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Prove $lim_to infty hat f(alpha )=0$
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Let $fin L^1(mathbb R)$. Set $$hat f(alpha )=int_mathbb Rf(x)e^-2ipi xalpha dx.$$
Prove $$lim_to infty int_mathbb Rf(x)e^-2ipi xalpha dx=0.$$
The thing is even if I put that limit in the integral, it doesn't work, so I don't know how to conclude this part (and it looks strange to me...).
fourier-analysis fourier-transform
asked Jul 20 at 20:40
MSE
1,471315
add a comment |Â
up vote
2
down vote
favorite
Let $fin L^1(mathbb R)$. Set $$hat f(alpha )=int_mathbb Rf(x)e^-2ipi xalpha dx.$$
Prove $$lim_to infty int_mathbb Rf(x)e^-2ipi xalpha dx=0.$$
The thing is even if I put that limit in the integral, it doesn't work, so I don't know how to conclude this part (and it looks strange to me...).
fourier-analysis fourier-transform
asked Jul 20 at 20:40
MSE
1,471315
1
This is the Riemann-Lebesgue lemma. The typical thing to do is to prove it for $C^infty_c(mathbb R)$ functions (where you can just integrate by parts and explicitly take the limit) and pass to $L^1(mathbb R)$ functions by density.
– User8128
Jul 20 at 20:43
This is the Riemann–Lebesgue lemma. Remember that step functions are dense in $L^1$.
– Chappers
Jul 20 at 20:44
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Let $fin L^1(mathbb R)$. Set $$hat f(alpha )=int_mathbb Rf(x)e^-2ipi xalpha dx.$$
Prove $$lim_to infty int_mathbb Rf(x)e^-2ipi xalpha dx=0.$$
The thing is even if I put that limit in the integral, it doesn't work, so I don't know how to conclude this part (and it looks strange to me...).
fourier-analysis fourier-transform
asked Jul 20 at 20:40
MSE
1,471315
Let $fin L^1(mathbb R)$. Set $$hat f(alpha )=int_mathbb Rf(x)e^-2ipi xalpha dx.$$
Prove $$lim_to infty int_mathbb Rf(x)e^-2ipi xalpha dx=0.$$
The thing is even if I put that limit in the integral, it doesn't work, so I don't know how to conclude this part (and it looks strange to me...).
fourier-analysis fourier-transform
asked Jul 20 at 20:40
MSE
1,471315
asked Jul 20 at 20:40
MSE
1,471315
asked Jul 20 at 20:40
MSE
1,471315
asked Jul 20 at 20:40
MSE
1,471315
1,471315
1
This is the Riemann-Lebesgue lemma. The typical thing to do is to prove it for $C^infty_c(mathbb R)$ functions (where you can just integrate by parts and explicitly take the limit) and pass to $L^1(mathbb R)$ functions by density.
– User8128
Jul 20 at 20:43
This is the Riemann–Lebesgue lemma. Remember that step functions are dense in $L^1$.
– Chappers
Jul 20 at 20:44
add a comment |Â
1
This is the Riemann-Lebesgue lemma. The typical thing to do is to prove it for $C^infty_c(mathbb R)$ functions (where you can just integrate by parts and explicitly take the limit) and pass to $L^1(mathbb R)$ functions by density.
– User8128
Jul 20 at 20:43
This is the Riemann–Lebesgue lemma. Remember that step functions are dense in $L^1$.
– Chappers
Jul 20 at 20:44
1
1
This is the Riemann-Lebesgue lemma. The typical thing to do is to prove it for $C^infty_c(mathbb R)$ functions (where you can just integrate by parts and explicitly take the limit) and pass to $L^1(mathbb R)$ functions by density.
– User8128
Jul 20 at 20:43
This is the Riemann-Lebesgue lemma. The typical thing to do is to prove it for $C^infty_c(mathbb R)$ functions (where you can just integrate by parts and explicitly take the limit) and pass to $L^1(mathbb R)$ functions by density.
