Mixing
Showing the existence of a subsequence that converge locally uniformly
I was reading a Navier-Stokes paper by Kato, and at the end of the paper he makes the following affirmation:
Let $f_i:(0,infty)timesmathbb{R^2}tomathbb{R}$ be a smooth function such that for all $ninmathbb{N}, qin(1,infty],kgeq0$, all the norms
$||partial^{n}_{t}D^kf_i(t)||_q$ are bounded on any compact interval $[a,b]subset(0,infty)$ uniformly in $i$. Then, we can extract a subsequence ${f_{i_k}}$ that converge locally uniformly to a smooth function $f:(0,infty)timesmathbb{R^2}tomathbb{R}$.
- here $D^k$ is the fractional derivative.
The only result I know about extracting subsequence that converge locally uniformly was Arzelá-Ascoli, but I could't show that the sequence ${f_i}$ was equicontínuous.
lp-spaces uniform-convergence
asked Oct 5 '18 at 22:52
Pires Dankan
186115
add a comment |
I was reading a Navier-Stokes paper by Kato, and at the end of the paper he makes the following affirmation:
Let $f_i:(0,infty)timesmathbb{R^2}tomathbb{R}$ be a smooth function such that for all $ninmathbb{N}, qin(1,infty],kgeq0$, all the norms
$||partial^{n}_{t}D^kf_i(t)||_q$ are bounded on any compact interval $[a,b]subset(0,infty)$ uniformly in $i$. Then, we can extract a subsequence ${f_{i_k}}$ that converge locally uniformly to a smooth function $f:(0,infty)timesmathbb{R^2}tomathbb{R}$.
- here $D^k$ is the fractional derivative.
The only result I know about extracting subsequence that converge locally uniformly was Arzelá-Ascoli, but I could't show that the sequence ${f_i}$ was equicontínuous.
lp-spaces uniform-convergence
asked Oct 5 '18 at 22:52
Pires Dankan
186115
Shouldn't it be $||partial_t^n D^k f_i(t,x,y)||_q$? $f_i$ takes in 3 inputs
– mathworker21
Oct 9 '18 at 6:30
@mathworker21 yes, but after you calculate the $L^q$ norm, it become just a function of $t$.
– Pires Dankan
Oct 9 '18 at 12:05
@mathworker21 It would be helpful if the function $f_i$ was a one variable function (at least I was able to deduce equicontinuity only in this case)
– Pires Dankan
Oct 16 '18 at 17:51
Please see my updated answer, which addresses the multivariable case.
– mathworker21
Nov 20 '18 at 21:49
add a comment |
I was reading a Navier-Stokes paper by Kato, and at the end of the paper he makes the following affirmation:
Let $f_i:(0,infty)timesmathbb{R^2}tomathbb{R}$ be a smooth function such that for all $ninmathbb{N}, qin(1,infty],kgeq0$, all the norms
$||partial^{n}_{t}D^kf_i(t)||_q$ are bounded on any compact interval $[a,b]subset(0,infty)$ uniformly in $i$. Then, we can extract a subsequence ${f_{i_k}}$ that converge locally uniformly to a smooth function $f:(0,infty)timesmathbb{R^2}tomathbb{R}$.
- here $D^k$ is the fractional derivative.
The only result I know about extracting subsequence that converge locally uniformly was Arzelá-Ascoli, but I could't show that the sequence ${f_i}$ was equicontínuous.
lp-spaces uniform-convergence
asked Oct 5 '18 at 22:52
Pires Dankan
186115
I was reading a Navier-Stokes paper by Kato, and at the end of the paper he makes the following affirmation:
Let $f_i:(0,infty)timesmathbb{R^2}tomathbb{R}$ be a smooth function such that for all $ninmathbb{N}, qin(1,infty],kgeq0$, all the norms
$||partial^{n}_{t}D^kf_i(t)||_q$ are bounded on any compact interval $[a,b]subset(0,infty)$ uniformly in $i$. Then, we can extract a subsequence ${f_{i_k}}$ that converge locally uniformly to a smooth function $f:(0,infty)timesmathbb{R^2}tomathbb{R}$.
- here $D^k$ is the fractional derivative.
