On the proof that a distribution with ${0}$ as support can be written as a sum of point masses
$begingroup$
We have the following Theorem and its accompanying proof from pages 46-47 of Hormander's The Analysis of Linear Partial Differential Operators.
Theorem 2.3.4 If $F$ is a distribution of order $k$ with support equal to ${0}$, then for $phi in C^k$, $F$ is of the form:
$$ F(phi) = sum_{|alpha| leq k}c_{alpha}partial^{alpha}phi(0) $$
Proof: Expanding $phi$ in a taylor series gives us:
$$phi(x) = sum_{|alpha| leq k}frac{partial^{alpha}phi(0)(x)^{alpha}}{alpha!} + psi(x) $$
We have that $partial^{alpha}psi(0) = 0$ when $|alpha| leq k$, so $F(psi) = 0$ by theorem 2.3.3. Hence, the result follows with $c_{alpha} = F(frac{(x)^{alpha}}{alpha!})$ $blacksquare$
As a note, theorem 2.3.3 mentioned in the proof is the statement that if all partials of a $C^k$ function vanish on a point in the support of $F$, then $F$ acting on that function is $0$.
I have a few questions about this proof.
1) It seems as though $psi$ is the remainder term of the taylor expansion. Is this true?
2) Why does it follow that the partials of $psi$ vanish at 0? Does this have to do with Taylor's theorem? It is not immediately clear.
3) Why does theorem 2.3.3 not apply to the function $frac{x^{alpha}}{alpha!}$? It seems to me that all partials of this function evaluated at 0, should be 0. So by theorem 2.3.3, $F(frac{x^{alpha}}{alpha!})$ should equal 0.
real-analysis distribution-theory
asked Jan 7 at 6:01
Nicholas RobertsNicholas Roberts
113112
$endgroup$
add a comment |
$begingroup$
We have the following Theorem and its accompanying proof from pages 46-47 of Hormander's The Analysis of Linear Partial Differential Operators.
Theorem 2.3.4 If $F$ is a distribution of order $k$ with support equal to ${0}$, then for $phi in C^k$, $F$ is of the form:
$$ F(phi) = sum_{|alpha| leq k}c_{alpha}partial^{alpha}phi(0) $$
Proof: Expanding $phi$ in a taylor series gives us:
$$phi(x) = sum_{|alpha| leq k}frac{partial^{alpha}phi(0)(x)^{alpha}}{alpha!} + psi(x) $$
We have that $partial^{alpha}psi(0) = 0$ when $|alpha| leq k$, so $F(psi) = 0$ by theorem 2.3.3. Hence, the result follows with $c_{alpha} = F(frac{(x)^{alpha}}{alpha!})$ $blacksquare$
As a note, theorem 2.3.3 mentioned in the proof is the statement that if all partials of a $C^k$ function vanish on a point in the support of $F$, then $F$ acting on that function is $0$.
I have a few questions about this proof.
1) It seems as though $psi$ is the remainder term of the taylor expansion. Is this true?
2) Why does it follow that the partials of $psi$ vanish at 0? Does this have to do with Taylor's theorem? It is not immediately clear.
3) Why does theorem 2.3.3 not apply to the function $frac{x^{alpha}}{alpha!}$? It seems to me that all partials of this function evaluated at 0, should be 0. So by theorem 2.3.3, $F(frac{x^{alpha}}{alpha!})$ should equal 0.
real-analysis distribution-theory
asked Jan 7 at 6:01
Nicholas RobertsNicholas Roberts
113112
$endgroup$
$begingroup$
If $phi(x)=x^n/n!$ then $partial^nphi(0)ne0$.
$endgroup$
– Did
Jan 7 at 7:12
$begingroup$
@Did Ah, I see. Can you provide any info for the second question I asked?
$endgroup$
– Nicholas Roberts
Jan 7 at 18:14
$begingroup$
Simply by differentiating $alpha$ times the identity $$ psi(x)=phi(x) - sum_{|beta| leq k}frac{partial^{beta}phi(0)(x)^{beta}}{beta!} $$
$endgroup$
– Did
Jan 7 at 18:16
$begingroup$
Thank you @Did I just have one more little question. Suppose that the distribution acted on $C^{infty}$ functions with compact support. In the context of this question, would the expression $F(frac{x^{alpha}}{alpha!})$ even make sense? Since the input does not have compact support.
$endgroup$
– Nicholas Roberts
Jan 7 at 18:48
$begingroup$
The answer is already in the question: if the distribution is only defined on functions with compact support then it cannot act on functions with noncompact support. Tautological.
