Mixing
Superposition of of bump functions identically equal to 1.
$begingroup$
I am trying to create a superposition of bump functions that adds identically to 1. Specifically I am looking to add two bump functions, say $f(x)$, $h(x)$ and $g(x)$ so that if $I,J,L subset mathbb{R}$ are the support of $f,h$ and $g$ respectively then I need $|I|=|J|=|L|$ and $Icap L = emptyset$. Furthermore, I need
$$
f(x)+h(x)+g(x) = 1
$$
for all $x in J$.
I guess I am getting stuck since the only bump function I know of is the usual
$$
b(x) = expbigg(1-frac{1}{1-big(frac{x-c}{r}big)^2}bigg)
$$
where $c$ is the peak and $r$ is the radius. I have been trying to combine these to get what I want but I haven't been having a lot of success.
Cheers in advance for any help.
real-analysis functional-analysis
asked Jan 18 at 6:54
Jandré SnymanJandré Snyman
17910
$endgroup$
add a comment |
$begingroup$
I am trying to create a superposition of bump functions that adds identically to 1. Specifically I am looking to add two bump functions, say $f(x)$, $h(x)$ and $g(x)$ so that if $I,J,L subset mathbb{R}$ are the support of $f,h$ and $g$ respectively then I need $|I|=|J|=|L|$ and $Icap L = emptyset$. Furthermore, I need
$$
f(x)+h(x)+g(x) = 1
$$
for all $x in J$.
I guess I am getting stuck since the only bump function I know of is the usual
$$
b(x) = expbigg(1-frac{1}{1-big(frac{x-c}{r}big)^2}bigg)
$$
where $c$ is the peak and $r$ is the radius. I have been trying to combine these to get what I want but I haven't been having a lot of success.
Cheers in advance for any help.
real-analysis functional-analysis
asked Jan 18 at 6:54
Jandré SnymanJandré Snyman
17910
$endgroup$
add a comment |
$begingroup$
I am trying to create a superposition of bump functions that adds identically to 1. Specifically I am looking to add two bump functions, say $f(x)$, $h(x)$ and $g(x)$ so that if $I,J,L subset mathbb{R}$ are the support of $f,h$ and $g$ respectively then I need $|I|=|J|=|L|$ and $Icap L = emptyset$. Furthermore, I need
$$
f(x)+h(x)+g(x) = 1
$$
for all $x in J$.
I guess I am getting stuck since the only bump function I know of is the usual
$$
b(x) = expbigg(1-frac{1}{1-big(frac{x-c}{r}big)^2}bigg)
$$
where $c$ is the peak and $r$ is the radius. I have been trying to combine these to get what I want but I haven't been having a lot of success.
Cheers in advance for any help.
real-analysis functional-analysis
asked Jan 18 at 6:54
Jandré SnymanJandré Snyman
17910
$endgroup$
I am trying to create a superposition of bump functions that adds identically to 1. Specifically I am looking to add two bump functions, say $f(x)$, $h(x)$ and $g(x)$ so that if $I,J,L subset mathbb{R}$ are the support of $f,h$ and $g$ respectively then I need $|I|=|J|=|L|$ and $Icap L = emptyset$. Furthermore, I need
$$
f(x)+h(x)+g(x) = 1
$$
for all $x in J$.
I guess I am getting stuck since the only bump function I know of is the usual
$$
b(x) = expbigg(1-frac{1}{1-big(frac{x-c}{r}big)^2}bigg)
$$
where $c$ is the peak and $r$ is the radius. I have been trying to combine these to get what I want but I haven't been having a lot of success.
