$x_i frac{partial f(x)}{partial x_i} = f(x)g_i(x)$ for all $i$
$begingroup$
What can we say about the function $f:mathbb{R}^k rightarrow mathbb{R}$ if for all $i=1,...,k$ and all $x in mathbb{R}^k$ we have
$x_ifrac{partial f(x)}{partial x_i} = f(x)g_i(x)$
Is there actually any restriction we can make other than that $f$ is differentiable?
ordinary-differential-equations functional-equations
edited Dec 5 '18 at 7:36
Martin Rosenau
1,156139
asked Dec 5 '18 at 7:21
HRSEHRSE
233110
$endgroup$
add a comment |
$begingroup$
What can we say about the function $f:mathbb{R}^k rightarrow mathbb{R}$ if for all $i=1,...,k$ and all $x in mathbb{R}^k$ we have
$x_ifrac{partial f(x)}{partial x_i} = f(x)g_i(x)$
Is there actually any restriction we can make other than that $f$ is differentiable?
ordinary-differential-equations functional-equations
edited Dec 5 '18 at 7:36
Martin Rosenau
1,156139
asked Dec 5 '18 at 7:21
HRSEHRSE
233110
$endgroup$
$begingroup$
$f(x)=0$ implies $x_ifrac {partial f (x)} {partial x_i}=0$.
$endgroup$
– Kavi Rama Murthy
Dec 5 '18 at 7:31
add a comment |
$begingroup$
What can we say about the function $f:mathbb{R}^k rightarrow mathbb{R}$ if for all $i=1,...,k$ and all $x in mathbb{R}^k$ we have
$x_ifrac{partial f(x)}{partial x_i} = f(x)g_i(x)$
Is there actually any restriction we can make other than that $f$ is differentiable?
ordinary-differential-equations functional-equations
edited Dec 5 '18 at 7:36
Martin Rosenau
1,156139
asked Dec 5 '18 at 7:21
HRSEHRSE
233110
$endgroup$
What can we say about the function $f:mathbb{R}^k rightarrow mathbb{R}$ if for all $i=1,...,k$ and all $x in mathbb{R}^k$ we have
$x_ifrac{partial f(x)}{partial x_i} = f(x)g_i(x)$
Is there actually any restriction we can make other than that $f$ is differentiable?
ordinary-differential-equations functional-equations
ordinary-differential-equations functional-equations
edited Dec 5 '18 at 7:36
Martin Rosenau
1,156139
asked Dec 5 '18 at 7:21
HRSEHRSE
233110
edited Dec 5 '18 at 7:36
Martin Rosenau
1,156139
asked Dec 5 '18 at 7:21
HRSEHRSE
233110
edited Dec 5 '18 at 7:36
Martin Rosenau
1,156139
edited Dec 5 '18 at 7:36
Martin Rosenau
1,156139
edited Dec 5 '18 at 7:36
Martin Rosenau
1,156139
1,156139
asked Dec 5 '18 at 7:21
HRSEHRSE
233110
asked Dec 5 '18 at 7:21
HRSEHRSE
233110
asked Dec 5 '18 at 7:21
HRSEHRSE
233110
233110
$begingroup$
$f(x)=0$ implies $x_ifrac {partial f (x)} {partial x_i}=0$.
$endgroup$
– Kavi Rama Murthy
Dec 5 '18 at 7:31
add a comment |
$begingroup$
$f(x)=0$ implies $x_ifrac {partial f (x)} {partial x_i}=0$.
$endgroup$
– Kavi Rama Murthy
Dec 5 '18 at 7:31
$begingroup$
$f(x)=0$ implies $x_ifrac {partial f (x)} {partial x_i}=0$.
$endgroup$
– Kavi Rama Murthy
Dec 5 '18 at 7:31
$begingroup$
$f(x)=0$ implies $x_ifrac {partial f (x)} {partial x_i}=0$.
$endgroup$
– Kavi Rama Murthy
Dec 5 '18 at 7:31
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Consider $phi=ln|f|$, then $$frac{∂phi(x)}{∂x_i}=frac{g_i(x)}{x_i}.$$ By the Schwarz lemma, the mixed second derivatives do not depend on the order of derivation, thus
$$
frac1{x_i}frac{∂g_i(x)}{∂x_j}=frac1{x_j}frac{∂g_j(x)}{∂x_i}
$$
is a necessary condition for $f$ to exist.
answered Dec 5 '18 at 10:20
LutzLLutzL
56.9k42055
$endgroup$
add a comment |
$begingroup$
If $g_i(x)=-1$ then you have that
$frac{partial}{partial x_i}(x_i f(x))=0$
So a solution is
$f(x)=frac{1}{prod_{k=1}^n x_k}$
answered Dec 5 '18 at 7:52
Federico FalluccaFederico Fallucca
1,85018
$endgroup$
add a comment |
$begingroup$
We can say nearly nothing. Let $f(x)$ never equals to zero. Then we can take $g_i(x)=frac{x_ifrac{partial f(x)}{partial x_i}}{f(x)}$
answered Dec 5 '18 at 9:10
MinzMinz
951127
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Consider $phi=ln|f|$, then $$frac{∂phi(x)}{∂x_i}=frac{g_i(x)}{x_i}.$$ By the Schwarz lemma, the mixed second derivatives do not depend on the order of derivation, thus
$$
frac1{x_i}frac{∂g_i(x)}{∂x_j}=frac1{x_j}frac{∂g_j(x)}{∂x_i}
$$
is a necessary condition for $f$ to exist.