The only result I know about extracting subsequence that converge locally uniformly was Arzelá-Ascoli, but I could't show that the sequence ${f_i}$ was equicontínuous.
lp-spaces uniform-convergence
lp-spaces uniform-convergence
asked Oct 5 '18 at 22:52
Pires Dankan
186115
asked Oct 5 '18 at 22:52
Pires Dankan
186115
asked Oct 5 '18 at 22:52
Pires Dankan
186115
asked Oct 5 '18 at 22:52
Pires Dankan
186115
asked Oct 5 '18 at 22:52
Pires Dankan
186115
186115
Shouldn't it be $||partial_t^n D^k f_i(t,x,y)||_q$? $f_i$ takes in 3 inputs
– mathworker21
Oct 9 '18 at 6:30
@mathworker21 yes, but after you calculate the $L^q$ norm, it become just a function of $t$.
– Pires Dankan
Oct 9 '18 at 12:05
@mathworker21 It would be helpful if the function $f_i$ was a one variable function (at least I was able to deduce equicontinuity only in this case)
– Pires Dankan
Oct 16 '18 at 17:51
Please see my updated answer, which addresses the multivariable case.
– mathworker21
Nov 20 '18 at 21:49
add a comment |
Shouldn't it be $||partial_t^n D^k f_i(t,x,y)||_q$? $f_i$ takes in 3 inputs
– mathworker21
Oct 9 '18 at 6:30
@mathworker21 yes, but after you calculate the $L^q$ norm, it become just a function of $t$.
– Pires Dankan
Oct 9 '18 at 12:05
@mathworker21 It would be helpful if the function $f_i$ was a one variable function (at least I was able to deduce equicontinuity only in this case)
– Pires Dankan
Oct 16 '18 at 17:51
Please see my updated answer, which addresses the multivariable case.
– mathworker21
Nov 20 '18 at 21:49
Shouldn't it be $||partial_t^n D^k f_i(t,x,y)||_q$? $f_i$ takes in 3 inputs
– mathworker21
Oct 9 '18 at 6:30
Shouldn't it be $||partial_t^n D^k f_i(t,x,y)||_q$? $f_i$ takes in 3 inputs
– mathworker21
Oct 9 '18 at 6:30
@mathworker21 yes, but after you calculate the $L^q$ norm, it become just a function of $t$.
– Pires Dankan
Oct 9 '18 at 12:05
@mathworker21 yes, but after you calculate the $L^q$ norm, it become just a function of $t$.
– Pires Dankan
Oct 9 '18 at 12:05
@mathworker21 It would be helpful if the function $f_i$ was a one variable function (at least I was able to deduce equicontinuity only in this case)
– Pires Dankan
Oct 16 '18 at 17:51
@mathworker21 It would be helpful if the function $f_i$ was a one variable function (at least I was able to deduce equicontinuity only in this case)
– Pires Dankan
Oct 16 '18 at 17:51
Please see my updated answer, which addresses the multivariable case.
– mathworker21
Nov 20 '18 at 21:49
Please see my updated answer, which addresses the multivariable case.
– mathworker21
Nov 20 '18 at 21:49
add a comment |
1 Answer
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You can deduce equicontinuity from a uniform bound on the derivative.
In general, let $(f_n)_n$ be a sequence of functions with $||f_n'||_infty le C$ for all $n$. Then for any $x,y$ and any $n ge 1$, $|f_n(x)-f_n(y)| le C|x-y|$, which of course in particular gives equicontinuity.
This also extends to functions of more than one variable. Indeed, if $(f_n)_n$ is a sequence of functions from $mathbb{R}^m$ with $||frac{partial}{partial x_j}f_n||_infty le C$ for each $1 le j le m$ and each $n$, then given any $(x_1,dots,x_m),(y_1,dots,y_m) in mathbb{R}^m$ and any $n ge 1$, we can bound $|f_n(x_1,dots,x_m)-f_n(y_1,dots,y_m)| le C_m C |x-y|$, where $C_m$ is just to switch between norms.