Cheers in advance for any help.
real-analysis functional-analysis
real-analysis functional-analysis
asked Jan 18 at 6:54
Jandré SnymanJandré Snyman
17910
asked Jan 18 at 6:54
Jandré SnymanJandré Snyman
17910
asked Jan 18 at 6:54
Jandré SnymanJandré Snyman
17910
asked Jan 18 at 6:54
Jandré SnymanJandré Snyman
17910
asked Jan 18 at 6:54
Jandré SnymanJandré Snyman
17910
17910
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
So, what you need are some tools for creating a larger variety of bump functions. The most important tool? Convolution $f*g(x)=int_{mathbb{R}}f(t)g(x-t),dt$; if $f$ is $C^{infty}$ and $g$ is anything at all reasonable, $f*g$ is also $C^{infty}$. We don't need even one derivative in $f$ to make that work. A bump function is also compactly supported; for that, note that the support of $f*g$ is contained in the sum of the support of $f$ and the support of $g$ - so if $f$ is a bump function and $g$ is compactly supported, $f*g$ is a bump function.
OK, now to the problem. We want three bump functions $f,h,g$ that sum to $1$ on an interval, such that the outer two have disjoint support. Their sum $s$ is, of course, also a bump function, so let's find that first.
So, how do we get a constant out of the convolution? We convolve with a constant; $b*1(x)=int_mathbb{R}b$, independent of $x$. Obviously, we don't want this on the whole real line - but if, instead, we convolve with the characteristic function of an interval $U=(rho-gamma,rho+gamma)$ larger than the support $B=(r-c,r+c)$ of $b$, we'll get a bump function which is constant on some smaller interval, of length $|U|-|B|$. So then, let $$u(x)=begin{cases}left(int_{mathbb{R}}bright)^{-1}&xin U\0&text{otherwise}end{cases}$$
and $s=b*u$. This $s$ is supported on $[r+rho-c-gamma,r+rho+c+gamma]$ and is equal to $1$ on $[r+rho-gamma+c,r+rho+gamma-c]$.
Now we need to break that $s$ up into a sum $f+h+g$, supported on three sets $I,J,L$ respectively of equal size, with $I$ and $L$ disjoint, and $J$ contained in that middle set where the sum is $1$. The obvious thing to do is to split $u$ into three equal parts, each supported on an interval of length $frac23gamma$. The first, leading to $f$ and $I$, will be $left(int_{mathbb{R}}bright)^{-1}$ times the characteristic function of $(rho-gamma,rho-frac13gamma]$, so we will have $I=[r+rho-gamma-c,r+rho-frac13gamma+c]$. Similarly, $J=[r+rho-frac13gamma-c,r+rho+frac13gamma+c]$ and $L=[r+rho+frac13gamma-c,r+rho+gamma+c]$. For this all to work out, we need ($I$ and $L$ disjoint)
$$r+rho-frac13gamma+c < r+rho+frac13gamma-c$$
$$2c<frac23gamma$$
and (lower endpoint of $J$ in the region where the sum is $1$)
$$r+rho-gamma+c < r+rho-frac13gamma-c$$
$$2c < frac23gamma$$
and another for the upper endpoint of $J$, which will simplify to the same condition as the two we already have. Basically, this three-part decomposition works as long as each of the individual pieces is large enough to have its own region where the convolution is constant.
answered Jan 18 at 9:54
jmerryjmerry
10.8k1225
$endgroup$
$begingroup$
Thanks for the incredibly detailed answer, more than I thought I would get. Tell me, will this method be able to be extended to as many of these functions as I need on a given interval? Say I wanted to fill the interval (0,2pi) with functions like this each with support on an interval of length 1/k for some real k, and no function has support overlapping with more than two others?
$endgroup$
– Jandré Snyman
Jan 18 at 10:08
$begingroup$
Yes, the method would work for that as well.
$endgroup$
– jmerry
Jan 18 at 10:11
$begingroup$
Cheers mate!!! I wish I could double upvote your answer.