$endgroup$
– Did
Jan 7 at 22:12
add a comment |
$begingroup$
We have the following Theorem and its accompanying proof from pages 46-47 of Hormander's The Analysis of Linear Partial Differential Operators.
Theorem 2.3.4 If $F$ is a distribution of order $k$ with support equal to ${0}$, then for $phi in C^k$, $F$ is of the form:
$$ F(phi) = sum_{|alpha| leq k}c_{alpha}partial^{alpha}phi(0) $$
Proof: Expanding $phi$ in a taylor series gives us:
$$phi(x) = sum_{|alpha| leq k}frac{partial^{alpha}phi(0)(x)^{alpha}}{alpha!} + psi(x) $$
We have that $partial^{alpha}psi(0) = 0$ when $|alpha| leq k$, so $F(psi) = 0$ by theorem 2.3.3. Hence, the result follows with $c_{alpha} = F(frac{(x)^{alpha}}{alpha!})$ $blacksquare$
As a note, theorem 2.3.3 mentioned in the proof is the statement that if all partials of a $C^k$ function vanish on a point in the support of $F$, then $F$ acting on that function is $0$.
I have a few questions about this proof.
1) It seems as though $psi$ is the remainder term of the taylor expansion. Is this true?
2) Why does it follow that the partials of $psi$ vanish at 0? Does this have to do with Taylor's theorem? It is not immediately clear.
3) Why does theorem 2.3.3 not apply to the function $frac{x^{alpha}}{alpha!}$? It seems to me that all partials of this function evaluated at 0, should be 0. So by theorem 2.3.3, $F(frac{x^{alpha}}{alpha!})$ should equal 0.
real-analysis distribution-theory
asked Jan 7 at 6:01
Nicholas RobertsNicholas Roberts
113112
$endgroup$
We have the following Theorem and its accompanying proof from pages 46-47 of Hormander's The Analysis of Linear Partial Differential Operators.
Theorem 2.3.4 If $F$ is a distribution of order $k$ with support equal to ${0}$, then for $phi in C^k$, $F$ is of the form:
$$ F(phi) = sum_{|alpha| leq k}c_{alpha}partial^{alpha}phi(0) $$
Proof: Expanding $phi$ in a taylor series gives us:
$$phi(x) = sum_{|alpha| leq k}frac{partial^{alpha}phi(0)(x)^{alpha}}{alpha!} + psi(x) $$
We have that $partial^{alpha}psi(0) = 0$ when $|alpha| leq k$, so $F(psi) = 0$ by theorem 2.3.3. Hence, the result follows with $c_{alpha} = F(frac{(x)^{alpha}}{alpha!})$ $blacksquare$
As a note, theorem 2.3.3 mentioned in the proof is the statement that if all partials of a $C^k$ function vanish on a point in the support of $F$, then $F$ acting on that function is $0$.
I have a few questions about this proof.
1) It seems as though $psi$ is the remainder term of the taylor expansion. Is this true?
2) Why does it follow that the partials of $psi$ vanish at 0? Does this have to do with Taylor's theorem? It is not immediately clear.
3) Why does theorem 2.3.3 not apply to the function $frac{x^{alpha}}{alpha!}$? It seems to me that all partials of this function evaluated at 0, should be 0. So by theorem 2.3.3, $F(frac{x^{alpha}}{alpha!})$ should equal 0.
real-analysis distribution-theory
real-analysis distribution-theory
asked Jan 7 at 6:01
Nicholas RobertsNicholas Roberts
113112
asked Jan 7 at 6:01
Nicholas RobertsNicholas Roberts
113112
asked Jan 7 at 6:01
Nicholas RobertsNicholas Roberts
113112
asked Jan 7 at 6:01
Nicholas RobertsNicholas Roberts
113112
asked Jan 7 at 6:01
Nicholas RobertsNicholas Roberts
113112
113112
$begingroup$
If $phi(x)=x^n/n!$ then $partial^nphi(0)ne0$.
$endgroup$
– Did
Jan 7 at 7:12
$begingroup$
@Did Ah, I see. Can you provide any info for the second question I asked?
$endgroup$
– Nicholas Roberts
Jan 7 at 18:14
$begingroup$
Simply by differentiating $alpha$ times the identity $$ psi(x)=phi(x) - sum_{|beta| leq k}frac{partial^{beta}phi(0)(x)^{beta}}{beta!} $$
$endgroup$
– Did
Jan 7 at 18:16
$begingroup$
Thank you @Did I just have one more little question. Suppose that the distribution acted on $C^{infty}$ functions with compact support. In the context of this question, would the expression $F(frac{x^{alpha}}{alpha!})$ even make sense? Since the input does not have compact support.
