Computing the inverse of the operator $L:= f - Delta f$.
$begingroup$
Here is an argument I read for computing the inverse of the differential operator $L:S(Bbb R^n) rightarrow S(Bbb R^n)$ on Schwartz spaces where
$$Lf:= f - Delta f = f- sum_{i=1}^n (D^i)^2 f$$
By the basic rules for the Fourier transform,
$$ widehat{Lf}(xi) = (1+|xi|^2) hat{f}(xi)$$
Therefore, $L$ is invertible with inverse $M$ given by
$$ widehat{Mf} (xi):= (1+|xi|^2)^{-1}hat{f}$$
I get how we got the formula $widehat{Lf}$ and why $M$ exists - using Fourier inversion.
But how is $M$ the inverse of $L$?
functional-analysis fourier-analysis fourier-transform
edited Dec 22 '18 at 15:44
David C. Ullrich
61.8k44095
asked Dec 22 '18 at 9:43
CL.CL.
2,3123925
$endgroup$
add a comment |
$begingroup$
Here is an argument I read for computing the inverse of the differential operator $L:S(Bbb R^n) rightarrow S(Bbb R^n)$ on Schwartz spaces where
$$Lf:= f - Delta f = f- sum_{i=1}^n (D^i)^2 f$$
By the basic rules for the Fourier transform,
$$ widehat{Lf}(xi) = (1+|xi|^2) hat{f}(xi)$$
Therefore, $L$ is invertible with inverse $M$ given by
$$ widehat{Mf} (xi):= (1+|xi|^2)^{-1}hat{f}$$
I get how we got the formula $widehat{Lf}$ and why $M$ exists - using Fourier inversion.
But how is $M$ the inverse of $L$?
functional-analysis fourier-analysis fourier-transform
edited Dec 22 '18 at 15:44
David C. Ullrich
61.8k44095
asked Dec 22 '18 at 9:43
CL.CL.
2,3123925
$endgroup$
$begingroup$
Apply $L$ to $Mf$, use the Fourier transform to see that happens. Or just solve the equation $Lu=f$ using again FT.
$endgroup$
– Andrew
Dec 22 '18 at 9:53
add a comment |
$begingroup$
Here is an argument I read for computing the inverse of the differential operator $L:S(Bbb R^n) rightarrow S(Bbb R^n)$ on Schwartz spaces where
$$Lf:= f - Delta f = f- sum_{i=1}^n (D^i)^2 f$$
By the basic rules for the Fourier transform,
$$ widehat{Lf}(xi) = (1+|xi|^2) hat{f}(xi)$$
Therefore, $L$ is invertible with inverse $M$ given by
$$ widehat{Mf} (xi):= (1+|xi|^2)^{-1}hat{f}$$
I get how we got the formula $widehat{Lf}$ and why $M$ exists - using Fourier inversion.
But how is $M$ the inverse of $L$?
functional-analysis fourier-analysis fourier-transform
edited Dec 22 '18 at 15:44
David C. Ullrich
61.8k44095
asked Dec 22 '18 at 9:43
CL.CL.
2,3123925
$endgroup$
Here is an argument I read for computing the inverse of the differential operator $L:S(Bbb R^n) rightarrow S(Bbb R^n)$ on Schwartz spaces where
$$Lf:= f - Delta f = f- sum_{i=1}^n (D^i)^2 f$$
By the basic rules for the Fourier transform,
$$ widehat{Lf}(xi) = (1+|xi|^2) hat{f}(xi)$$
Therefore, $L$ is invertible with inverse $M$ given by
$$ widehat{Mf} (xi):= (1+|xi|^2)^{-1}hat{f}$$
I get how we got the formula $widehat{Lf}$ and why $M$ exists - using Fourier inversion.
But how is $M$ the inverse of $L$?
functional-analysis fourier-analysis fourier-transform
functional-analysis fourier-analysis fourier-transform
edited Dec 22 '18 at 15:44
David C. Ullrich
61.8k44095
asked Dec 22 '18 at 9:43
CL.CL.
2,3123925
edited Dec 22 '18 at 15:44
David C. Ullrich
61.8k44095
asked Dec 22 '18 at 9:43
CL.CL.
2,3123925
edited Dec 22 '18 at 15:44
David C. Ullrich
61.8k44095
edited Dec 22 '18 at 15:44
David C. Ullrich
61.8k44095
edited Dec 22 '18 at 15:44
David C. Ullrich
61.8k44095
61.8k44095
asked Dec 22 '18 at 9:43
CL.CL.
