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If directional derivatives are bounded, then function is continuous
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Suppose we have $f:Sto mathbb R$ where $S$ is the open unit circle:
$S=lbrace (x,y)in mathbb R ^2 mid x^2 +y^2 <1rbrace $, and suppose
$vec v, vec w$ are two independent unit vectors.
Suppose $exists C>0,forall vec zin S,Bigl|frac partial fpartial v(z)Bigr|<C,Bigl|frac partial fpartial w(z)Bigr|<C$.
Show that $f$ is continuous in $S$.
Now, I know that if the partial derivatives exist and bounded, then $f$ is continous, but I'm having trouble expressing the partial derivatives in terms of the given directional derivatives. Of course $e_x=(1,0)$ can be expressed as a linear combination of $v,w$, but how can I formally apply it to $fracpartial fpartial x$?
multivariable-calculus partial-derivative
asked Jul 15 at 18:27
gbi1977
1247
add a comment |Â
up vote
1
down vote
favorite
Suppose we have $f:Sto mathbb R$ where $S$ is the open unit circle:
$S=lbrace (x,y)in mathbb R ^2 mid x^2 +y^2 <1rbrace $, and suppose
$vec v, vec w$ are two independent unit vectors.
Suppose $exists C>0,forall vec zin S,Bigl|frac partial fpartial v(z)Bigr|<C,Bigl|frac partial fpartial w(z)Bigr|<C$.
Show that $f$ is continuous in $S$.
Now, I know that if the partial derivatives exist and bounded, then $f$ is continous, but I'm having trouble expressing the partial derivatives in terms of the given directional derivatives. Of course $e_x=(1,0)$ can be expressed as a linear combination of $v,w$, but how can I formally apply it to $fracpartial fpartial x$?
multivariable-calculus partial-derivative
asked Jul 15 at 18:27
gbi1977
1247
Between every two points in $S$ there is a two segment path that joins them and the segments are parallel to the axes. Use this to estimate the difference in $f$.
– copper.hat
Jul 15 at 18:32
@copper.hat I'm not sure how to construct this two segment path, I could only come up with a one-segment path $phi(t)=tcdot(alpha vec w +beta vec v)+(1-t)cdot (lambda vec w +gamma vec v) $ for $tin [0,1]$, where $z_1 ,z_2 in S$ are represented by these two linear combinations.
– gbi1977
Jul 15 at 20:13
Draw a picture. Take two points in the disc. Join them with a Manhattan like path.
– copper.hat
Jul 15 at 20:15
@copper.hat oh okay I think I see what you mean. Then the idea is that each point on this L path can be represented as a linear comb. of $v,w$. So along the path, the directional derivative is of course bounded, depending on the $v$ or $w$ component, but since it applies to any point on a line pararllel to the axes - it applies to the partial derivatives themselves?
– gbi1977
Jul 15 at 20:34
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Suppose we have $f:Sto mathbb R$ where $S$ is the open unit circle:
$S=lbrace (x,y)in mathbb R ^2 mid x^2 +y^2 <1rbrace $, and suppose
$vec v, vec w$ are two independent unit vectors.
Suppose $exists C>0,forall vec zin S,Bigl|frac partial fpartial v(z)Bigr|<C,Bigl|frac partial fpartial w(z)Bigr|<C$.
Show that $f$ is continuous in $S$.
Now, I know that if the partial derivatives exist and bounded, then $f$ is continous, but I'm having trouble expressing the partial derivatives in terms of the given directional derivatives. Of course $e_x=(1,0)$ can be expressed as a linear combination of $v,w$, but how can I formally apply it to $fracpartial fpartial x$?
multivariable-calculus partial-derivative
asked Jul 15 at 18:27
gbi1977
1247
Suppose we have $f:Sto mathbb R$ where $S$ is the open unit circle:
$S=lbrace (x,y)in mathbb R ^2 mid x^2 +y^2 <1rbrace $, and suppose
$vec v, vec w$ are two independent unit vectors.
