Mixing
Which theorem is this? If $phi(x)$ is differentiable at $0$, then there exists $psi(x)$, continuous at $0$,...
$begingroup$
I'm self studying generalized functions, and it's been the third time or so that I come across the following statement :
If $phi(x)$ is differentiable at $0$, then there exists a function $psi(x)$ continuous at $0$ such that $phi(x) = phi(0) + xpsi(x)$.
Which theorem or result is used here? Is there a higher order version (i.e. in the case where $phi$ is twice differentiable or more).
At first I used to tell myself it's kind of like a taylor expansion, say :
$$phi(x) = phi(0)+xphi'(0) + x^2epsilon(x)$$
but then $phi'(0)+xepsilon(x)$ is not guaranteed to be defined at $0$; well, if it were, then we'd have the continuity at $0$ due to $lim_{x to 0} epsilon(x) = 0$.
real-analysis soft-question distribution-theory
edited Jan 27 at 23:08
Blue
49.2k870157
asked Jan 27 at 22:55
rapidracimrapidracim
1,7241419
$endgroup$
add a comment |
$begingroup$
I'm self studying generalized functions, and it's been the third time or so that I come across the following statement :
If $phi(x)$ is differentiable at $0$, then there exists a function $psi(x)$ continuous at $0$ such that $phi(x) = phi(0) + xpsi(x)$.
Which theorem or result is used here? Is there a higher order version (i.e. in the case where $phi$ is twice differentiable or more).
At first I used to tell myself it's kind of like a taylor expansion, say :
$$phi(x) = phi(0)+xphi'(0) + x^2epsilon(x)$$
but then $phi'(0)+xepsilon(x)$ is not guaranteed to be defined at $0$; well, if it were, then we'd have the continuity at $0$ due to $lim_{x to 0} epsilon(x) = 0$.
real-analysis soft-question distribution-theory
edited Jan 27 at 23:08
Blue
49.2k870157
asked Jan 27 at 22:55
rapidracimrapidracim
1,7241419
$endgroup$
$begingroup$
By generalized functions do you mean distributions?
$endgroup$
– N. S.
Jan 27 at 22:58
$begingroup$
@N.S. yes. the book I use is called 'Generalized Functions : Theory and Applications'
$endgroup$
– rapidracim
Jan 27 at 22:59
$begingroup$
$phi$ bounded (continuous) means $psi(x)=frac{phi(x)-phi(0)}{x}$ is bounded (continuous) for $x ne 0$ and $phi$ differentiable at $0$ means $psi(x)$ is continuous at $0$.
$endgroup$
– reuns
Jan 27 at 23:00
$begingroup$
I believe that's Caratheodory's formulation of differentiability( at 0). From my undergrad classes, I know its equivalent to being differentiable (at 0) but I can't find a source. There is some discussion here reddit.com/r/math/comments/60c9z7/…
$endgroup$
– Calvin Khor
Jan 27 at 23:07
$begingroup$
This has further details and references as well as sketches of e.g. how this is equivalent to a Frechet derivative in Banach spaces math.stackexchange.com/questions/2321000/…
$endgroup$
– Calvin Khor
Jan 28 at 11:43
add a comment |
$begingroup$
I'm self studying generalized functions, and it's been the third time or so that I come across the following statement :
If $phi(x)$ is differentiable at $0$, then there exists a function $psi(x)$ continuous at $0$ such that $phi(x) = phi(0) + xpsi(x)$.
Which theorem or result is used here? Is there a higher order version (i.e. in the case where $phi$ is twice differentiable or more).
At first I used to tell myself it's kind of like a taylor expansion, say :
$$phi(x) = phi(0)+xphi'(0) + x^2epsilon(x)$$
but then $phi'(0)+xepsilon(x)$ is not guaranteed to be defined at $0$; well, if it were, then we'd have the continuity at $0$ due to $lim_{x to 0} epsilon(x) = 0$.
real-analysis soft-question distribution-theory
edited Jan 27 at 23:08
Blue
49.2k870157
asked Jan 27 at 22:55
rapidracimrapidracim
1,7241419
$endgroup$
I'm self studying generalized functions, and it's been the third time or so that I come across the following statement :
If $phi(x)$ is differentiable at $0$, then there exists a function $psi(x)$ continuous at $0$ such that $phi(x) = phi(0) + xpsi(x)$.
