Mixing
Asymptotics of a function defined by an integral
Clash Royale CLAN TAG#URR8PPP
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This arises in trying to understand the Uncertainty Principle. Suppose $psi(x)$ is a Schwartz function on $mathbbR^n$ such that $|psi|_2=1$. Let $R_1$ denote the rectangle $[-1,1]^n$ in $mathbbR ^n$. How can one show that $f(h),$ for $h$ a small parameter and $epsilon > 0$ defined as
$$f(h):= h^-n epsilonint_mathbbR^n setminus R_1 left(psileft(fracxh^epsilonright)right)^2 dx$$
belongs to $O(h^infty)$ as $h to 0?$
real-analysis integration fourier-analysis
edited Aug 2 at 15:37
Adrian Keister
3,49321533
asked Aug 2 at 15:30
Memeboy Inc.
499
add a comment |Â
up vote
1
down vote
favorite
This arises in trying to understand the Uncertainty Principle. Suppose $psi(x)$ is a Schwartz function on $mathbbR^n$ such that $|psi|_2=1$. Let $R_1$ denote the rectangle $[-1,1]^n$ in $mathbbR ^n$. How can one show that $f(h),$ for $h$ a small parameter and $epsilon > 0$ defined as
$$f(h):= h^-n epsilonint_mathbbR^n setminus R_1 left(psileft(fracxh^epsilonright)right)^2 dx$$
belongs to $O(h^infty)$ as $h to 0?$
real-analysis integration fourier-analysis
edited Aug 2 at 15:37
Adrian Keister
3,49321533
asked Aug 2 at 15:30
Memeboy Inc.
499
What's $O(h^infty)$??
– mathworker21
2 days ago
Letting $y = fracxh^epsilon$ gives that $f(h) = int_mathbbR^n setminus I_h psi(y)^2 dy$, where $I_h = [-frac1h^epsilon,frac1h^epsilon]$. As $h to 0$, $f(h) to 0$.
– mathworker21
2 days ago
$f(h) in O(h^infty)$ means that $f(h) in O(h^n)$ for every $n in mathbbZ_+$ as $h to 0$. I know $f(h) to 0$ but I don't know how to show the order.
– Memeboy Inc.
2 days ago
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
This arises in trying to understand the Uncertainty Principle. Suppose $psi(x)$ is a Schwartz function on $mathbbR^n$ such that $|psi|_2=1$. Let $R_1$ denote the rectangle $[-1,1]^n$ in $mathbbR ^n$. How can one show that $f(h),$ for $h$ a small parameter and $epsilon > 0$ defined as
$$f(h):= h^-n epsilonint_mathbbR^n setminus R_1 left(psileft(fracxh^epsilonright)right)^2 dx$$
belongs to $O(h^infty)$ as $h to 0?$
real-analysis integration fourier-analysis
edited Aug 2 at 15:37
Adrian Keister
3,49321533
asked Aug 2 at 15:30
Memeboy Inc.
499
This arises in trying to understand the Uncertainty Principle. Suppose $psi(x)$ is a Schwartz function on $mathbbR^n$ such that $|psi|_2=1$. Let $R_1$ denote the rectangle $[-1,1]^n$ in $mathbbR ^n$. How can one show that $f(h),$ for $h$ a small parameter and $epsilon > 0$ defined as
$$f(h):= h^-n epsilonint_mathbbR^n setminus R_1 left(psileft(fracxh^epsilonright)right)^2 dx$$
belongs to $O(h^infty)$ as $h to 0?$
real-analysis integration fourier-analysis
edited Aug 2 at 15:37
Adrian Keister
3,49321533
asked Aug 2 at 15:30
Memeboy Inc.
499
edited Aug 2 at 15:37
Adrian Keister
3,49321533
edited Aug 2 at 15:37
Adrian Keister
3,49321533
edited Aug 2 at 15:37
Adrian Keister
3,49321533
3,49321533
asked Aug 2 at 15:30
Memeboy Inc.
499
asked Aug 2 at 15:30
Memeboy Inc.