Suppose $exists C>0,forall vec zin S,Bigl|frac partial fpartial v(z)Bigr|<C,Bigl|frac partial fpartial w(z)Bigr|<C$.
Show that $f$ is continuous in $S$.
Now, I know that if the partial derivatives exist and bounded, then $f$ is continous, but I'm having trouble expressing the partial derivatives in terms of the given directional derivatives. Of course $e_x=(1,0)$ can be expressed as a linear combination of $v,w$, but how can I formally apply it to $fracpartial fpartial x$?
multivariable-calculus partial-derivative
asked Jul 15 at 18:27
gbi1977
1247
edited Jul 15 at 19:43
asked Jul 15 at 18:27
gbi1977
1247
asked Jul 15 at 18:27
gbi1977
1247
asked Jul 15 at 18:27
gbi1977
1247
1247
Between every two points in $S$ there is a two segment path that joins them and the segments are parallel to the axes. Use this to estimate the difference in $f$.
– copper.hat
Jul 15 at 18:32
@copper.hat I'm not sure how to construct this two segment path, I could only come up with a one-segment path $phi(t)=tcdot(alpha vec w +beta vec v)+(1-t)cdot (lambda vec w +gamma vec v) $ for $tin [0,1]$, where $z_1 ,z_2 in S$ are represented by these two linear combinations.
– gbi1977
Jul 15 at 20:13
Draw a picture. Take two points in the disc. Join them with a Manhattan like path.
– copper.hat
Jul 15 at 20:15
@copper.hat oh okay I think I see what you mean. Then the idea is that each point on this L path can be represented as a linear comb. of $v,w$. So along the path, the directional derivative is of course bounded, depending on the $v$ or $w$ component, but since it applies to any point on a line pararllel to the axes - it applies to the partial derivatives themselves?
– gbi1977
Jul 15 at 20:34
add a comment |Â
Between every two points in $S$ there is a two segment path that joins them and the segments are parallel to the axes. Use this to estimate the difference in $f$.
– copper.hat
Jul 15 at 18:32
@copper.hat I'm not sure how to construct this two segment path, I could only come up with a one-segment path $phi(t)=tcdot(alpha vec w +beta vec v)+(1-t)cdot (lambda vec w +gamma vec v) $ for $tin [0,1]$, where $z_1 ,z_2 in S$ are represented by these two linear combinations.
– gbi1977
Jul 15 at 20:13
Draw a picture. Take two points in the disc. Join them with a Manhattan like path.
– copper.hat
Jul 15 at 20:15
@copper.hat oh okay I think I see what you mean. Then the idea is that each point on this L path can be represented as a linear comb. of $v,w$. So along the path, the directional derivative is of course bounded, depending on the $v$ or $w$ component, but since it applies to any point on a line pararllel to the axes - it applies to the partial derivatives themselves?
– gbi1977
Jul 15 at 20:34
Between every two points in $S$ there is a two segment path that joins them and the segments are parallel to the axes. Use this to estimate the difference in $f$.
– copper.hat
Jul 15 at 18:32
Between every two points in $S$ there is a two segment path that joins them and the segments are parallel to the axes. Use this to estimate the difference in $f$.
– copper.hat
Jul 15 at 18:32
@copper.hat I'm not sure how to construct this two segment path, I could only come up with a one-segment path $phi(t)=tcdot(alpha vec w +beta vec v)+(1-t)cdot (lambda vec w +gamma vec v) $ for $tin [0,1]$, where $z_1 ,z_2 in S$ are represented by these two linear combinations.
– gbi1977
Jul 15 at 20:13
@copper.hat I'm not sure how to construct this two segment path, I could only come up with a one-segment path $phi(t)=tcdot(alpha vec w +beta vec v)+(1-t)cdot (lambda vec w +gamma vec v) $ for $tin [0,1]$, where $z_1 ,z_2 in S$ are represented by these two linear combinations.