Which theorem or result is used here? Is there a higher order version (i.e. in the case where $phi$ is twice differentiable or more).
At first I used to tell myself it's kind of like a taylor expansion, say :
$$phi(x) = phi(0)+xphi'(0) + x^2epsilon(x)$$
but then $phi'(0)+xepsilon(x)$ is not guaranteed to be defined at $0$; well, if it were, then we'd have the continuity at $0$ due to $lim_{x to 0} epsilon(x) = 0$.
real-analysis soft-question distribution-theory
real-analysis soft-question distribution-theory
edited Jan 27 at 23:08
Blue
49.2k870157
asked Jan 27 at 22:55
rapidracimrapidracim
1,7241419
edited Jan 27 at 23:08
Blue
49.2k870157
asked Jan 27 at 22:55
rapidracimrapidracim
1,7241419
edited Jan 27 at 23:08
Blue
49.2k870157
edited Jan 27 at 23:08
Blue
49.2k870157
edited Jan 27 at 23:08
Blue
49.2k870157
49.2k870157
asked Jan 27 at 22:55
rapidracimrapidracim
1,7241419
asked Jan 27 at 22:55
rapidracimrapidracim
1,7241419
asked Jan 27 at 22:55
rapidracimrapidracim
1,7241419
1,7241419
$begingroup$
By generalized functions do you mean distributions?
$endgroup$
– N. S.
Jan 27 at 22:58
$begingroup$
@N.S. yes. the book I use is called 'Generalized Functions : Theory and Applications'
$endgroup$
– rapidracim
Jan 27 at 22:59
$begingroup$
$phi$ bounded (continuous) means $psi(x)=frac{phi(x)-phi(0)}{x}$ is bounded (continuous) for $x ne 0$ and $phi$ differentiable at $0$ means $psi(x)$ is continuous at $0$.
$endgroup$
– reuns
Jan 27 at 23:00
$begingroup$
I believe that's Caratheodory's formulation of differentiability( at 0). From my undergrad classes, I know its equivalent to being differentiable (at 0) but I can't find a source. There is some discussion here reddit.com/r/math/comments/60c9z7/…
$endgroup$
– Calvin Khor
Jan 27 at 23:07
$begingroup$
This has further details and references as well as sketches of e.g. how this is equivalent to a Frechet derivative in Banach spaces math.stackexchange.com/questions/2321000/…
$endgroup$
– Calvin Khor
Jan 28 at 11:43
add a comment |
$begingroup$
By generalized functions do you mean distributions?
$endgroup$
– N. S.
Jan 27 at 22:58
$begingroup$
@N.S. yes. the book I use is called 'Generalized Functions : Theory and Applications'
$endgroup$
– rapidracim
Jan 27 at 22:59
$begingroup$
$phi$ bounded (continuous) means $psi(x)=frac{phi(x)-phi(0)}{x}$ is bounded (continuous) for $x ne 0$ and $phi$ differentiable at $0$ means $psi(x)$ is continuous at $0$.
$endgroup$
– reuns
Jan 27 at 23:00
$begingroup$
I believe that's Caratheodory's formulation of differentiability( at 0). From my undergrad classes, I know its equivalent to being differentiable (at 0) but I can't find a source. There is some discussion here reddit.com/r/math/comments/60c9z7/…
$endgroup$
– Calvin Khor
Jan 27 at 23:07
$begingroup$
This has further details and references as well as sketches of e.g. how this is equivalent to a Frechet derivative in Banach spaces math.stackexchange.com/questions/2321000/…
$endgroup$
– Calvin Khor
Jan 28 at 11:43
$begingroup$
By generalized functions do you mean distributions?
$endgroup$
– N. S.
Jan 27 at 22:58
$begingroup$
By generalized functions do you mean distributions?
$endgroup$
– N. S.
Jan 27 at 22:58
$begingroup$
@N.S. yes. the book I use is called 'Generalized Functions : Theory and Applications'
$endgroup$
– rapidracim
Jan 27 at 22:59
$begingroup$
@N.S. yes. the book I use is called 'Generalized Functions : Theory and Applications'
$endgroup$
– rapidracim
Jan 27 at 22:59
$begingroup$
$phi$ bounded (continuous) means $psi(x)=frac{phi(x)-phi(0)}{x}$ is bounded (continuous) for $x ne 0$ and $phi$ differentiable at $0$ means $psi(x)$ is continuous at $0$.