answered Dec 5 '18 at 10:20
LutzLLutzL
56.9k42055
$endgroup$
add a comment |
$begingroup$
Consider $phi=ln|f|$, then $$frac{∂phi(x)}{∂x_i}=frac{g_i(x)}{x_i}.$$ By the Schwarz lemma, the mixed second derivatives do not depend on the order of derivation, thus
$$
frac1{x_i}frac{∂g_i(x)}{∂x_j}=frac1{x_j}frac{∂g_j(x)}{∂x_i}
$$
is a necessary condition for $f$ to exist.
answered Dec 5 '18 at 10:20
LutzLLutzL
56.9k42055
$endgroup$
add a comment |
$begingroup$
Consider $phi=ln|f|$, then $$frac{∂phi(x)}{∂x_i}=frac{g_i(x)}{x_i}.$$ By the Schwarz lemma, the mixed second derivatives do not depend on the order of derivation, thus
$$
frac1{x_i}frac{∂g_i(x)}{∂x_j}=frac1{x_j}frac{∂g_j(x)}{∂x_i}
$$
is a necessary condition for $f$ to exist.
answered Dec 5 '18 at 10:20
LutzLLutzL
56.9k42055
$endgroup$
Consider $phi=ln|f|$, then $$frac{∂phi(x)}{∂x_i}=frac{g_i(x)}{x_i}.$$ By the Schwarz lemma, the mixed second derivatives do not depend on the order of derivation, thus
$$
frac1{x_i}frac{∂g_i(x)}{∂x_j}=frac1{x_j}frac{∂g_j(x)}{∂x_i}
$$
is a necessary condition for $f$ to exist.
answered Dec 5 '18 at 10:20
LutzLLutzL
56.9k42055
answered Dec 5 '18 at 10:20
LutzLLutzL
56.9k42055
answered Dec 5 '18 at 10:20
LutzLLutzL
56.9k42055
answered Dec 5 '18 at 10:20
LutzLLutzL
56.9k42055
56.9k42055
add a comment |
add a comment |
$begingroup$
If $g_i(x)=-1$ then you have that
$frac{partial}{partial x_i}(x_i f(x))=0$
So a solution is
$f(x)=frac{1}{prod_{k=1}^n x_k}$
answered Dec 5 '18 at 7:52
Federico FalluccaFederico Fallucca
1,85018
$endgroup$
add a comment |
$begingroup$
If $g_i(x)=-1$ then you have that
$frac{partial}{partial x_i}(x_i f(x))=0$
So a solution is
$f(x)=frac{1}{prod_{k=1}^n x_k}$
answered Dec 5 '18 at 7:52
Federico FalluccaFederico Fallucca
1,85018
$endgroup$
add a comment |
$begingroup$
If $g_i(x)=-1$ then you have that
$frac{partial}{partial x_i}(x_i f(x))=0$
So a solution is
$f(x)=frac{1}{prod_{k=1}^n x_k}$
answered Dec 5 '18 at 7:52
Federico FalluccaFederico Fallucca
1,85018
$endgroup$
If $g_i(x)=-1$ then you have that
$frac{partial}{partial x_i}(x_i f(x))=0$
So a solution is
$f(x)=frac{1}{prod_{k=1}^n x_k}$
answered Dec 5 '18 at 7:52
Federico FalluccaFederico Fallucca
1,85018
edited Dec 5 '18 at 7:59
answered Dec 5 '18 at 7:52
Federico FalluccaFederico Fallucca
1,85018
answered Dec 5 '18 at 7:52
Federico FalluccaFederico Fallucca
1,85018
answered Dec 5 '18 at 7:52
Federico FalluccaFederico Fallucca
1,85018
1,85018
add a comment |
add a comment |
$begingroup$
We can say nearly nothing. Let $f(x)$ never equals to zero. Then we can take $g_i(x)=frac{x_ifrac{partial f(x)}{partial x_i}}{f(x)}$
answered Dec 5 '18 at 9:10
MinzMinz
951127
$endgroup$
add a comment |
$begingroup$
We can say nearly nothing. Let $f(x)$ never equals to zero. Then we can take $g_i(x)=frac{x_ifrac{partial f(x)}{partial x_i}}{f(x)}$
answered Dec 5 '18 at 9:10
MinzMinz
951127
$endgroup$
add a comment |
$begingroup$
We can say nearly nothing. Let $f(x)$ never equals to zero. Then we can take $g_i(x)=frac{x_ifrac{partial f(x)}{partial x_i}}{f(x)}$
answered Dec 5 '18 at 9:10
MinzMinz
951127
$endgroup$
We can say nearly nothing. Let $f(x)$ never equals to zero. Then we can take $g_i(x)=frac{x_ifrac{partial f(x)}{partial x_i}}{f(x)}$
answered Dec 5 '18 at 9:10
MinzMinz
951127
answered Dec 5 '18 at 9:10
MinzMinz
951127
answered Dec 5 '18 at 9:10
MinzMinz
951127
answered Dec 5 '18 at 9:10
MinzMinz
951127
951127
add a comment |
add a comment |
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$begingroup$
$f(x)=0$ implies $x_ifrac {partial f (x)} {partial x_i}=0$.
$endgroup$
– Kavi Rama Murthy
Dec 5 '18 at 7:31