For example, for $m=3$, we do $$|f(x_1,x_2,x_3)-f(y_1,y_2,y_3)| le$$ $$|f(x_1,x_2,x_3)-f(x_1,x_2,y_3)|+|f(x_1,x_2,y_3)-f(y_1,y_2,y_3)|$$ $$ le C|x_3-y_3|+|f(x_1,x_2,y_3)-f(x_1,y_2,y_3)|+|f(x_1,y_2,y_3)-f(y_1,y_2,y_3)|$$ $$le C|x_3-y_3|+C|x_2-y_2|+C|x_1-y_1| le C_3 C |x-y|.$$
answered Oct 9 '18 at 6:32
mathworker21
8,6521928
Thanks @mathworker21! Very nice idea to generalize to the $n$ dimension.
– Pires Dankan
Nov 22 '18 at 14:12
add a comment |
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You can deduce equicontinuity from a uniform bound on the derivative.
In general, let $(f_n)_n$ be a sequence of functions with $||f_n'||_infty le C$ for all $n$. Then for any $x,y$ and any $n ge 1$, $|f_n(x)-f_n(y)| le C|x-y|$, which of course in particular gives equicontinuity.
This also extends to functions of more than one variable. Indeed, if $(f_n)_n$ is a sequence of functions from $mathbb{R}^m$ with $||frac{partial}{partial x_j}f_n||_infty le C$ for each $1 le j le m$ and each $n$, then given any $(x_1,dots,x_m),(y_1,dots,y_m) in mathbb{R}^m$ and any $n ge 1$, we can bound $|f_n(x_1,dots,x_m)-f_n(y_1,dots,y_m)| le C_m C |x-y|$, where $C_m$ is just to switch between norms.
For example, for $m=3$, we do $$|f(x_1,x_2,x_3)-f(y_1,y_2,y_3)| le$$ $$|f(x_1,x_2,x_3)-f(x_1,x_2,y_3)|+|f(x_1,x_2,y_3)-f(y_1,y_2,y_3)|$$ $$ le C|x_3-y_3|+|f(x_1,x_2,y_3)-f(x_1,y_2,y_3)|+|f(x_1,y_2,y_3)-f(y_1,y_2,y_3)|$$ $$le C|x_3-y_3|+C|x_2-y_2|+C|x_1-y_1| le C_3 C |x-y|.$$
answered Oct 9 '18 at 6:32
mathworker21
8,6521928
Thanks @mathworker21! Very nice idea to generalize to the $n$ dimension.
– Pires Dankan
Nov 22 '18 at 14:12
add a comment |
You can deduce equicontinuity from a uniform bound on the derivative.
In general, let $(f_n)_n$ be a sequence of functions with $||f_n'||_infty le C$ for all $n$. Then for any $x,y$ and any $n ge 1$, $|f_n(x)-f_n(y)| le C|x-y|$, which of course in particular gives equicontinuity.
This also extends to functions of more than one variable. Indeed, if $(f_n)_n$ is a sequence of functions from $mathbb{R}^m$ with $||frac{partial}{partial x_j}f_n||_infty le C$ for each $1 le j le m$ and each $n$, then given any $(x_1,dots,x_m),(y_1,dots,y_m) in mathbb{R}^m$ and any $n ge 1$, we can bound $|f_n(x_1,dots,x_m)-f_n(y_1,dots,y_m)| le C_m C |x-y|$, where $C_m$ is just to switch between norms.
For example, for $m=3$, we do $$|f(x_1,x_2,x_3)-f(y_1,y_2,y_3)| le$$ $$|f(x_1,x_2,x_3)-f(x_1,x_2,y_3)|+|f(x_1,x_2,y_3)-f(y_1,y_2,y_3)|$$ $$ le C|x_3-y_3|+|f(x_1,x_2,y_3)-f(x_1,y_2,y_3)|+|f(x_1,y_2,y_3)-f(y_1,y_2,y_3)|$$ $$le C|x_3-y_3|+C|x_2-y_2|+C|x_1-y_1| le C_3 C |x-y|.$$
answered Oct 9 '18 at 6:32
mathworker21
8,6521928
Thanks @mathworker21! Very nice idea to generalize to the $n$ dimension.
– Pires Dankan
Nov 22 '18 at 14:12
add a comment |
You can deduce equicontinuity from a uniform bound on the derivative.
In general, let $(f_n)_n$ be a sequence of functions with $||f_n'||_infty le C$ for all $n$. Then for any $x,y$ and any $n ge 1$, $|f_n(x)-f_n(y)| le C|x-y|$, which of course in particular gives equicontinuity.
