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Integrability of Lipschitz constant of multivariate function
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Suppose $f in C(U, mathbbR^n)$, where $U$ is an open subset of $mathbbR times mathbbR^n$, and $f$ is locally Lipschitz continuous in the second argument, meaning that, for each $t$,
$$
sup_x neq y in V fracf(t, x) - f(t, y)x - y
$$
is finite for every compact subset $V$ of $mathbbR^n$ (such that $t times V subset U$, of course). Now, suppose $(t_0, x_0) subset U$ and choose $delta, T > 0$ such that $[t_0, t_0 + T] times overlineB_delta(x_0) subset U$. Set
$$
L(t) = sup_x neq y in B_delta(x_0) fracf(t, x) - f(t, y)x - y.
$$
and
$$
L_1(t) = int_t_0^t L(s) ds
$$
Now, I'm reading a book where one condition of a theorem is that $L_1(T_0) < infty$, where $0 < T_0 le T$, and along the proof the author assumes that $L_1'(t) = L(t)$.
I was thinking that $L$ must be continuous (so one can use the Fundamental Theorem of Calculus to have $L_1'(t) = L(t)$), but then integrability comes for free and it wouldn't make sense for the author to put $L_1(T_0) < infty$ as a condition in the theorem (it would already be true). So if $L$ is not continuous, why is it integrable (at least on a small interval arounnd $t_0$) and why is $L_1'(t) = L(t)$?
real-analysis integration
asked Jul 25 at 14:32
Doughnut Pump
439312
add a comment |Â
up vote
0
down vote
favorite
Suppose $f in C(U, mathbbR^n)$, where $U$ is an open subset of $mathbbR times mathbbR^n$, and $f$ is locally Lipschitz continuous in the second argument, meaning that, for each $t$,
$$
sup_x neq y in V fracf(t, x) - f(t, y)x - y
$$
is finite for every compact subset $V$ of $mathbbR^n$ (such that $t times V subset U$, of course). Now, suppose $(t_0, x_0) subset U$ and choose $delta, T > 0$ such that $[t_0, t_0 + T] times overlineB_delta(x_0) subset U$. Set
$$
L(t) = sup_x neq y in B_delta(x_0) fracf(t, x) - f(t, y)x - y.
$$
and
$$
L_1(t) = int_t_0^t L(s) ds
$$
Now, I'm reading a book where one condition of a theorem is that $L_1(T_0) < infty$, where $0 < T_0 le T$, and along the proof the author assumes that $L_1'(t) = L(t)$.
I was thinking that $L$ must be continuous (so one can use the Fundamental Theorem of Calculus to have $L_1'(t) = L(t)$), but then integrability comes for free and it wouldn't make sense for the author to put $L_1(T_0) < infty$ as a condition in the theorem (it would already be true). So if $L$ is not continuous, why is it integrable (at least on a small interval arounnd $t_0$) and why is $L_1'(t) = L(t)$?
real-analysis integration
asked Jul 25 at 14:32
Doughnut Pump
439312
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Suppose $f in C(U, mathbbR^n)$, where $U$ is an open subset of $mathbbR times mathbbR^n$, and $f$ is locally Lipschitz continuous in the second argument, meaning that, for each $t$,
$$
sup_x neq y in V fracf(t, x) - f(t, y)x - y
$$
is finite for every compact subset $V$ of $mathbbR^n$ (such that $t times V subset U$, of course). Now, suppose $(t_0, x_0) subset U$ and choose $delta, T > 0$ such that $[t_0, t_0 + T] times overlineB_delta(x_0) subset U$. Set
$$
L(t) = sup_x neq y in B_delta(x_0) fracf(t, x) - f(t, y)x - y.
$$
and
$$
L_1(t) = int_t_0^t L(s) ds
$$
Now, I'm reading a book where one condition of a theorem is that $L_1(T_0) < infty$, where $0 < T_0 le T$, and along the proof the author assumes that $L_1'(t) = L(t)$.
I was thinking that $L$ must be continuous (so one can use the Fundamental Theorem of Calculus to have $L_1'(t) = L(t)$), but then integrability comes for free and it wouldn't make sense for the author to put $L_1(T_0) < infty$ as a condition in the theorem (it would already be true). So if $L$ is not continuous, why is it integrable (at least on a small interval arounnd $t_0$) and why is $L_1'(t) = L(t)$?
real-analysis integration
asked Jul 25 at 14:32
Doughnut Pump
439312
Suppose $f in C(U, mathbbR^n)$, where $U$ is an open subset of $mathbbR times mathbbR^n$, and $f$ is locally Lipschitz continuous in the second argument, meaning that, for each $t$,
$$
sup_x neq y in V fracf(t, x) - f(t, y)x - y
$$
is finite for every compact subset $V$ of $mathbbR^n$ (such that $t times V subset U$, of course). Now, suppose $(t_0, x_0) subset U$ and choose $delta, T > 0$ such that $[t_0, t_0 + T] times overlineB_delta(x_0) subset U$. Set
$$
L(t) = sup_x neq y in B_delta(x_0) fracf(t, x) - f(t, y)x - y.
$$
and
$$
L_1(t) = int_t_0^t L(s) ds
$$
Now, I'm reading a book where one condition of a theorem is that $L_1(T_0) < infty$, where $0 < T_0 le T$, and along the proof the author assumes that $L_1'(t) = L(t)$.
I was thinking that $L$ must be continuous (so one can use the Fundamental Theorem of Calculus to have $L_1'(t) = L(t)$), but then integrability comes for free and it wouldn't make sense for the author to put $L_1(T_0) < infty$ as a condition in the theorem (it would already be true). So if $L$ is not continuous, why is it integrable (at least on a small interval arounnd $t_0$) and why is $L_1'(t) = L(t)$?
real-analysis integration
asked Jul 25 at 14:32
Doughnut Pump
439312
asked Jul 25 at 14:32
Doughnut Pump
439312
asked Jul 25 at 14:32
Doughnut Pump
439312
asked Jul 25 at 14:32
Doughnut Pump
439312
439312
add a comment |Â
add a comment |Â
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