499
asked Aug 2 at 15:30
Memeboy Inc.
499
499
What's $O(h^infty)$??
– mathworker21
2 days ago
Letting $y = fracxh^epsilon$ gives that $f(h) = int_mathbbR^n setminus I_h psi(y)^2 dy$, where $I_h = [-frac1h^epsilon,frac1h^epsilon]$. As $h to 0$, $f(h) to 0$.
– mathworker21
2 days ago
$f(h) in O(h^infty)$ means that $f(h) in O(h^n)$ for every $n in mathbbZ_+$ as $h to 0$. I know $f(h) to 0$ but I don't know how to show the order.
– Memeboy Inc.
2 days ago
add a comment |Â
What's $O(h^infty)$??
– mathworker21
2 days ago
Letting $y = fracxh^epsilon$ gives that $f(h) = int_mathbbR^n setminus I_h psi(y)^2 dy$, where $I_h = [-frac1h^epsilon,frac1h^epsilon]$. As $h to 0$, $f(h) to 0$.
– mathworker21
2 days ago
$f(h) in O(h^infty)$ means that $f(h) in O(h^n)$ for every $n in mathbbZ_+$ as $h to 0$. I know $f(h) to 0$ but I don't know how to show the order.
– Memeboy Inc.
2 days ago
What's $O(h^infty)$??
– mathworker21
2 days ago
What's $O(h^infty)$??
– mathworker21
2 days ago
Letting $y = fracxh^epsilon$ gives that $f(h) = int_mathbbR^n setminus I_h psi(y)^2 dy$, where $I_h = [-frac1h^epsilon,frac1h^epsilon]$. As $h to 0$, $f(h) to 0$.
– mathworker21
2 days ago
Letting $y = fracxh^epsilon$ gives that $f(h) = int_mathbbR^n setminus I_h psi(y)^2 dy$, where $I_h = [-frac1h^epsilon,frac1h^epsilon]$. As $h to 0$, $f(h) to 0$.
– mathworker21
2 days ago
$f(h) in O(h^infty)$ means that $f(h) in O(h^n)$ for every $n in mathbbZ_+$ as $h to 0$. I know $f(h) to 0$ but I don't know how to show the order.
– Memeboy Inc.
2 days ago
$f(h) in O(h^infty)$ means that $f(h) in O(h^n)$ for every $n in mathbbZ_+$ as $h to 0$. I know $f(h) to 0$ but I don't know how to show the order.
– Memeboy Inc.
2 days ago
add a comment |Â
1 Answer
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I think this just follows from the definition of a Schwartz function. Fix $epsilon > 0$ and let $I_h = [-frac1h^epsilon,frac1h^epsilon]$. For $M ge 2$, $|psi(y)| lesssim_M y$ on the range $|y| ge 1$ (say). Then, $int_mathbbR^nsetminus I_h psi(y)^2 lesssim_M int_mathbbR^nsetminus I_h frac1ydy lesssim_M h^epsilon(2M-1)$. As $M$ can be arbitrarily large, this shows $f(h) in O(h^infty)$.
answered 2 days ago
mathworker21
6,4231727
Ah, of course! For clarification your notation means $leq C_M |y|^M$ where $C_M$ is a constant depending on $M$?
– Memeboy Inc.
2 days ago
@MemeboyInc. Yes
– mathworker21
2 days ago
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
I think this just follows from the definition of a Schwartz function. Fix $epsilon > 0$ and let $I_h = [-frac1h^epsilon,frac1h^epsilon]$. For $M ge 2$, $|psi(y)| lesssim_M y$ on the range $|y| ge 1$ (say). Then, $int_mathbbR^nsetminus I_h psi(y)^2 lesssim_M int_mathbbR^nsetminus I_h frac1ydy lesssim_M h^epsilon(2M-1)$. As $M$ can be arbitrarily large, this shows $f(h) in O(h^infty)$.
answered 2 days ago
mathworker21
6,4231727
Ah, of course! For clarification your notation means $leq C_M |y|^M$ where $C_M$ is a constant depending on $M$?