– gbi1977
Jul 15 at 20:13
Draw a picture. Take two points in the disc. Join them with a Manhattan like path.
– copper.hat
Jul 15 at 20:15
Draw a picture. Take two points in the disc. Join them with a Manhattan like path.
– copper.hat
Jul 15 at 20:15
@copper.hat oh okay I think I see what you mean. Then the idea is that each point on this L path can be represented as a linear comb. of $v,w$. So along the path, the directional derivative is of course bounded, depending on the $v$ or $w$ component, but since it applies to any point on a line pararllel to the axes - it applies to the partial derivatives themselves?
– gbi1977
Jul 15 at 20:34
@copper.hat oh okay I think I see what you mean. Then the idea is that each point on this L path can be represented as a linear comb. of $v,w$. So along the path, the directional derivative is of course bounded, depending on the $v$ or $w$ component, but since it applies to any point on a line pararllel to the axes - it applies to the partial derivatives themselves?
– gbi1977
Jul 15 at 20:34
add a comment |Â
1 Answer
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Here is a more complete proof. However, I think the basic idea gets lost in
the details.
Let $A= beginbmatrixv & w endbmatrix$ and define $phi(z) = f(A z)$ on the open convex set $Sigma=A^-1 S$. Note that
$partial phi(z) over partial z_1 = partial f(Az) over partial u$
and $partial phi(z) over partial z_2 = partial f(Az) over partial w$.
The key element of the proof is that for any $z_1,z_2 in Sigma$, we can choose (because $Sigma$ is open & convex) a path $y_1=z_1 to y_2to cdots to y_n-1 to y_n = z_2$ such that each segment of the path is parallel to either axis and the direction of each segment is the same (with respect to the corresponding axis).
In particular, if $V_j(y) = [y]_j$ we have
$V_j(z_2-z_1) = sum_k V_j(y_k-y_k-1)$, for a given $j$, all of the $V_j(y_k-y_k-1)$ have the same sign and at most one of
$V_1(y_k-y_k-1) $ or $V_2(y_k-y_k-1)$ are
non zero. Hence $|V_j(z_2-z_1)| = sum_k |V_j(y_k-y_k-1)|$.
Note that on each segment, we can apply the mean value theorem to get
$|phi(y_k)-phi(y_k-1)| le C (|V_1(y_k-y_k-1) | + |V_2(y_k-y_k-1)| )$ and
so
begineqnarray
|phi(z_2)-phi(z_1)| &le& sum_k |phi(y_k)-phi(y_k-1)| \
&le & C sum_k(|V_1(y_k-y_k-1) | + |V_2(y_k-y_k-1)| ) \
&=& C(|V_1(z_2-z_1)| + |V_2(z_2-z_1)|) \
&=& C |z_2-z_1|_1
endeqnarray
Finally, we return to '$f$' space:
$|f(z_2)-f(z_1)| = |phi(A^-1 z_2) - phi(A^-1 z_1)| le C |A^-1(z_2-z_1)|_1 le C|A^-1|_1 |z_2-z_1|_1$
Hence $f$ is Lipschitz continuous with rank at most $C|A^-1|_1$.
answered Jul 15 at 21:30
copper.hat
122k557156
1
Doesn’t this argument assume that $vec v=langle1,0rangle$ and $vec w=langle0,1rangle$ (or vice versa)?
– Steve Kass
Jul 15 at 21:42
@SteveKass: It does, but exactly the same approach works except for the $sqrt 2$ part and the $C need to be adjusted.
– copper.hat
Jul 16 at 0:06
Yes, it just seemed to me that the details of how to make that approach work were at the heart of the OP’s question. (Or a little more bluntly, if I’d assigned this as a class exercise, I would expect those details to be worked out in a submitted answer.)