– User8128
Jul 20 at 20:43
This is the Riemann–Lebesgue lemma. Remember that step functions are dense in $L^1$.
– Chappers
Jul 20 at 20:44
This is the Riemann–Lebesgue lemma. Remember that step functions are dense in $L^1$.
– Chappers
Jul 20 at 20:44
add a comment |Â
2 Answers
2
active
oldest
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1
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We prove the lemma first for $C^infty_c(mathbb R)$. Suppose that $f in C^infty_c(mathbb R)$. Then we see using integration by parts $$left lvert int_mathbb R f(x) e^ixi x dx right rvert = left lvertfrac1ixi int_mathbb R f'(x) e^i xi x dx right rvert le frac1lvertxi rvert int_mathbb R lvert f'(x)rvert dx$$ where the boundary term in the integration by parts goes to zero since $f$ is compactly supported. Thus taking $lvert xi rvert to infty$, the above inequality shows that $$int_mathbb R f(x) e^i xi x dx to 0.$$ Thus the lemma holds for $f in C^infty_c(mathbb R)$.
For general $f in L^1(mathbb R)$ and arbitrary $epsilon > 0$, find $g in C^infty_c(mathbb R)$ with $$ int_mathbb R lvert f(x) - g(x) rvert dx < epsilon $$ (we can do this since $C_c^infty(mathbb R)$ is dense in $L^1(mathbb R)$. By the above, there is $M > 0$ such that $lvert xi rvert> M$ gives $$left lvert int_mathbb R g(x) e^ixi x dxright rvert < epsilon. $$ But then $$left lvert int_mathbb R f(x) e^ixi x dx right rvert le left lvert int_mathbb R g(x) e^ixi x dxright rvert + left lvert int_mathbb R [f(x) - g(x)] e^ixi xdx right rvert le epsilon + int_mathbb R lvert f(x) - g(x) rvert dx < 2epsilon$$ for all $xi in mathbb R$ with $lvert xirvert > M$. Thus $$int_mathbb R f(x) e^i xi x dx to 0$$ as $lvert xirvert toinfty$ for any $f in L^1(mathbb R)$.
EDIT: I typed up this entire answer using $xi$ rather than $alpha$, and I forgot the $-2pi$ in the exponent, but the substance of the proof is the same.
answered Jul 20 at 20:54
User8128
10.2k1522
It's a very good answer :-) (+1) (and indeed no need to know that the fourier transform is a bijection $S(mathbb R)to S(mathbb R)$).
– Surb
Jul 20 at 21:03
add a comment |Â
up vote
0
down vote
By density of Schwartz function, there are $(f_n)$ Schwartz s.t. $f_nto f$ in $L^1$. Set $hat f_n$ Fourier transform of $f_n$. Then
$$|hat f(alpha )|leq |hat f(alpha) -hat f_n(alpha )|+|hat f_n(alpha )|leq |f_n-f|_L^1+|hat f_n(alpha )|undersetsubstackalpha longrightarrow |f_n-f|_L^1undersetnto infty longrightarrow 0.$$
answered Jul 20 at 20:47
Surb
36.3k84274
Of course, this requires one to know that $hat f$ is a Schwartz function when $f$ is a Schwartz function, which is a much more advanced fact.
– User8128
Jul 20 at 20:52
@User8128: Thank you for your comment. In fact I just answer to the OP to an other question here where I just mentioned the fact that Fourier transform was a bijection Schwartz to Schwartz. I'm not sure it's an advanced fact, but indeed it require some background in functional analysis :-)
– Surb
Jul 20 at 20:55
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
We prove the lemma first for $C^infty_c(mathbb R)$. Suppose that $f in C^infty_c(mathbb R)$. Then we see using integration by parts $$left lvert int_mathbb R f(x) e^ixi x dx right rvert = left lvertfrac1ixi int_mathbb R f'(x) e^i xi x dx right rvert le frac1lvertxi rvert int_mathbb R lvert f'(x)rvert dx$$ where the boundary term in the integration by parts goes to zero since $f$ is compactly supported. Thus taking $lvert xi rvert to infty$, the above inequality shows that $$int_mathbb R f(x) e^i xi x dx to 0.$$ Thus the lemma holds for $f in C^infty_c(mathbb R)$.