This also extends to functions of more than one variable. Indeed, if $(f_n)_n$ is a sequence of functions from $mathbb{R}^m$ with $||frac{partial}{partial x_j}f_n||_infty le C$ for each $1 le j le m$ and each $n$, then given any $(x_1,dots,x_m),(y_1,dots,y_m) in mathbb{R}^m$ and any $n ge 1$, we can bound $|f_n(x_1,dots,x_m)-f_n(y_1,dots,y_m)| le C_m C |x-y|$, where $C_m$ is just to switch between norms.
For example, for $m=3$, we do $$|f(x_1,x_2,x_3)-f(y_1,y_2,y_3)| le$$ $$|f(x_1,x_2,x_3)-f(x_1,x_2,y_3)|+|f(x_1,x_2,y_3)-f(y_1,y_2,y_3)|$$ $$ le C|x_3-y_3|+|f(x_1,x_2,y_3)-f(x_1,y_2,y_3)|+|f(x_1,y_2,y_3)-f(y_1,y_2,y_3)|$$ $$le C|x_3-y_3|+C|x_2-y_2|+C|x_1-y_1| le C_3 C |x-y|.$$
answered Oct 9 '18 at 6:32
mathworker21
8,6521928
You can deduce equicontinuity from a uniform bound on the derivative.
In general, let $(f_n)_n$ be a sequence of functions with $||f_n'||_infty le C$ for all $n$. Then for any $x,y$ and any $n ge 1$, $|f_n(x)-f_n(y)| le C|x-y|$, which of course in particular gives equicontinuity.
This also extends to functions of more than one variable. Indeed, if $(f_n)_n$ is a sequence of functions from $mathbb{R}^m$ with $||frac{partial}{partial x_j}f_n||_infty le C$ for each $1 le j le m$ and each $n$, then given any $(x_1,dots,x_m),(y_1,dots,y_m) in mathbb{R}^m$ and any $n ge 1$, we can bound $|f_n(x_1,dots,x_m)-f_n(y_1,dots,y_m)| le C_m C |x-y|$, where $C_m$ is just to switch between norms.
For example, for $m=3$, we do $$|f(x_1,x_2,x_3)-f(y_1,y_2,y_3)| le$$ $$|f(x_1,x_2,x_3)-f(x_1,x_2,y_3)|+|f(x_1,x_2,y_3)-f(y_1,y_2,y_3)|$$ $$ le C|x_3-y_3|+|f(x_1,x_2,y_3)-f(x_1,y_2,y_3)|+|f(x_1,y_2,y_3)-f(y_1,y_2,y_3)|$$ $$le C|x_3-y_3|+C|x_2-y_2|+C|x_1-y_1| le C_3 C |x-y|.$$
answered Oct 9 '18 at 6:32
mathworker21
8,6521928
edited Nov 22 '18 at 14:15
answered Oct 9 '18 at 6:32
mathworker21
8,6521928
answered Oct 9 '18 at 6:32
mathworker21
8,6521928
answered Oct 9 '18 at 6:32
mathworker21
8,6521928
8,6521928
Thanks @mathworker21! Very nice idea to generalize to the $n$ dimension.
– Pires Dankan
Nov 22 '18 at 14:12
add a comment |
Thanks @mathworker21! Very nice idea to generalize to the $n$ dimension.
– Pires Dankan
Nov 22 '18 at 14:12
Thanks @mathworker21! Very nice idea to generalize to the $n$ dimension.
– Pires Dankan
Nov 22 '18 at 14:12
Thanks @mathworker21! Very nice idea to generalize to the $n$ dimension.
– Pires Dankan
Nov 22 '18 at 14:12
add a comment |
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Shouldn't it be $||partial_t^n D^k f_i(t,x,y)||_q$? $f_i$ takes in 3 inputs
– mathworker21
Oct 9 '18 at 6:30
@mathworker21 yes, but after you calculate the $L^q$ norm, it become just a function of $t$.
– Pires Dankan
Oct 9 '18 at 12:05
@mathworker21 It would be helpful if the function $f_i$ was a one variable function (at least I was able to deduce equicontinuity only in this case)
– Pires Dankan
Oct 16 '18 at 17:51
Please see my updated answer, which addresses the multivariable case.
– mathworker21
Nov 20 '18 at 21:49