Evaluating convolution $fstar g:=int_{mathbb{R}^d}chi_{[-c,c]}(x-y)chi_{[0,infty]}(y)e^{-y} dlambda_d(y)$
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$f(x)=chi_{[-c,c]}(x), c>0$
$g(x)=chi_{[0,infty)}(x)e^{-x}$
where $chi$ denotes the indicator function.
I want to determine the convolution $fstar g:=int_{mathbb{R}^d}chi_{[-c,c]}(x-y)chi_{[0,infty]}(y)e^{-y} dlambda_d(y)$
I looked at both indicator functions:
$chi_{[-c,c]}(x-y)=chi_{[x-c,x+c]}(y)$ and
$chi_{[0,infty)}(y)$
So if I didn't make any mistake the integrand $chi_{[-c,c]}(x-y)chi_{[0,infty]}(y)e^{-y}neq0 Leftrightarrow y in [max(x-c,0),x+c]$.
Then $f star g=int_{max(x-c,0)}^{x+c}e^{-y}dy$ and otherwise $fstar g=0$
Is that correct? Thanks for your help!
calculus real-analysis complex-analysis analysis
asked Nov 17 at 19:35
conrad
707
add a comment |
up vote
0
down vote
favorite
$f(x)=chi_{[-c,c]}(x), c>0$
$g(x)=chi_{[0,infty)}(x)e^{-x}$
where $chi$ denotes the indicator function.
I want to determine the convolution $fstar g:=int_{mathbb{R}^d}chi_{[-c,c]}(x-y)chi_{[0,infty]}(y)e^{-y} dlambda_d(y)$
I looked at both indicator functions:
$chi_{[-c,c]}(x-y)=chi_{[x-c,x+c]}(y)$ and
$chi_{[0,infty)}(y)$
So if I didn't make any mistake the integrand $chi_{[-c,c]}(x-y)chi_{[0,infty]}(y)e^{-y}neq0 Leftrightarrow y in [max(x-c,0),x+c]$.
Then $f star g=int_{max(x-c,0)}^{x+c}e^{-y}dy$ and otherwise $fstar g=0$
Is that correct? Thanks for your help!
calculus real-analysis complex-analysis analysis
asked Nov 17 at 19:35
conrad
707
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
$f(x)=chi_{[-c,c]}(x), c>0$
$g(x)=chi_{[0,infty)}(x)e^{-x}$
where $chi$ denotes the indicator function.
I want to determine the convolution $fstar g:=int_{mathbb{R}^d}chi_{[-c,c]}(x-y)chi_{[0,infty]}(y)e^{-y} dlambda_d(y)$
I looked at both indicator functions:
$chi_{[-c,c]}(x-y)=chi_{[x-c,x+c]}(y)$ and
$chi_{[0,infty)}(y)$
So if I didn't make any mistake the integrand $chi_{[-c,c]}(x-y)chi_{[0,infty]}(y)e^{-y}neq0 Leftrightarrow y in [max(x-c,0),x+c]$.
Then $f star g=int_{max(x-c,0)}^{x+c}e^{-y}dy$ and otherwise $fstar g=0$
Is that correct? Thanks for your help!
calculus real-analysis complex-analysis analysis
asked Nov 17 at 19:35
conrad
707
$f(x)=chi_{[-c,c]}(x), c>0$
$g(x)=chi_{[0,infty)}(x)e^{-x}$
where $chi$ denotes the indicator function.
I want to determine the convolution $fstar g:=int_{mathbb{R}^d}chi_{[-c,c]}(x-y)chi_{[0,infty]}(y)e^{-y} dlambda_d(y)$
I looked at both indicator functions:
$chi_{[-c,c]}(x-y)=chi_{[x-c,x+c]}(y)$ and
$chi_{[0,infty)}(y)$
So if I didn't make any mistake the integrand $chi_{[-c,c]}(x-y)chi_{[0,infty]}(y)e^{-y}neq0 Leftrightarrow y in [max(x-c,0),x+c]$.
Then $f star g=int_{max(x-c,0)}^{x+c}e^{-y}dy$ and otherwise $fstar g=0$
Is that correct? Thanks for your help!
calculus real-analysis complex-analysis analysis
calculus real-analysis complex-analysis analysis
asked Nov 17 at 19:35
conrad
707
asked Nov 17 at 19:35
conrad
707
asked Nov 17 at 19:35
conrad
707
asked Nov 17 at 19:35
conrad
707
asked Nov 17 at 19:35
conrad
707
707
add a comment |
add a comment |
1 Answer
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Well what about values of $x$ and $c$ such that $x+c<0$? The convolution would certainly be $0$.
Otherwise, this looks right to me. We have that $chi_{[-c,c]}(x-y)=1$ iff $-cleq x-yleq c$ (or $ x-cleq yleq x+c$). Let $M=mathrm{max}{0,x-c}$. Given that we don't have the situation mentioned above, taking $M$ as the lower bound guarantees that $chi_{[0,infty]}(y)=1$ for $yin [M,x+c]$.
answered Nov 18 at 2:37
gHem
583
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Well what about values of $x$ and $c$ such that $x+c<0$? The convolution would certainly be $0$.
Otherwise, this looks right to me. We have that $chi_{[-c,c]}(x-y)=1$ iff $-cleq x-yleq c$ (or $ x-cleq yleq x+c$). Let $M=mathrm{max}{0,x-c}$. Given that we don't have the situation mentioned above, taking $M$ as the lower bound guarantees that $chi_{[0,infty]}(y)=1$ for $yin [M,x+c]$.
answered Nov 18 at 2:37
gHem
583
add a comment |
up vote
0
down vote
Well what about values of $x$ and $c$ such that $x+c<0$? The convolution would certainly be $0$.
Otherwise, this looks right to me. We have that $chi_{[-c,c]}(x-y)=1$ iff $-cleq x-yleq c$ (or $ x-cleq yleq x+c$). Let $M=mathrm{max}{0,x-c}$. Given that we don't have the situation mentioned above, taking $M$ as the lower bound guarantees that $chi_{[0,infty]}(y)=1$ for $yin [M,x+c]$.
answered Nov 18 at 2:37
gHem
583
add a comment |
up vote
0
down vote
up vote
0
down vote
Well what about values of $x$ and $c$ such that $x+c<0$? The convolution would certainly be $0$.
Otherwise, this looks right to me. We have that $chi_{[-c,c]}(x-y)=1$ iff $-cleq x-yleq c$ (or $ x-cleq yleq x+c$). Let $M=mathrm{max}{0,x-c}$. Given that we don't have the situation mentioned above, taking $M$ as the lower bound guarantees that $chi_{[0,infty]}(y)=1$ for $yin [M,x+c]$.
answered Nov 18 at 2:37
gHem
583
Well what about values of $x$ and $c$ such that $x+c<0$? The convolution would certainly be $0$.
Otherwise, this looks right to me. We have that $chi_{[-c,c]}(x-y)=1$ iff $-cleq x-yleq c$ (or $ x-cleq yleq x+c$). Let $M=mathrm{max}{0,x-c}$. Given that we don't have the situation mentioned above, taking $M$ as the lower bound guarantees that $chi_{[0,infty]}(y)=1$ for $yin [M,x+c]$.
answered Nov 18 at 2:37
gHem
583
answered Nov 18 at 2:37
gHem
583
answered Nov 18 at 2:37
gHem
583
answered Nov 18 at 2:37
gHem
583
583
add a comment |
add a comment |
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