$endgroup$
– Jandré Snyman
Jan 18 at 10:11
add a comment |
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$begingroup$
So, what you need are some tools for creating a larger variety of bump functions. The most important tool? Convolution $f*g(x)=int_{mathbb{R}}f(t)g(x-t),dt$; if $f$ is $C^{infty}$ and $g$ is anything at all reasonable, $f*g$ is also $C^{infty}$. We don't need even one derivative in $f$ to make that work. A bump function is also compactly supported; for that, note that the support of $f*g$ is contained in the sum of the support of $f$ and the support of $g$ - so if $f$ is a bump function and $g$ is compactly supported, $f*g$ is a bump function.
OK, now to the problem. We want three bump functions $f,h,g$ that sum to $1$ on an interval, such that the outer two have disjoint support. Their sum $s$ is, of course, also a bump function, so let's find that first.
So, how do we get a constant out of the convolution? We convolve with a constant; $b*1(x)=int_mathbb{R}b$, independent of $x$. Obviously, we don't want this on the whole real line - but if, instead, we convolve with the characteristic function of an interval $U=(rho-gamma,rho+gamma)$ larger than the support $B=(r-c,r+c)$ of $b$, we'll get a bump function which is constant on some smaller interval, of length $|U|-|B|$. So then, let $$u(x)=begin{cases}left(int_{mathbb{R}}bright)^{-1}&xin U\0&text{otherwise}end{cases}$$
and $s=b*u$. This $s$ is supported on $[r+rho-c-gamma,r+rho+c+gamma]$ and is equal to $1$ on $[r+rho-gamma+c,r+rho+gamma-c]$.
Now we need to break that $s$ up into a sum $f+h+g$, supported on three sets $I,J,L$ respectively of equal size, with $I$ and $L$ disjoint, and $J$ contained in that middle set where the sum is $1$. The obvious thing to do is to split $u$ into three equal parts, each supported on an interval of length $frac23gamma$. The first, leading to $f$ and $I$, will be $left(int_{mathbb{R}}bright)^{-1}$ times the characteristic function of $(rho-gamma,rho-frac13gamma]$, so we will have $I=[r+rho-gamma-c,r+rho-frac13gamma+c]$. Similarly, $J=[r+rho-frac13gamma-c,r+rho+frac13gamma+c]$ and $L=[r+rho+frac13gamma-c,r+rho+gamma+c]$. For this all to work out, we need ($I$ and $L$ disjoint)
$$r+rho-frac13gamma+c < r+rho+frac13gamma-c$$
$$2c<frac23gamma$$
and (lower endpoint of $J$ in the region where the sum is $1$)
$$r+rho-gamma+c < r+rho-frac13gamma-c$$
$$2c < frac23gamma$$
and another for the upper endpoint of $J$, which will simplify to the same condition as the two we already have. Basically, this three-part decomposition works as long as each of the individual pieces is large enough to have its own region where the convolution is constant.
answered Jan 18 at 9:54
jmerryjmerry
10.8k1225
$endgroup$
$begingroup$
Thanks for the incredibly detailed answer, more than I thought I would get. Tell me, will this method be able to be extended to as many of these functions as I need on a given interval? Say I wanted to fill the interval (0,2pi) with functions like this each with support on an interval of length 1/k for some real k, and no function has support overlapping with more than two others?
$endgroup$
– Jandré Snyman
Jan 18 at 10:08
$begingroup$
Yes, the method would work for that as well.
$endgroup$
– jmerry
Jan 18 at 10:11
$begingroup$
Cheers mate!!! I wish I could double upvote your answer.
$endgroup$
– Jandré Snyman
Jan 18 at 10:11
add a comment |
$begingroup$
So, what you need are some tools for creating a larger variety of bump functions. The most important tool? Convolution $f*g(x)=int_{mathbb{R}}f(t)g(x-t),dt$; if $f$ is $C^{infty}$ and $g$ is anything at all reasonable, $f*g$ is also $C^{infty}$. We don't need even one derivative in $f$ to make that work. A bump function is also compactly supported; for that, note that the support of $f*g$ is contained in the sum of the support of $f$ and the support of $g$ - so if $f$ is a bump function and $g$ is compactly supported, $f*g$ is a bump function.
