$. Then, $Tphi = xT(psi) = 0$ by assumption.
Fix $chi in mathcal{S}(mathbb{R})$ such that $chi(0) = 1$. Let $Tchi = a$. Then, for any $phi in mathcal{S}(mathbb{R})$, $$Tphi = T(phi - phi(0) chi + phi(0) chi) = T(phi - phi(0) chi) + T(phi(0) chi).$$
This is where I've gotten stuf. I'm not sure how, in the linked solution, the answer reduces from $T(phi - phi(0) chi) + T(phi(0) chi)$ to $0 + a phi(0)$ (my primary confusion is $T(phi - phi(0) chi) = 0$) nor how to adapt that for my own question.
Any suggestions?
functional-analysis distribution-theory schwartz-space
asked Jan 4 at 5:41
kkckkc
1058
$endgroup$
add a comment |
$begingroup$
I'm modeling my solution after this answer to a similar question. This is as far as I've gotten:
Every $phi in mathcal{S}(mathbb{R})$ that vanishes at $0$ can be expressed as $phi(x) = x psi (x)$. Then, $Tphi = xT(psi) = 0$ by assumption.
Fix $chi in mathcal{S}(mathbb{R})$ such that $chi(0) = 1$. Let $Tchi = a$. Then, for any $phi in mathcal{S}(mathbb{R})$, $$Tphi = T(phi - phi(0) chi + phi(0) chi) = T(phi - phi(0) chi) + T(phi(0) chi).$$
This is where I've gotten stuf. I'm not sure how, in the linked solution, the answer reduces from $T(phi - phi(0) chi) + T(phi(0) chi)$ to $0 + a phi(0)$ (my primary confusion is $T(phi - phi(0) chi) = 0$) nor how to adapt that for my own question.
Any suggestions?
functional-analysis distribution-theory schwartz-space
asked Jan 4 at 5:41
kkckkc
1058
$endgroup$
1
$begingroup$
You know that the distributions with $xT=0$ are the multiples of the Dirac delta distribution?
$endgroup$
– Lord Shark the Unknown
Jan 4 at 5:45
$begingroup$
@LordSharktheUnknown I don't know that. Could you explain further or refer me to the appropriate resource? I'm fairly new to distributions
$endgroup$
– kkc
Jan 4 at 5:49
$begingroup$
You have $[phi-phi(0)chi](0) = 0 $ obviously, and $phi-phi(0)chiinmathcal{S}(mathbb{R})$. So by the first argument $T(phi-phi(0)chi)=0$.
$endgroup$
– Vobo
Jan 6 at 20:02
add a comment |
$begingroup$
I'm modeling my solution after this answer to a similar question. This is as far as I've gotten:
Every $phi in mathcal{S}(mathbb{R})$ that vanishes at $0$ can be expressed as $phi(x) = x psi (x)$. Then, $Tphi = xT(psi) = 0$ by assumption.
Fix $chi in mathcal{S}(mathbb{R})$ such that $chi(0) = 1$. Let $Tchi = a$. Then, for any $phi in mathcal{S}(mathbb{R})$, $$Tphi = T(phi - phi(0) chi + phi(0) chi) = T(phi - phi(0) chi) + T(phi(0) chi).$$
This is where I've gotten stuf. I'm not sure how, in the linked solution, the answer reduces from $T(phi - phi(0) chi) + T(phi(0) chi)$ to $0 + a phi(0)$ (my primary confusion is $T(phi - phi(0) chi) = 0$) nor how to adapt that for my own question.
Any suggestions?
functional-analysis distribution-theory schwartz-space
asked Jan 4 at 5:41
kkckkc
1058
$endgroup$
I'm modeling my solution after this answer to a similar question. This is as far as I've gotten:
Every $phi in mathcal{S}(mathbb{R})$ that vanishes at $0$ can be expressed as $phi(x) = x psi (x)$. Then, $Tphi = xT(psi) = 0$ by assumption.