$endgroup$
– Nicholas Roberts
Jan 7 at 18:48
$begingroup$
The answer is already in the question: if the distribution is only defined on functions with compact support then it cannot act on functions with noncompact support. Tautological.
$endgroup$
– Did
Jan 7 at 22:12
add a comment |
$begingroup$
If $phi(x)=x^n/n!$ then $partial^nphi(0)ne0$.
$endgroup$
– Did
Jan 7 at 7:12
$begingroup$
@Did Ah, I see. Can you provide any info for the second question I asked?
$endgroup$
– Nicholas Roberts
Jan 7 at 18:14
$begingroup$
Simply by differentiating $alpha$ times the identity $$ psi(x)=phi(x) - sum_{|beta| leq k}frac{partial^{beta}phi(0)(x)^{beta}}{beta!} $$
$endgroup$
– Did
Jan 7 at 18:16
$begingroup$
Thank you @Did I just have one more little question. Suppose that the distribution acted on $C^{infty}$ functions with compact support. In the context of this question, would the expression $F(frac{x^{alpha}}{alpha!})$ even make sense? Since the input does not have compact support.
$endgroup$
– Nicholas Roberts
Jan 7 at 18:48
$begingroup$
The answer is already in the question: if the distribution is only defined on functions with compact support then it cannot act on functions with noncompact support. Tautological.
$endgroup$
– Did
Jan 7 at 22:12
$begingroup$
If $phi(x)=x^n/n!$ then $partial^nphi(0)ne0$.
$endgroup$
– Did
Jan 7 at 7:12
$begingroup$
If $phi(x)=x^n/n!$ then $partial^nphi(0)ne0$.
$endgroup$
– Did
Jan 7 at 7:12
$begingroup$
@Did Ah, I see. Can you provide any info for the second question I asked?
$endgroup$
– Nicholas Roberts
Jan 7 at 18:14
$begingroup$
@Did Ah, I see. Can you provide any info for the second question I asked?
$endgroup$
– Nicholas Roberts
Jan 7 at 18:14
$begingroup$
Simply by differentiating $alpha$ times the identity $$ psi(x)=phi(x) - sum_{|beta| leq k}frac{partial^{beta}phi(0)(x)^{beta}}{beta!} $$
$endgroup$
– Did
Jan 7 at 18:16
$begingroup$
Simply by differentiating $alpha$ times the identity $$ psi(x)=phi(x) - sum_{|beta| leq k}frac{partial^{beta}phi(0)(x)^{beta}}{beta!} $$
$endgroup$
– Did
Jan 7 at 18:16
$begingroup$
Thank you @Did I just have one more little question. Suppose that the distribution acted on $C^{infty}$ functions with compact support. In the context of this question, would the expression $F(frac{x^{alpha}}{alpha!})$ even make sense? Since the input does not have compact support.
$endgroup$
– Nicholas Roberts
Jan 7 at 18:48
$begingroup$
Thank you @Did I just have one more little question. Suppose that the distribution acted on $C^{infty}$ functions with compact support. In the context of this question, would the expression $F(frac{x^{alpha}}{alpha!})$ even make sense? Since the input does not have compact support.
$endgroup$
– Nicholas Roberts
Jan 7 at 18:48
$begingroup$
The answer is already in the question: if the distribution is only defined on functions with compact support then it cannot act on functions with noncompact support. Tautological.
$endgroup$
– Did
Jan 7 at 22:12
$begingroup$
The answer is already in the question: if the distribution is only defined on functions with compact support then it cannot act on functions with noncompact support. Tautological.
$endgroup$
– Did
Jan 7 at 22:12
add a comment |
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$begingroup$
If $phi(x)=x^n/n!$ then $partial^nphi(0)ne0$.
$endgroup$
– Did
Jan 7 at 7:12
$begingroup$
@Did Ah, I see. Can you provide any info for the second question I asked?
$endgroup$
– Nicholas Roberts
Jan 7 at 18:14
$begingroup$
Simply by differentiating $alpha$ times the identity $$ psi(x)=phi(x) - sum_{|beta| leq k}frac{partial^{beta}phi(0)(x)^{beta}}{beta!} $$
$endgroup$
– Did
Jan 7 at 18:16
$begingroup$
Thank you @Did I just have one more little question. Suppose that the distribution acted on $C^{infty}$ functions with compact support. In the context of this question, would the expression $F(frac{x^{alpha}}{alpha!})$ even make sense? Since the input does not have compact support.
$endgroup$
– Nicholas Roberts
Jan 7 at 18:48
$begingroup$
The answer is already in the question: if the distribution is only defined on functions with compact support then it cannot act on functions with noncompact support. Tautological.
$endgroup$
– Did
Jan 7 at 22:12