OK, now to the problem. We want three bump functions $f,h,g$ that sum to $1$ on an interval, such that the outer two have disjoint support. Their sum $s$ is, of course, also a bump function, so let's find that first.
So, how do we get a constant out of the convolution? We convolve with a constant; $b*1(x)=int_mathbb{R}b$, independent of $x$. Obviously, we don't want this on the whole real line - but if, instead, we convolve with the characteristic function of an interval $U=(rho-gamma,rho+gamma)$ larger than the support $B=(r-c,r+c)$ of $b$, we'll get a bump function which is constant on some smaller interval, of length $|U|-|B|$. So then, let $$u(x)=begin{cases}left(int_{mathbb{R}}bright)^{-1}&xin U\0&text{otherwise}end{cases}$$
and $s=b*u$. This $s$ is supported on $[r+rho-c-gamma,r+rho+c+gamma]$ and is equal to $1$ on $[r+rho-gamma+c,r+rho+gamma-c]$.
Now we need to break that $s$ up into a sum $f+h+g$, supported on three sets $I,J,L$ respectively of equal size, with $I$ and $L$ disjoint, and $J$ contained in that middle set where the sum is $1$. The obvious thing to do is to split $u$ into three equal parts, each supported on an interval of length $frac23gamma$. The first, leading to $f$ and $I$, will be $left(int_{mathbb{R}}bright)^{-1}$ times the characteristic function of $(rho-gamma,rho-frac13gamma]$, so we will have $I=[r+rho-gamma-c,r+rho-frac13gamma+c]$. Similarly, $J=[r+rho-frac13gamma-c,r+rho+frac13gamma+c]$ and $L=[r+rho+frac13gamma-c,r+rho+gamma+c]$. For this all to work out, we need ($I$ and $L$ disjoint)
$$r+rho-frac13gamma+c < r+rho+frac13gamma-c$$
$$2c<frac23gamma$$
and (lower endpoint of $J$ in the region where the sum is $1$)
$$r+rho-gamma+c < r+rho-frac13gamma-c$$
$$2c < frac23gamma$$
and another for the upper endpoint of $J$, which will simplify to the same condition as the two we already have. Basically, this three-part decomposition works as long as each of the individual pieces is large enough to have its own region where the convolution is constant.
answered Jan 18 at 9:54
jmerryjmerry
10.8k1225
$endgroup$
$begingroup$
Thanks for the incredibly detailed answer, more than I thought I would get. Tell me, will this method be able to be extended to as many of these functions as I need on a given interval? Say I wanted to fill the interval (0,2pi) with functions like this each with support on an interval of length 1/k for some real k, and no function has support overlapping with more than two others?
$endgroup$
– Jandré Snyman
Jan 18 at 10:08
$begingroup$
Yes, the method would work for that as well.
$endgroup$
– jmerry
Jan 18 at 10:11
$begingroup$
Cheers mate!!! I wish I could double upvote your answer.
$endgroup$
– Jandré Snyman
Jan 18 at 10:11
add a comment |
$begingroup$
So, what you need are some tools for creating a larger variety of bump functions. The most important tool? Convolution $f*g(x)=int_{mathbb{R}}f(t)g(x-t),dt$; if $f$ is $C^{infty}$ and $g$ is anything at all reasonable, $f*g$ is also $C^{infty}$. We don't need even one derivative in $f$ to make that work. A bump function is also compactly supported; for that, note that the support of $f*g$ is contained in the sum of the support of $f$ and the support of $g$ - so if $f$ is a bump function and $g$ is compactly supported, $f*g$ is a bump function.
OK, now to the problem. We want three bump functions $f,h,g$ that sum to $1$ on an interval, such that the outer two have disjoint support. Their sum $s$ is, of course, also a bump function, so let's find that first.