For general $f in L^1(mathbb R)$ and arbitrary $epsilon > 0$, find $g in C^infty_c(mathbb R)$ with $$ int_mathbb R lvert f(x) - g(x) rvert dx < epsilon $$ (we can do this since $C_c^infty(mathbb R)$ is dense in $L^1(mathbb R)$. By the above, there is $M > 0$ such that $lvert xi rvert> M$ gives $$left lvert int_mathbb R g(x) e^ixi x dxright rvert < epsilon. $$ But then $$left lvert int_mathbb R f(x) e^ixi x dx right rvert le left lvert int_mathbb R g(x) e^ixi x dxright rvert + left lvert int_mathbb R [f(x) - g(x)] e^ixi xdx right rvert le epsilon + int_mathbb R lvert f(x) - g(x) rvert dx < 2epsilon$$ for all $xi in mathbb R$ with $lvert xirvert > M$. Thus $$int_mathbb R f(x) e^i xi x dx to 0$$ as $lvert xirvert toinfty$ for any $f in L^1(mathbb R)$.
EDIT: I typed up this entire answer using $xi$ rather than $alpha$, and I forgot the $-2pi$ in the exponent, but the substance of the proof is the same.
answered Jul 20 at 20:54
User8128
10.2k1522
It's a very good answer :-) (+1) (and indeed no need to know that the fourier transform is a bijection $S(mathbb R)to S(mathbb R)$).
– Surb
Jul 20 at 21:03
add a comment |Â
up vote
1
down vote
We prove the lemma first for $C^infty_c(mathbb R)$. Suppose that $f in C^infty_c(mathbb R)$. Then we see using integration by parts $$left lvert int_mathbb R f(x) e^ixi x dx right rvert = left lvertfrac1ixi int_mathbb R f'(x) e^i xi x dx right rvert le frac1lvertxi rvert int_mathbb R lvert f'(x)rvert dx$$ where the boundary term in the integration by parts goes to zero since $f$ is compactly supported. Thus taking $lvert xi rvert to infty$, the above inequality shows that $$int_mathbb R f(x) e^i xi x dx to 0.$$ Thus the lemma holds for $f in C^infty_c(mathbb R)$.
For general $f in L^1(mathbb R)$ and arbitrary $epsilon > 0$, find $g in C^infty_c(mathbb R)$ with $$ int_mathbb R lvert f(x) - g(x) rvert dx < epsilon $$ (we can do this since $C_c^infty(mathbb R)$ is dense in $L^1(mathbb R)$. By the above, there is $M > 0$ such that $lvert xi rvert> M$ gives $$left lvert int_mathbb R g(x) e^ixi x dxright rvert < epsilon. $$ But then $$left lvert int_mathbb R f(x) e^ixi x dx right rvert le left lvert int_mathbb R g(x) e^ixi x dxright rvert + left lvert int_mathbb R [f(x) - g(x)] e^ixi xdx right rvert le epsilon + int_mathbb R lvert f(x) - g(x) rvert dx < 2epsilon$$ for all $xi in mathbb R$ with $lvert xirvert > M$. Thus $$int_mathbb R f(x) e^i xi x dx to 0$$ as $lvert xirvert toinfty$ for any $f in L^1(mathbb R)$.
EDIT: I typed up this entire answer using $xi$ rather than $alpha$, and I forgot the $-2pi$ in the exponent, but the substance of the proof is the same.
answered Jul 20 at 20:54
User8128
10.2k1522
It's a very good answer :-) (+1) (and indeed no need to know that the fourier transform is a bijection $S(mathbb R)to S(mathbb R)$).