Fix $chi in mathcal{S}(mathbb{R})$ such that $chi(0) = 1$. Let $Tchi = a$. Then, for any $phi in mathcal{S}(mathbb{R})$, $$Tphi = T(phi - phi(0) chi + phi(0) chi) = T(phi - phi(0) chi) + T(phi(0) chi).$$
This is where I've gotten stuf. I'm not sure how, in the linked solution, the answer reduces from $T(phi - phi(0) chi) + T(phi(0) chi)$ to $0 + a phi(0)$ (my primary confusion is $T(phi - phi(0) chi) = 0$) nor how to adapt that for my own question.
Any suggestions?
functional-analysis distribution-theory schwartz-space
functional-analysis distribution-theory schwartz-space
asked Jan 4 at 5:41
kkckkc
1058
asked Jan 4 at 5:41
kkckkc
1058
asked Jan 4 at 5:41
kkckkc
1058
asked Jan 4 at 5:41
kkckkc
1058
asked Jan 4 at 5:41
kkckkc
1058
1058
1
$begingroup$
You know that the distributions with $xT=0$ are the multiples of the Dirac delta distribution?
$endgroup$
– Lord Shark the Unknown
Jan 4 at 5:45
$begingroup$
@LordSharktheUnknown I don't know that. Could you explain further or refer me to the appropriate resource? I'm fairly new to distributions
$endgroup$
– kkc
Jan 4 at 5:49
$begingroup$
You have $[phi-phi(0)chi](0) = 0 $ obviously, and $phi-phi(0)chiinmathcal{S}(mathbb{R})$. So by the first argument $T(phi-phi(0)chi)=0$.
$endgroup$
– Vobo
Jan 6 at 20:02
add a comment |
1
$begingroup$
You know that the distributions with $xT=0$ are the multiples of the Dirac delta distribution?
$endgroup$
– Lord Shark the Unknown
Jan 4 at 5:45
$begingroup$
@LordSharktheUnknown I don't know that. Could you explain further or refer me to the appropriate resource? I'm fairly new to distributions
$endgroup$
– kkc
Jan 4 at 5:49
$begingroup$
You have $[phi-phi(0)chi](0) = 0 $ obviously, and $phi-phi(0)chiinmathcal{S}(mathbb{R})$. So by the first argument $T(phi-phi(0)chi)=0$.
$endgroup$
– Vobo
Jan 6 at 20:02
1
1
$begingroup$
You know that the distributions with $xT=0$ are the multiples of the Dirac delta distribution?
$endgroup$
– Lord Shark the Unknown
Jan 4 at 5:45
$begingroup$
You know that the distributions with $xT=0$ are the multiples of the Dirac delta distribution?
$endgroup$
– Lord Shark the Unknown
Jan 4 at 5:45
$begingroup$
@LordSharktheUnknown I don't know that. Could you explain further or refer me to the appropriate resource? I'm fairly new to distributions
$endgroup$
– kkc
Jan 4 at 5:49
$begingroup$
@LordSharktheUnknown I don't know that. Could you explain further or refer me to the appropriate resource? I'm fairly new to distributions
$endgroup$
– kkc
Jan 4 at 5:49
$begingroup$
You have $[phi-phi(0)chi](0) = 0 $ obviously, and $phi-phi(0)chiinmathcal{S}(mathbb{R})$. So by the first argument $T(phi-phi(0)chi)=0$.
$endgroup$
– Vobo
Jan 6 at 20:02
$begingroup$
You have $[phi-phi(0)chi](0) = 0 $ obviously, and $phi-phi(0)chiinmathcal{S}(mathbb{R})$. So by the first argument $T(phi-phi(0)chi)=0$.
$endgroup$
– Vobo
Jan 6 at 20:02
add a comment |
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$begingroup$
You know that the distributions with $xT=0$ are the multiples of the Dirac delta distribution?
$endgroup$
– Lord Shark the Unknown
Jan 4 at 5:45
$begingroup$
@LordSharktheUnknown I don't know that. Could you explain further or refer me to the appropriate resource? I'm fairly new to distributions
$endgroup$
– kkc
Jan 4 at 5:49
$begingroup$
You have $[phi-phi(0)chi](0) = 0 $ obviously, and $phi-phi(0)chiinmathcal{S}(mathbb{R})$. So by the first argument $T(phi-phi(0)chi)=0$.
$endgroup$
– Vobo
Jan 6 at 20:02