– Memeboy Inc.
2 days ago
@MemeboyInc. Yes
– mathworker21
2 days ago
add a comment |Â
up vote
2
down vote
I think this just follows from the definition of a Schwartz function. Fix $epsilon > 0$ and let $I_h = [-frac1h^epsilon,frac1h^epsilon]$. For $M ge 2$, $|psi(y)| lesssim_M y$ on the range $|y| ge 1$ (say). Then, $int_mathbbR^nsetminus I_h psi(y)^2 lesssim_M int_mathbbR^nsetminus I_h frac1ydy lesssim_M h^epsilon(2M-1)$. As $M$ can be arbitrarily large, this shows $f(h) in O(h^infty)$.
answered 2 days ago
mathworker21
6,4231727
Ah, of course! For clarification your notation means $leq C_M |y|^M$ where $C_M$ is a constant depending on $M$?
– Memeboy Inc.
2 days ago
@MemeboyInc. Yes
– mathworker21
2 days ago
add a comment |Â
up vote
2
down vote
up vote
2
down vote
I think this just follows from the definition of a Schwartz function. Fix $epsilon > 0$ and let $I_h = [-frac1h^epsilon,frac1h^epsilon]$. For $M ge 2$, $|psi(y)| lesssim_M y$ on the range $|y| ge 1$ (say). Then, $int_mathbbR^nsetminus I_h psi(y)^2 lesssim_M int_mathbbR^nsetminus I_h frac1ydy lesssim_M h^epsilon(2M-1)$. As $M$ can be arbitrarily large, this shows $f(h) in O(h^infty)$.
answered 2 days ago
mathworker21
6,4231727
I think this just follows from the definition of a Schwartz function. Fix $epsilon > 0$ and let $I_h = [-frac1h^epsilon,frac1h^epsilon]$. For $M ge 2$, $|psi(y)| lesssim_M y$ on the range $|y| ge 1$ (say). Then, $int_mathbbR^nsetminus I_h psi(y)^2 lesssim_M int_mathbbR^nsetminus I_h frac1ydy lesssim_M h^epsilon(2M-1)$. As $M$ can be arbitrarily large, this shows $f(h) in O(h^infty)$.
answered 2 days ago
mathworker21
6,4231727
answered 2 days ago
mathworker21
6,4231727
answered 2 days ago
mathworker21
6,4231727
answered 2 days ago
mathworker21
6,4231727
6,4231727
Ah, of course! For clarification your notation means $leq C_M |y|^M$ where $C_M$ is a constant depending on $M$?
– Memeboy Inc.
2 days ago
@MemeboyInc. Yes
– mathworker21
2 days ago
add a comment |Â
Ah, of course! For clarification your notation means $leq C_M |y|^M$ where $C_M$ is a constant depending on $M$?
– Memeboy Inc.
2 days ago
@MemeboyInc. Yes
– mathworker21
2 days ago
Ah, of course! For clarification your notation means $leq C_M |y|^M$ where $C_M$ is a constant depending on $M$?
– Memeboy Inc.
2 days ago
Ah, of course! For clarification your notation means $leq C_M |y|^M$ where $C_M$ is a constant depending on $M$?
– Memeboy Inc.
2 days ago
@MemeboyInc. Yes
– mathworker21
2 days ago
@MemeboyInc. Yes
– mathworker21
2 days ago
add a comment |Â
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What's $O(h^infty)$??
– mathworker21
2 days ago
Letting $y = fracxh^epsilon$ gives that $f(h) = int_mathbbR^n setminus I_h psi(y)^2 dy$, where $I_h = [-frac1h^epsilon,frac1h^epsilon]$. As $h to 0$, $f(h) to 0$.
– mathworker21
2 days ago
$f(h) in O(h^infty)$ means that $f(h) in O(h^n)$ for every $n in mathbbZ_+$ as $h to 0$. I know $f(h) to 0$ but I don't know how to show the order.
– Memeboy Inc.
2 days ago