– Surb
Jul 20 at 21:03
add a comment |Â
up vote
1
down vote
up vote
1
down vote
We prove the lemma first for $C^infty_c(mathbb R)$. Suppose that $f in C^infty_c(mathbb R)$. Then we see using integration by parts $$left lvert int_mathbb R f(x) e^ixi x dx right rvert = left lvertfrac1ixi int_mathbb R f'(x) e^i xi x dx right rvert le frac1lvertxi rvert int_mathbb R lvert f'(x)rvert dx$$ where the boundary term in the integration by parts goes to zero since $f$ is compactly supported. Thus taking $lvert xi rvert to infty$, the above inequality shows that $$int_mathbb R f(x) e^i xi x dx to 0.$$ Thus the lemma holds for $f in C^infty_c(mathbb R)$.
For general $f in L^1(mathbb R)$ and arbitrary $epsilon > 0$, find $g in C^infty_c(mathbb R)$ with $$ int_mathbb R lvert f(x) - g(x) rvert dx < epsilon $$ (we can do this since $C_c^infty(mathbb R)$ is dense in $L^1(mathbb R)$. By the above, there is $M > 0$ such that $lvert xi rvert> M$ gives $$left lvert int_mathbb R g(x) e^ixi x dxright rvert < epsilon. $$ But then $$left lvert int_mathbb R f(x) e^ixi x dx right rvert le left lvert int_mathbb R g(x) e^ixi x dxright rvert + left lvert int_mathbb R [f(x) - g(x)] e^ixi xdx right rvert le epsilon + int_mathbb R lvert f(x) - g(x) rvert dx < 2epsilon$$ for all $xi in mathbb R$ with $lvert xirvert > M$. Thus $$int_mathbb R f(x) e^i xi x dx to 0$$ as $lvert xirvert toinfty$ for any $f in L^1(mathbb R)$.
EDIT: I typed up this entire answer using $xi$ rather than $alpha$, and I forgot the $-2pi$ in the exponent, but the substance of the proof is the same.
answered Jul 20 at 20:54
User8128
10.2k1522
We prove the lemma first for $C^infty_c(mathbb R)$. Suppose that $f in C^infty_c(mathbb R)$. Then we see using integration by parts $$left lvert int_mathbb R f(x) e^ixi x dx right rvert = left lvertfrac1ixi int_mathbb R f'(x) e^i xi x dx right rvert le frac1lvertxi rvert int_mathbb R lvert f'(x)rvert dx$$ where the boundary term in the integration by parts goes to zero since $f$ is compactly supported. Thus taking $lvert xi rvert to infty$, the above inequality shows that $$int_mathbb R f(x) e^i xi x dx to 0.$$ Thus the lemma holds for $f in C^infty_c(mathbb R)$.
For general $f in L^1(mathbb R)$ and arbitrary $epsilon > 0$, find $g in C^infty_c(mathbb R)$ with $$ int_mathbb R lvert f(x) - g(x) rvert dx < epsilon $$ (we can do this since $C_c^infty(mathbb R)$ is dense in $L^1(mathbb R)$. By the above, there is $M > 0$ such that $lvert xi rvert> M$ gives $$left lvert int_mathbb R g(x) e^ixi x dxright rvert < epsilon. $$ But then $$left lvert int_mathbb R f(x) e^ixi x dx right rvert le left lvert int_mathbb R g(x) e^ixi x dxright rvert + left lvert int_mathbb R [f(x) - g(x)] e^ixi xdx right rvert le epsilon + int_mathbb R lvert f(x) - g(x) rvert dx < 2epsilon$$ for all $xi in mathbb R$ with $lvert xirvert > M$. Thus $$int_mathbb R f(x) e^i xi x dx to 0$$ as $lvert xirvert toinfty$ for any $f in L^1(mathbb R)$.