– Steve Kass
Jul 16 at 0:27
@SteveKass: I guess my take was that the OP just needed to see the idea and could take it from there. I will sketch out the details.
– copper.hat
Jul 16 at 0:50
@SteveKass: I elaborated the detail, but I think it has become a technical mess.
– copper.hat
Jul 16 at 2:28
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Here is a more complete proof. However, I think the basic idea gets lost in
the details.
Let $A= beginbmatrixv & w endbmatrix$ and define $phi(z) = f(A z)$ on the open convex set $Sigma=A^-1 S$. Note that
$partial phi(z) over partial z_1 = partial f(Az) over partial u$
and $partial phi(z) over partial z_2 = partial f(Az) over partial w$.
The key element of the proof is that for any $z_1,z_2 in Sigma$, we can choose (because $Sigma$ is open & convex) a path $y_1=z_1 to y_2to cdots to y_n-1 to y_n = z_2$ such that each segment of the path is parallel to either axis and the direction of each segment is the same (with respect to the corresponding axis).
In particular, if $V_j(y) = [y]_j$ we have
$V_j(z_2-z_1) = sum_k V_j(y_k-y_k-1)$, for a given $j$, all of the $V_j(y_k-y_k-1)$ have the same sign and at most one of
$V_1(y_k-y_k-1) $ or $V_2(y_k-y_k-1)$ are
non zero. Hence $|V_j(z_2-z_1)| = sum_k |V_j(y_k-y_k-1)|$.
Note that on each segment, we can apply the mean value theorem to get
$|phi(y_k)-phi(y_k-1)| le C (|V_1(y_k-y_k-1) | + |V_2(y_k-y_k-1)| )$ and
so
begineqnarray
|phi(z_2)-phi(z_1)| &le& sum_k |phi(y_k)-phi(y_k-1)| \
&le & C sum_k(|V_1(y_k-y_k-1) | + |V_2(y_k-y_k-1)| ) \
&=& C(|V_1(z_2-z_1)| + |V_2(z_2-z_1)|) \
&=& C |z_2-z_1|_1
endeqnarray
Finally, we return to '$f$' space:
$|f(z_2)-f(z_1)| = |phi(A^-1 z_2) - phi(A^-1 z_1)| le C |A^-1(z_2-z_1)|_1 le C|A^-1|_1 |z_2-z_1|_1$
Hence $f$ is Lipschitz continuous with rank at most $C|A^-1|_1$.
answered Jul 15 at 21:30
copper.hat
122k557156
1
Doesn’t this argument assume that $vec v=langle1,0rangle$ and $vec w=langle0,1rangle$ (or vice versa)?
– Steve Kass
Jul 15 at 21:42
@SteveKass: It does, but exactly the same approach works except for the $sqrt 2$ part and the $C need to be adjusted.
– copper.hat
Jul 16 at 0:06
Yes, it just seemed to me that the details of how to make that approach work were at the heart of the OP’s question. (Or a little more bluntly, if I’d assigned this as a class exercise, I would expect those details to be worked out in a submitted answer.)
– Steve Kass
Jul 16 at 0:27
@SteveKass: I guess my take was that the OP just needed to see the idea and could take it from there. I will sketch out the details.
– copper.hat
Jul 16 at 0:50
@SteveKass: I elaborated the detail, but I think it has become a technical mess.
– copper.hat
Jul 16 at 2:28
add a comment |Â
up vote
1
down vote
Here is a more complete proof. However, I think the basic idea gets lost in
the details.
Let $A= beginbmatrixv & w endbmatrix$ and define $phi(z) = f(A z)$ on the open convex set $Sigma=A^-1 S$. Note that
$partial phi(z) over partial z_1 = partial f(Az) over partial u$
and $partial phi(z) over partial z_2 = partial f(Az) over partial w$.