2,3123925
asked Dec 22 '18 at 9:43
CL.CL.
2,3123925
asked Dec 22 '18 at 9:43
CL.CL.
2,3123925
2,3123925
$begingroup$
Apply $L$ to $Mf$, use the Fourier transform to see that happens. Or just solve the equation $Lu=f$ using again FT.
$endgroup$
– Andrew
Dec 22 '18 at 9:53
add a comment |
$begingroup$
Apply $L$ to $Mf$, use the Fourier transform to see that happens. Or just solve the equation $Lu=f$ using again FT.
$endgroup$
– Andrew
Dec 22 '18 at 9:53
$begingroup$
Apply $L$ to $Mf$, use the Fourier transform to see that happens. Or just solve the equation $Lu=f$ using again FT.
$endgroup$
– Andrew
Dec 22 '18 at 9:53
$begingroup$
Apply $L$ to $Mf$, use the Fourier transform to see that happens. Or just solve the equation $Lu=f$ using again FT.
$endgroup$
– Andrew
Dec 22 '18 at 9:53
add a comment |
1 Answer
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$begingroup$
$widehat {MLf} (xi) =(1+|xi|^{2})^{-1}) widehat {Lf} (xi)=(1+|xi|^{2})^{-1})(1+|xi|^{2})hat {f} (xi)=hat {f} (xi)$ So ML=I. Similarly, $LM=I$.
edited Dec 22 '18 at 15:45
David C. Ullrich
61.8k44095
answered Dec 22 '18 at 11:39
Kavi Rama MurthyKavi Rama Murthy
74.9k53270
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
$widehat {MLf} (xi) =(1+|xi|^{2})^{-1}) widehat {Lf} (xi)=(1+|xi|^{2})^{-1})(1+|xi|^{2})hat {f} (xi)=hat {f} (xi)$ So ML=I. Similarly, $LM=I$.
edited Dec 22 '18 at 15:45
David C. Ullrich
61.8k44095
answered Dec 22 '18 at 11:39
Kavi Rama MurthyKavi Rama Murthy
74.9k53270
$endgroup$
add a comment |
$begingroup$
$widehat {MLf} (xi) =(1+|xi|^{2})^{-1}) widehat {Lf} (xi)=(1+|xi|^{2})^{-1})(1+|xi|^{2})hat {f} (xi)=hat {f} (xi)$ So ML=I. Similarly, $LM=I$.
edited Dec 22 '18 at 15:45
David C. Ullrich
61.8k44095
answered Dec 22 '18 at 11:39
Kavi Rama MurthyKavi Rama Murthy
74.9k53270
$endgroup$
add a comment |
$begingroup$
$widehat {MLf} (xi) =(1+|xi|^{2})^{-1}) widehat {Lf} (xi)=(1+|xi|^{2})^{-1})(1+|xi|^{2})hat {f} (xi)=hat {f} (xi)$ So ML=I. Similarly, $LM=I$.
edited Dec 22 '18 at 15:45
David C. Ullrich
61.8k44095
answered Dec 22 '18 at 11:39
Kavi Rama MurthyKavi Rama Murthy
74.9k53270
$endgroup$
$widehat {MLf} (xi) =(1+|xi|^{2})^{-1}) widehat {Lf} (xi)=(1+|xi|^{2})^{-1})(1+|xi|^{2})hat {f} (xi)=hat {f} (xi)$ So ML=I. Similarly, $LM=I$.
edited Dec 22 '18 at 15:45
David C. Ullrich
61.8k44095
answered Dec 22 '18 at 11:39
Kavi Rama MurthyKavi Rama Murthy
74.9k53270
edited Dec 22 '18 at 15:45
David C. Ullrich
61.8k44095
edited Dec 22 '18 at 15:45
David C. Ullrich
61.8k44095
edited Dec 22 '18 at 15:45
David C. Ullrich
61.8k44095
61.8k44095
answered Dec 22 '18 at 11:39
Kavi Rama MurthyKavi Rama Murthy
74.9k53270
answered Dec 22 '18 at 11:39
Kavi Rama MurthyKavi Rama Murthy
74.9k53270
answered Dec 22 '18 at 11:39
Kavi Rama MurthyKavi Rama Murthy
74.9k53270
74.9k53270
add a comment |
add a comment |
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$begingroup$
Apply $L$ to $Mf$, use the Fourier transform to see that happens. Or just solve the equation $Lu=f$ using again FT.
$endgroup$
– Andrew
Dec 22 '18 at 9:53