So, how do we get a constant out of the convolution? We convolve with a constant; $b*1(x)=int_mathbb{R}b$, independent of $x$. Obviously, we don't want this on the whole real line - but if, instead, we convolve with the characteristic function of an interval $U=(rho-gamma,rho+gamma)$ larger than the support $B=(r-c,r+c)$ of $b$, we'll get a bump function which is constant on some smaller interval, of length $|U|-|B|$. So then, let $$u(x)=begin{cases}left(int_{mathbb{R}}bright)^{-1}&xin U\0&text{otherwise}end{cases}$$
and $s=b*u$. This $s$ is supported on $[r+rho-c-gamma,r+rho+c+gamma]$ and is equal to $1$ on $[r+rho-gamma+c,r+rho+gamma-c]$.
Now we need to break that $s$ up into a sum $f+h+g$, supported on three sets $I,J,L$ respectively of equal size, with $I$ and $L$ disjoint, and $J$ contained in that middle set where the sum is $1$. The obvious thing to do is to split $u$ into three equal parts, each supported on an interval of length $frac23gamma$. The first, leading to $f$ and $I$, will be $left(int_{mathbb{R}}bright)^{-1}$ times the characteristic function of $(rho-gamma,rho-frac13gamma]$, so we will have $I=[r+rho-gamma-c,r+rho-frac13gamma+c]$. Similarly, $J=[r+rho-frac13gamma-c,r+rho+frac13gamma+c]$ and $L=[r+rho+frac13gamma-c,r+rho+gamma+c]$. For this all to work out, we need ($I$ and $L$ disjoint)
$$r+rho-frac13gamma+c < r+rho+frac13gamma-c$$
$$2c<frac23gamma$$
and (lower endpoint of $J$ in the region where the sum is $1$)
$$r+rho-gamma+c < r+rho-frac13gamma-c$$
$$2c < frac23gamma$$
and another for the upper endpoint of $J$, which will simplify to the same condition as the two we already have. Basically, this three-part decomposition works as long as each of the individual pieces is large enough to have its own region where the convolution is constant.
answered Jan 18 at 9:54
jmerryjmerry
10.8k1225
$endgroup$
So, what you need are some tools for creating a larger variety of bump functions. The most important tool? Convolution $f*g(x)=int_{mathbb{R}}f(t)g(x-t),dt$; if $f$ is $C^{infty}$ and $g$ is anything at all reasonable, $f*g$ is also $C^{infty}$. We don't need even one derivative in $f$ to make that work. A bump function is also compactly supported; for that, note that the support of $f*g$ is contained in the sum of the support of $f$ and the support of $g$ - so if $f$ is a bump function and $g$ is compactly supported, $f*g$ is a bump function.
OK, now to the problem. We want three bump functions $f,h,g$ that sum to $1$ on an interval, such that the outer two have disjoint support. Their sum $s$ is, of course, also a bump function, so let's find that first.
So, how do we get a constant out of the convolution? We convolve with a constant; $b*1(x)=int_mathbb{R}b$, independent of $x$. Obviously, we don't want this on the whole real line - but if, instead, we convolve with the characteristic function of an interval $U=(rho-gamma,rho+gamma)$ larger than the support $B=(r-c,r+c)$ of $b$, we'll get a bump function which is constant on some smaller interval, of length $|U|-|B|$. So then, let $$u(x)=begin{cases}left(int_{mathbb{R}}bright)^{-1}&xin U\0&text{otherwise}end{cases}$$
and $s=b*u$. This $s$ is supported on $[r+rho-c-gamma,r+rho+c+gamma]$ and is equal to $1$ on $[r+rho-gamma+c,r+rho+gamma-c]$.