$endgroup$
– reuns
Jan 27 at 23:00
$begingroup$
$phi$ bounded (continuous) means $psi(x)=frac{phi(x)-phi(0)}{x}$ is bounded (continuous) for $x ne 0$ and $phi$ differentiable at $0$ means $psi(x)$ is continuous at $0$.
$endgroup$
– reuns
Jan 27 at 23:00
$begingroup$
I believe that's Caratheodory's formulation of differentiability( at 0). From my undergrad classes, I know its equivalent to being differentiable (at 0) but I can't find a source. There is some discussion here reddit.com/r/math/comments/60c9z7/…
$endgroup$
– Calvin Khor
Jan 27 at 23:07
$begingroup$
I believe that's Caratheodory's formulation of differentiability( at 0). From my undergrad classes, I know its equivalent to being differentiable (at 0) but I can't find a source. There is some discussion here reddit.com/r/math/comments/60c9z7/…
$endgroup$
– Calvin Khor
Jan 27 at 23:07
$begingroup$
This has further details and references as well as sketches of e.g. how this is equivalent to a Frechet derivative in Banach spaces math.stackexchange.com/questions/2321000/…
$endgroup$
– Calvin Khor
Jan 28 at 11:43
$begingroup$
This has further details and references as well as sketches of e.g. how this is equivalent to a Frechet derivative in Banach spaces math.stackexchange.com/questions/2321000/…
$endgroup$
– Calvin Khor
Jan 28 at 11:43
add a comment |
1 Answer
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$begingroup$
Hint
$$phi(x) = phi(0) + xpsi(x) Leftrightarrow \
psi(x)= frac{phi(x)-phi(0)}{x-0} forall x neq 0$$
$phi$ is differentiable at $x=0$ if and only if
$$lim_{x to 0} psi(x)= phi'(0)$$
In order to make $psi$ continuous at $x=0$ you need to define $psi(0)$ such that
$$lim_{x to 0} psi(x)= psi(0)$$
SOOO You know what $psi(x)$ should be for $x neq 0$ and what $psi(0)$ should be.
answered Jan 27 at 23:01
N. S.N. S.
105k7114210
$endgroup$
$begingroup$
oh I see, extension by continuity. so we could've also defined $psi$ for non-zero $x$'s the way I thought in the 1st place using taylor expansion and at $0$ we'd just put the limit of that same function and then we'd get the same $phi'(0)$. thanks.
$endgroup$
– rapidracim
Jan 27 at 23:06
add a comment |
Your Answer
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1 Answer
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1 Answer
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$begingroup$
Hint
$$phi(x) = phi(0) + xpsi(x) Leftrightarrow \
psi(x)= frac{phi(x)-phi(0)}{x-0} forall x neq 0$$
$phi$ is differentiable at $x=0$ if and only if
$$lim_{x to 0} psi(x)= phi'(0)$$
In order to make $psi$ continuous at $x=0$ you need to define $psi(0)$ such that
$$lim_{x to 0} psi(x)= psi(0)$$
SOOO You know what $psi(x)$ should be for $x neq 0$ and what $psi(0)$ should be.
answered Jan 27 at 23:01
N. S.N. S.
105k7114210
$endgroup$
$begingroup$
oh I see, extension by continuity. so we could've also defined $psi$ for non-zero $x$'s the way I thought in the 1st place using taylor expansion and at $0$ we'd just put the limit of that same function and then we'd get the same $phi'(0)$. thanks.
$endgroup$
– rapidracim
Jan 27 at 23:06
add a comment |
$begingroup$
Hint
$$phi(x) = phi(0) + xpsi(x) Leftrightarrow \
psi(x)= frac{phi(x)-phi(0)}{x-0} forall x neq 0$$
$phi$ is differentiable at $x=0$ if and only if
$$lim_{x to 0} psi(x)= phi'(0)$$
In order to make $psi$ continuous at $x=0$ you need to define $psi(0)$ such that
$$lim_{x to 0} psi(x)= psi(0)$$
SOOO You know what $psi(x)$ should be for $x neq 0$ and what $psi(0)$ should be.
answered Jan 27 at 23:01
N. S.N. S.
105k7114210
$endgroup$
$begingroup$
oh I see, extension by continuity. so we could've also defined $psi$ for non-zero $x$'s the way I thought in the 1st place using taylor expansion and at $0$ we'd just put the limit of that same function and then we'd get the same $phi'(0)$. thanks.