The key element of the proof is that for any $z_1,z_2 in Sigma$, we can choose (because $Sigma$ is open & convex) a path $y_1=z_1 to y_2to cdots to y_n-1 to y_n = z_2$ such that each segment of the path is parallel to either axis and the direction of each segment is the same (with respect to the corresponding axis).
In particular, if $V_j(y) = [y]_j$ we have
$V_j(z_2-z_1) = sum_k V_j(y_k-y_k-1)$, for a given $j$, all of the $V_j(y_k-y_k-1)$ have the same sign and at most one of
$V_1(y_k-y_k-1) $ or $V_2(y_k-y_k-1)$ are
non zero. Hence $|V_j(z_2-z_1)| = sum_k |V_j(y_k-y_k-1)|$.
Note that on each segment, we can apply the mean value theorem to get
$|phi(y_k)-phi(y_k-1)| le C (|V_1(y_k-y_k-1) | + |V_2(y_k-y_k-1)| )$ and
so
begineqnarray
|phi(z_2)-phi(z_1)| &le& sum_k |phi(y_k)-phi(y_k-1)| \
&le & C sum_k(|V_1(y_k-y_k-1) | + |V_2(y_k-y_k-1)| ) \
&=& C(|V_1(z_2-z_1)| + |V_2(z_2-z_1)|) \
&=& C |z_2-z_1|_1
endeqnarray
Finally, we return to '$f$' space:
$|f(z_2)-f(z_1)| = |phi(A^-1 z_2) - phi(A^-1 z_1)| le C |A^-1(z_2-z_1)|_1 le C|A^-1|_1 |z_2-z_1|_1$
Hence $f$ is Lipschitz continuous with rank at most $C|A^-1|_1$.
answered Jul 15 at 21:30
copper.hat
122k557156
1
Doesn’t this argument assume that $vec v=langle1,0rangle$ and $vec w=langle0,1rangle$ (or vice versa)?
– Steve Kass
Jul 15 at 21:42
@SteveKass: It does, but exactly the same approach works except for the $sqrt 2$ part and the $C need to be adjusted.
– copper.hat
Jul 16 at 0:06
Yes, it just seemed to me that the details of how to make that approach work were at the heart of the OP’s question. (Or a little more bluntly, if I’d assigned this as a class exercise, I would expect those details to be worked out in a submitted answer.)
– Steve Kass
Jul 16 at 0:27
@SteveKass: I guess my take was that the OP just needed to see the idea and could take it from there. I will sketch out the details.
– copper.hat
Jul 16 at 0:50
@SteveKass: I elaborated the detail, but I think it has become a technical mess.
– copper.hat
Jul 16 at 2:28
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Here is a more complete proof. However, I think the basic idea gets lost in
the details.
Let $A= beginbmatrixv & w endbmatrix$ and define $phi(z) = f(A z)$ on the open convex set $Sigma=A^-1 S$. Note that
$partial phi(z) over partial z_1 = partial f(Az) over partial u$
and $partial phi(z) over partial z_2 = partial f(Az) over partial w$.
The key element of the proof is that for any $z_1,z_2 in Sigma$, we can choose (because $Sigma$ is open & convex) a path $y_1=z_1 to y_2to cdots to y_n-1 to y_n = z_2$ such that each segment of the path is parallel to either axis and the direction of each segment is the same (with respect to the corresponding axis).
In particular, if $V_j(y) = [y]_j$ we have
$V_j(z_2-z_1) = sum_k V_j(y_k-y_k-1)$, for a given $j$, all of the $V_j(y_k-y_k-1)$ have the same sign and at most one of
$V_1(y_k-y_k-1) $ or $V_2(y_k-y_k-1)$ are
non zero. Hence $|V_j(z_2-z_1)| = sum_k |V_j(y_k-y_k-1)|$.