Now we need to break that $s$ up into a sum $f+h+g$, supported on three sets $I,J,L$ respectively of equal size, with $I$ and $L$ disjoint, and $J$ contained in that middle set where the sum is $1$. The obvious thing to do is to split $u$ into three equal parts, each supported on an interval of length $frac23gamma$. The first, leading to $f$ and $I$, will be $left(int_{mathbb{R}}bright)^{-1}$ times the characteristic function of $(rho-gamma,rho-frac13gamma]$, so we will have $I=[r+rho-gamma-c,r+rho-frac13gamma+c]$. Similarly, $J=[r+rho-frac13gamma-c,r+rho+frac13gamma+c]$ and $L=[r+rho+frac13gamma-c,r+rho+gamma+c]$. For this all to work out, we need ($I$ and $L$ disjoint)
$$r+rho-frac13gamma+c < r+rho+frac13gamma-c$$
$$2c<frac23gamma$$
and (lower endpoint of $J$ in the region where the sum is $1$)
$$r+rho-gamma+c < r+rho-frac13gamma-c$$
$$2c < frac23gamma$$
and another for the upper endpoint of $J$, which will simplify to the same condition as the two we already have. Basically, this three-part decomposition works as long as each of the individual pieces is large enough to have its own region where the convolution is constant.
answered Jan 18 at 9:54
jmerryjmerry
10.8k1225
answered Jan 18 at 9:54
jmerryjmerry
10.8k1225
answered Jan 18 at 9:54
jmerryjmerry
10.8k1225
answered Jan 18 at 9:54
jmerryjmerry
10.8k1225
10.8k1225
$begingroup$
Thanks for the incredibly detailed answer, more than I thought I would get. Tell me, will this method be able to be extended to as many of these functions as I need on a given interval? Say I wanted to fill the interval (0,2pi) with functions like this each with support on an interval of length 1/k for some real k, and no function has support overlapping with more than two others?
$endgroup$
– Jandré Snyman
Jan 18 at 10:08
$begingroup$
Yes, the method would work for that as well.
$endgroup$
– jmerry
Jan 18 at 10:11
$begingroup$
Cheers mate!!! I wish I could double upvote your answer.
$endgroup$
– Jandré Snyman
Jan 18 at 10:11
add a comment |
$begingroup$
Thanks for the incredibly detailed answer, more than I thought I would get. Tell me, will this method be able to be extended to as many of these functions as I need on a given interval? Say I wanted to fill the interval (0,2pi) with functions like this each with support on an interval of length 1/k for some real k, and no function has support overlapping with more than two others?
$endgroup$
– Jandré Snyman
Jan 18 at 10:08
$begingroup$
Yes, the method would work for that as well.
$endgroup$
– jmerry
Jan 18 at 10:11
$begingroup$
Cheers mate!!! I wish I could double upvote your answer.
$endgroup$
– Jandré Snyman
Jan 18 at 10:11
$begingroup$
Thanks for the incredibly detailed answer, more than I thought I would get. Tell me, will this method be able to be extended to as many of these functions as I need on a given interval? Say I wanted to fill the interval (0,2pi) with functions like this each with support on an interval of length 1/k for some real k, and no function has support overlapping with more than two others?
$endgroup$
– Jandré Snyman
Jan 18 at 10:08
$begingroup$
Thanks for the incredibly detailed answer, more than I thought I would get. Tell me, will this method be able to be extended to as many of these functions as I need on a given interval? Say I wanted to fill the interval (0,2pi) with functions like this each with support on an interval of length 1/k for some real k, and no function has support overlapping with more than two others?
$endgroup$
– Jandré Snyman
Jan 18 at 10:08
$begingroup$
Yes, the method would work for that as well.
$endgroup$
– jmerry
Jan 18 at 10:11
$begingroup$
Yes, the method would work for that as well.
$endgroup$
– jmerry
Jan 18 at 10:11
$begingroup$
Cheers mate!!! I wish I could double upvote your answer.
$endgroup$
– Jandré Snyman
Jan 18 at 10:11
$begingroup$
Cheers mate!!! I wish I could double upvote your answer.
$endgroup$
– Jandré Snyman
Jan 18 at 10:11
add a comment |
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StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
