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How to prove that $ ||T_lambdaf||_L^q(mathbbR^n)leq C_p, qlambda^n(1/p-1/q)||f||_L^p(mathbbR^n),quad lambda>0. $
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Suppose $ hin C_0^infty(mathbbR) $ supported in $ (0, infty) $ and for $
lambda >0 $
$$
T_lambdaf(x)=(2pi)^-nint_mathbbR^n e^ixcdotxihleft(fraclambdaright)hatf(xi)dxi
$$
Where $$ hatf(xi)=int_mathbbR^ne^-ixcdot xif(x)dx .$$
Prove that if $ 1leq pleq qleq infty $ there is a uniform constant $ C_p, q $ so that
$$
||T_lambdaf||_L^q(mathbbR^n)leq C_p, qlambda^n(1/p-1/q)||f||_L^p(mathbbR^n),quad lambda>0.
$$
I think it maybe needs some interpolation theorems, but how to do so?
functional-analysis analysis fourier-analysis lp-spaces harmonic-analysis
edited Jul 23 at 17:33
Davide Giraudo
121k15146249
asked Jul 23 at 5:45
A beginer
235
add a comment |Â
up vote
4
down vote
favorite
3
Suppose $ hin C_0^infty(mathbbR) $ supported in $ (0, infty) $ and for $
lambda >0 $
$$
T_lambdaf(x)=(2pi)^-nint_mathbbR^n e^ixcdotxihleft(fraclambdaright)hatf(xi)dxi
$$
Where $$ hatf(xi)=int_mathbbR^ne^-ixcdot xif(x)dx .$$
Prove that if $ 1leq pleq qleq infty $ there is a uniform constant $ C_p, q $ so that
$$
||T_lambdaf||_L^q(mathbbR^n)leq C_p, qlambda^n(1/p-1/q)||f||_L^p(mathbbR^n),quad lambda>0.
$$
I think it maybe needs some interpolation theorems, but how to do so?
functional-analysis analysis fourier-analysis lp-spaces harmonic-analysis
edited Jul 23 at 17:33
Davide Giraudo
121k15146249
asked Jul 23 at 5:45
A beginer
235
If you start with an $mathitL^p$ function, the Fourier transform does not need to be a function again, but might be a distribution. In which sense is the above integral defined then?
– Jonas Lenz
Jul 23 at 5:49
Why? If $f$ is in $mathitL^p$ but not integrable, the right hand side is perfectly finite.
– Jonas Lenz
Jul 23 at 6:49
@JonasLenz I really don't know what you mean, I didn't meet any trouble integrating $ f $, you should know that integrable can refer to both finite and infinite situations and this need to be discussed in your proof. And you'd better compose an answer rather than querying me at comments.
– A beginer
Jul 23 at 6:55
Actually, I do not know the answer, that is why I am not writing an answer. Maybe we should first check terminology. For me, $f$ is called integrable if $int_mathbbR^n |f| mathrmdlambda$ is finite. Similarly, $fin mathitL^p$ if $int_mathbbR^n |f|^p mathrmdlambda $ is finite. I am not completely sure what you mean with finite and infinite situation, could you please clarify? For which functions do you want to show the estimate? From the above I would assume $fin mathitL^p$, right?
– Jonas Lenz
Jul 23 at 7:16
add a comment |Â
up vote
4
down vote
favorite
3
up vote
4
down vote
favorite
3
3
Suppose $ hin C_0^infty(mathbbR) $ supported in $ (0, infty) $ and for $
lambda >0 $
$$
T_lambdaf(x)=(2pi)^-nint_mathbbR^n e^ixcdotxihleft(fraclambdaright)hatf(xi)dxi
$$
Where $$ hatf(xi)=int_mathbbR^ne^-ixcdot xif(x)dx .$$
Prove that if $ 1leq pleq qleq infty $ there is a uniform constant $ C_p, q $ so that
$$
||T_lambdaf||_L^q(mathbbR^n)leq C_p, qlambda^n(1/p-1/q)||f||_L^p(mathbbR^n),quad lambda>0.
$$
I think it maybe needs some interpolation theorems, but how to do so?
functional-analysis analysis fourier-analysis lp-spaces harmonic-analysis
edited Jul 23 at 17:33
Davide Giraudo
121k15146249
asked Jul 23 at 5:45
A beginer
235
Suppose $ hin C_0^infty(mathbbR) $ supported in $ (0, infty) $ and for $
lambda >0 $
$$
T_lambdaf(x)=(2pi)^-nint_mathbbR^n e^ixcdotxihleft(fraclambdaright)hatf(xi)dxi
$$
Where $$ hatf(xi)=int_mathbbR^ne^-ixcdot xif(x)dx .$$
Prove that if $ 1leq pleq qleq infty $ there is a uniform constant $ C_p, q $ so that
$$
||T_lambdaf||_L^q(mathbbR^n)leq C_p, qlambda^n(1/p-1/q)||f||_L^p(mathbbR^n),quad lambda>0.
$$
I think it maybe needs some interpolation theorems, but how to do so?
functional-analysis analysis fourier-analysis lp-spaces harmonic-analysis
edited Jul 23 at 17:33
Davide Giraudo
121k15146249
asked Jul 23 at 5:45
A beginer
235
edited Jul 23 at 17:33
Davide Giraudo
121k15146249
edited Jul 23 at 17:33
Davide Giraudo
121k15146249
edited Jul 23 at 17:33
Davide Giraudo
121k15146249
121k15146249
asked Jul 23 at 5:45
A beginer
235
asked Jul 23 at 5:45
A beginer
235
asked Jul 23 at 5:45
A beginer
235
235
If you start with an $mathitL^p$ function, the Fourier transform does not need to be a function again, but might be a distribution. In which sense is the above integral defined then?