$endgroup$
– rapidracim
Jan 27 at 23:06
add a comment |
$begingroup$
Hint
$$phi(x) = phi(0) + xpsi(x) Leftrightarrow \
psi(x)= frac{phi(x)-phi(0)}{x-0} forall x neq 0$$
$phi$ is differentiable at $x=0$ if and only if
$$lim_{x to 0} psi(x)= phi'(0)$$
In order to make $psi$ continuous at $x=0$ you need to define $psi(0)$ such that
$$lim_{x to 0} psi(x)= psi(0)$$
SOOO You know what $psi(x)$ should be for $x neq 0$ and what $psi(0)$ should be.
answered Jan 27 at 23:01
N. S.N. S.
105k7114210
$endgroup$
Hint
$$phi(x) = phi(0) + xpsi(x) Leftrightarrow \
psi(x)= frac{phi(x)-phi(0)}{x-0} forall x neq 0$$
$phi$ is differentiable at $x=0$ if and only if
$$lim_{x to 0} psi(x)= phi'(0)$$
In order to make $psi$ continuous at $x=0$ you need to define $psi(0)$ such that
$$lim_{x to 0} psi(x)= psi(0)$$
SOOO You know what $psi(x)$ should be for $x neq 0$ and what $psi(0)$ should be.
answered Jan 27 at 23:01
N. S.N. S.
105k7114210
answered Jan 27 at 23:01
N. S.N. S.
105k7114210
answered Jan 27 at 23:01
N. S.N. S.
105k7114210
answered Jan 27 at 23:01
N. S.N. S.
105k7114210
105k7114210
$begingroup$
oh I see, extension by continuity. so we could've also defined $psi$ for non-zero $x$'s the way I thought in the 1st place using taylor expansion and at $0$ we'd just put the limit of that same function and then we'd get the same $phi'(0)$. thanks.
$endgroup$
– rapidracim
Jan 27 at 23:06
add a comment |
$begingroup$
oh I see, extension by continuity. so we could've also defined $psi$ for non-zero $x$'s the way I thought in the 1st place using taylor expansion and at $0$ we'd just put the limit of that same function and then we'd get the same $phi'(0)$. thanks.
$endgroup$
– rapidracim
Jan 27 at 23:06
$begingroup$
oh I see, extension by continuity. so we could've also defined $psi$ for non-zero $x$'s the way I thought in the 1st place using taylor expansion and at $0$ we'd just put the limit of that same function and then we'd get the same $phi'(0)$. thanks.
$endgroup$
– rapidracim
Jan 27 at 23:06
$begingroup$
oh I see, extension by continuity. so we could've also defined $psi$ for non-zero $x$'s the way I thought in the 1st place using taylor expansion and at $0$ we'd just put the limit of that same function and then we'd get the same $phi'(0)$. thanks.
$endgroup$
– rapidracim
Jan 27 at 23:06
add a comment |
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$begingroup$
By generalized functions do you mean distributions?
$endgroup$
– N. S.
Jan 27 at 22:58
$begingroup$
@N.S. yes. the book I use is called 'Generalized Functions : Theory and Applications'
$endgroup$
– rapidracim
Jan 27 at 22:59
$begingroup$
$phi$ bounded (continuous) means $psi(x)=frac{phi(x)-phi(0)}{x}$ is bounded (continuous) for $x ne 0$ and $phi$ differentiable at $0$ means $psi(x)$ is continuous at $0$.
$endgroup$
– reuns
Jan 27 at 23:00
$begingroup$
I believe that's Caratheodory's formulation of differentiability( at 0). From my undergrad classes, I know its equivalent to being differentiable (at 0) but I can't find a source. There is some discussion here reddit.com/r/math/comments/60c9z7/…
$endgroup$
– Calvin Khor
Jan 27 at 23:07
$begingroup$
This has further details and references as well as sketches of e.g. how this is equivalent to a Frechet derivative in Banach spaces math.stackexchange.com/questions/2321000/…
$endgroup$
– Calvin Khor
Jan 28 at 11:43