Note that on each segment, we can apply the mean value theorem to get
$|phi(y_k)-phi(y_k-1)| le C (|V_1(y_k-y_k-1) | + |V_2(y_k-y_k-1)| )$ and
so
begineqnarray
|phi(z_2)-phi(z_1)| &le& sum_k |phi(y_k)-phi(y_k-1)| \
&le & C sum_k(|V_1(y_k-y_k-1) | + |V_2(y_k-y_k-1)| ) \
&=& C(|V_1(z_2-z_1)| + |V_2(z_2-z_1)|) \
&=& C |z_2-z_1|_1
endeqnarray
Finally, we return to '$f$' space:
$|f(z_2)-f(z_1)| = |phi(A^-1 z_2) - phi(A^-1 z_1)| le C |A^-1(z_2-z_1)|_1 le C|A^-1|_1 |z_2-z_1|_1$
Hence $f$ is Lipschitz continuous with rank at most $C|A^-1|_1$.
answered Jul 15 at 21:30
copper.hat
122k557156
Here is a more complete proof. However, I think the basic idea gets lost in
the details.
Let $A= beginbmatrixv & w endbmatrix$ and define $phi(z) = f(A z)$ on the open convex set $Sigma=A^-1 S$. Note that
$partial phi(z) over partial z_1 = partial f(Az) over partial u$
and $partial phi(z) over partial z_2 = partial f(Az) over partial w$.
The key element of the proof is that for any $z_1,z_2 in Sigma$, we can choose (because $Sigma$ is open & convex) a path $y_1=z_1 to y_2to cdots to y_n-1 to y_n = z_2$ such that each segment of the path is parallel to either axis and the direction of each segment is the same (with respect to the corresponding axis).
In particular, if $V_j(y) = [y]_j$ we have
$V_j(z_2-z_1) = sum_k V_j(y_k-y_k-1)$, for a given $j$, all of the $V_j(y_k-y_k-1)$ have the same sign and at most one of
$V_1(y_k-y_k-1) $ or $V_2(y_k-y_k-1)$ are
non zero. Hence $|V_j(z_2-z_1)| = sum_k |V_j(y_k-y_k-1)|$.
Note that on each segment, we can apply the mean value theorem to get
$|phi(y_k)-phi(y_k-1)| le C (|V_1(y_k-y_k-1) | + |V_2(y_k-y_k-1)| )$ and
so
begineqnarray
|phi(z_2)-phi(z_1)| &le& sum_k |phi(y_k)-phi(y_k-1)| \
&le & C sum_k(|V_1(y_k-y_k-1) | + |V_2(y_k-y_k-1)| ) \
&=& C(|V_1(z_2-z_1)| + |V_2(z_2-z_1)|) \
&=& C |z_2-z_1|_1
endeqnarray
Finally, we return to '$f$' space:
$|f(z_2)-f(z_1)| = |phi(A^-1 z_2) - phi(A^-1 z_1)| le C |A^-1(z_2-z_1)|_1 le C|A^-1|_1 |z_2-z_1|_1$
Hence $f$ is Lipschitz continuous with rank at most $C|A^-1|_1$.
answered Jul 15 at 21:30
copper.hat
122k557156
edited Jul 16 at 2:25
answered Jul 15 at 21:30
copper.hat
122k557156
answered Jul 15 at 21:30
copper.hat
122k557156
answered Jul 15 at 21:30
copper.hat
122k557156
122k557156
1
Doesn’t this argument assume that $vec v=langle1,0rangle$ and $vec w=langle0,1rangle$ (or vice versa)?
– Steve Kass
Jul 15 at 21:42
@SteveKass: It does, but exactly the same approach works except for the $sqrt 2$ part and the $C need to be adjusted.
– copper.hat
Jul 16 at 0:06
Yes, it just seemed to me that the details of how to make that approach work were at the heart of the OP’s question. (Or a little more bluntly, if I’d assigned this as a class exercise, I would expect those details to be worked out in a submitted answer.)
– Steve Kass
Jul 16 at 0:27
@SteveKass: I guess my take was that the OP just needed to see the idea and could take it from there. I will sketch out the details.