– Jonas Lenz
Jul 23 at 5:49
Why? If $f$ is in $mathitL^p$ but not integrable, the right hand side is perfectly finite.
– Jonas Lenz
Jul 23 at 6:49
@JonasLenz I really don't know what you mean, I didn't meet any trouble integrating $ f $, you should know that integrable can refer to both finite and infinite situations and this need to be discussed in your proof. And you'd better compose an answer rather than querying me at comments.
– A beginer
Jul 23 at 6:55
Actually, I do not know the answer, that is why I am not writing an answer. Maybe we should first check terminology. For me, $f$ is called integrable if $int_mathbbR^n |f| mathrmdlambda$ is finite. Similarly, $fin mathitL^p$ if $int_mathbbR^n |f|^p mathrmdlambda $ is finite. I am not completely sure what you mean with finite and infinite situation, could you please clarify? For which functions do you want to show the estimate? From the above I would assume $fin mathitL^p$, right?
– Jonas Lenz
Jul 23 at 7:16
add a comment |Â
If you start with an $mathitL^p$ function, the Fourier transform does not need to be a function again, but might be a distribution. In which sense is the above integral defined then?
– Jonas Lenz
Jul 23 at 5:49
Why? If $f$ is in $mathitL^p$ but not integrable, the right hand side is perfectly finite.
– Jonas Lenz
Jul 23 at 6:49
@JonasLenz I really don't know what you mean, I didn't meet any trouble integrating $ f $, you should know that integrable can refer to both finite and infinite situations and this need to be discussed in your proof. And you'd better compose an answer rather than querying me at comments.
– A beginer
Jul 23 at 6:55
Actually, I do not know the answer, that is why I am not writing an answer. Maybe we should first check terminology. For me, $f$ is called integrable if $int_mathbbR^n |f| mathrmdlambda$ is finite. Similarly, $fin mathitL^p$ if $int_mathbbR^n |f|^p mathrmdlambda $ is finite. I am not completely sure what you mean with finite and infinite situation, could you please clarify? For which functions do you want to show the estimate? From the above I would assume $fin mathitL^p$, right?
– Jonas Lenz
Jul 23 at 7:16
If you start with an $mathitL^p$ function, the Fourier transform does not need to be a function again, but might be a distribution. In which sense is the above integral defined then?
– Jonas Lenz
Jul 23 at 5:49
If you start with an $mathitL^p$ function, the Fourier transform does not need to be a function again, but might be a distribution. In which sense is the above integral defined then?
– Jonas Lenz
Jul 23 at 5:49
Why? If $f$ is in $mathitL^p$ but not integrable, the right hand side is perfectly finite.
– Jonas Lenz
Jul 23 at 6:49
Why? If $f$ is in $mathitL^p$ but not integrable, the right hand side is perfectly finite.
– Jonas Lenz
Jul 23 at 6:49
@JonasLenz I really don't know what you mean, I didn't meet any trouble integrating $ f $, you should know that integrable can refer to both finite and infinite situations and this need to be discussed in your proof. And you'd better compose an answer rather than querying me at comments.
– A beginer
Jul 23 at 6:55
@JonasLenz I really don't know what you mean, I didn't meet any trouble integrating $ f $, you should know that integrable can refer to both finite and infinite situations and this need to be discussed in your proof. And you'd better compose an answer rather than querying me at comments.
– A beginer
Jul 23 at 6:55
Actually, I do not know the answer, that is why I am not writing an answer. Maybe we should first check terminology. For me, $f$ is called integrable if $int_mathbbR^n |f| mathrmdlambda$ is finite. Similarly, $fin mathitL^p$ if $int_mathbbR^n |f|^p mathrmdlambda $ is finite. I am not completely sure what you mean with finite and infinite situation, could you please clarify? For which functions do you want to show the estimate? From the above I would assume $fin mathitL^p$, right?
– Jonas Lenz
Jul 23 at 7:16
Actually, I do not know the answer, that is why I am not writing an answer. Maybe we should first check terminology. For me, $f$ is called integrable if $int_mathbbR^n |f| mathrmdlambda$ is finite. Similarly, $fin mathitL^p$ if $int_mathbbR^n |f|^p mathrmdlambda $ is finite. I am not completely sure what you mean with finite and infinite situation, could you please clarify? For which functions do you want to show the estimate? From the above I would assume $fin mathitL^p$, right?
– Jonas Lenz
Jul 23 at 7:16
add a comment |Â
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If you start with an $mathitL^p$ function, the Fourier transform does not need to be a function again, but might be a distribution. In which sense is the above integral defined then?
– Jonas Lenz
Jul 23 at 5:49
Why? If $f$ is in $mathitL^p$ but not integrable, the right hand side is perfectly finite.
– Jonas Lenz
Jul 23 at 6:49
@JonasLenz I really don't know what you mean, I didn't meet any trouble integrating $ f $, you should know that integrable can refer to both finite and infinite situations and this need to be discussed in your proof. And you'd better compose an answer rather than querying me at comments.
– A beginer
Jul 23 at 6:55
Actually, I do not know the answer, that is why I am not writing an answer. Maybe we should first check terminology. For me, $f$ is called integrable if $int_mathbbR^n |f| mathrmdlambda$ is finite. Similarly, $fin mathitL^p$ if $int_mathbbR^n |f|^p mathrmdlambda $ is finite. I am not completely sure what you mean with finite and infinite situation, could you please clarify? For which functions do you want to show the estimate? From the above I would assume $fin mathitL^p$, right?
– Jonas Lenz
Jul 23 at 7:16