– copper.hat
Jul 16 at 0:50
@SteveKass: I elaborated the detail, but I think it has become a technical mess.
– copper.hat
Jul 16 at 2:28
add a comment |Â
1
Doesn’t this argument assume that $vec v=langle1,0rangle$ and $vec w=langle0,1rangle$ (or vice versa)?
– Steve Kass
Jul 15 at 21:42
@SteveKass: It does, but exactly the same approach works except for the $sqrt 2$ part and the $C need to be adjusted.
– copper.hat
Jul 16 at 0:06
Yes, it just seemed to me that the details of how to make that approach work were at the heart of the OP’s question. (Or a little more bluntly, if I’d assigned this as a class exercise, I would expect those details to be worked out in a submitted answer.)
– Steve Kass
Jul 16 at 0:27
@SteveKass: I guess my take was that the OP just needed to see the idea and could take it from there. I will sketch out the details.
– copper.hat
Jul 16 at 0:50
@SteveKass: I elaborated the detail, but I think it has become a technical mess.
– copper.hat
Jul 16 at 2:28
1
1
Doesn’t this argument assume that $vec v=langle1,0rangle$ and $vec w=langle0,1rangle$ (or vice versa)?
– Steve Kass
Jul 15 at 21:42
Doesn’t this argument assume that $vec v=langle1,0rangle$ and $vec w=langle0,1rangle$ (or vice versa)?
– Steve Kass
Jul 15 at 21:42
@SteveKass: It does, but exactly the same approach works except for the $sqrt 2$ part and the $C need to be adjusted.
– copper.hat
Jul 16 at 0:06
@SteveKass: It does, but exactly the same approach works except for the $sqrt 2$ part and the $C need to be adjusted.
– copper.hat
Jul 16 at 0:06
Yes, it just seemed to me that the details of how to make that approach work were at the heart of the OP’s question. (Or a little more bluntly, if I’d assigned this as a class exercise, I would expect those details to be worked out in a submitted answer.)
– Steve Kass
Jul 16 at 0:27
Yes, it just seemed to me that the details of how to make that approach work were at the heart of the OP’s question. (Or a little more bluntly, if I’d assigned this as a class exercise, I would expect those details to be worked out in a submitted answer.)
– Steve Kass
Jul 16 at 0:27
@SteveKass: I guess my take was that the OP just needed to see the idea and could take it from there. I will sketch out the details.
– copper.hat
Jul 16 at 0:50
@SteveKass: I guess my take was that the OP just needed to see the idea and could take it from there. I will sketch out the details.
– copper.hat
Jul 16 at 0:50
@SteveKass: I elaborated the detail, but I think it has become a technical mess.
– copper.hat
Jul 16 at 2:28
@SteveKass: I elaborated the detail, but I think it has become a technical mess.
– copper.hat
Jul 16 at 2:28
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Between every two points in $S$ there is a two segment path that joins them and the segments are parallel to the axes. Use this to estimate the difference in $f$.
– copper.hat
Jul 15 at 18:32
@copper.hat I'm not sure how to construct this two segment path, I could only come up with a one-segment path $phi(t)=tcdot(alpha vec w +beta vec v)+(1-t)cdot (lambda vec w +gamma vec v) $ for $tin [0,1]$, where $z_1 ,z_2 in S$ are represented by these two linear combinations.
– gbi1977
Jul 15 at 20:13
Draw a picture. Take two points in the disc. Join them with a Manhattan like path.
– copper.hat
Jul 15 at 20:15
@copper.hat oh okay I think I see what you mean. Then the idea is that each point on this L path can be represented as a linear comb. of $v,w$. So along the path, the directional derivative is of course bounded, depending on the $v$ or $w$ component, but since it applies to any point on a line pararllel to the axes - it applies to the partial derivatives themselves?
– gbi1977
Jul 15 at 20:34