EDIT: I typed up this entire answer using $xi$ rather than $alpha$, and I forgot the $-2pi$ in the exponent, but the substance of the proof is the same.
answered Jul 20 at 20:54
User8128
10.2k1522
answered Jul 20 at 20:54
User8128
10.2k1522
answered Jul 20 at 20:54
User8128
10.2k1522
answered Jul 20 at 20:54
User8128
10.2k1522
10.2k1522
It's a very good answer :-) (+1) (and indeed no need to know that the fourier transform is a bijection $S(mathbb R)to S(mathbb R)$).
– Surb
Jul 20 at 21:03
add a comment |Â
It's a very good answer :-) (+1) (and indeed no need to know that the fourier transform is a bijection $S(mathbb R)to S(mathbb R)$).
– Surb
Jul 20 at 21:03
It's a very good answer :-) (+1) (and indeed no need to know that the fourier transform is a bijection $S(mathbb R)to S(mathbb R)$).
– Surb
Jul 20 at 21:03
It's a very good answer :-) (+1) (and indeed no need to know that the fourier transform is a bijection $S(mathbb R)to S(mathbb R)$).
– Surb
Jul 20 at 21:03
add a comment |Â
up vote
0
down vote
By density of Schwartz function, there are $(f_n)$ Schwartz s.t. $f_nto f$ in $L^1$. Set $hat f_n$ Fourier transform of $f_n$. Then
$$|hat f(alpha )|leq |hat f(alpha) -hat f_n(alpha )|+|hat f_n(alpha )|leq |f_n-f|_L^1+|hat f_n(alpha )|undersetsubstackalpha longrightarrow |f_n-f|_L^1undersetnto infty longrightarrow 0.$$
answered Jul 20 at 20:47
Surb
36.3k84274
Of course, this requires one to know that $hat f$ is a Schwartz function when $f$ is a Schwartz function, which is a much more advanced fact.
– User8128
Jul 20 at 20:52
@User8128: Thank you for your comment. In fact I just answer to the OP to an other question here where I just mentioned the fact that Fourier transform was a bijection Schwartz to Schwartz. I'm not sure it's an advanced fact, but indeed it require some background in functional analysis :-)
– Surb
Jul 20 at 20:55
add a comment |Â
up vote
0
down vote
By density of Schwartz function, there are $(f_n)$ Schwartz s.t. $f_nto f$ in $L^1$. Set $hat f_n$ Fourier transform of $f_n$. Then
$$|hat f(alpha )|leq |hat f(alpha) -hat f_n(alpha )|+|hat f_n(alpha )|leq |f_n-f|_L^1+|hat f_n(alpha )|undersetsubstackalpha longrightarrow |f_n-f|_L^1undersetnto infty longrightarrow 0.$$
answered Jul 20 at 20:47
Surb
36.3k84274
Of course, this requires one to know that $hat f$ is a Schwartz function when $f$ is a Schwartz function, which is a much more advanced fact.
– User8128
Jul 20 at 20:52
@User8128: Thank you for your comment. In fact I just answer to the OP to an other question here where I just mentioned the fact that Fourier transform was a bijection Schwartz to Schwartz. I'm not sure it's an advanced fact, but indeed it require some background in functional analysis :-)
– Surb
Jul 20 at 20:55
add a comment |Â
up vote
0
down vote
up vote
0
down vote
By density of Schwartz function, there are $(f_n)$ Schwartz s.t. $f_nto f$ in $L^1$. Set $hat f_n$ Fourier transform of $f_n$. Then
$$|hat f(alpha )|leq |hat f(alpha) -hat f_n(alpha )|+|hat f_n(alpha )|leq |f_n-f|_L^1+|hat f_n(alpha )|undersetsubstackalpha longrightarrow |f_n-f|_L^1undersetnto infty longrightarrow 0.$$
answered Jul 20 at 20:47
Surb
36.3k84274
By density of Schwartz function, there are $(f_n)$ Schwartz s.t. $f_nto f$ in $L^1$. Set $hat f_n$ Fourier transform of $f_n$. Then
$$|hat f(alpha )|leq |hat f(alpha) -hat f_n(alpha )|+|hat f_n(alpha )|leq |f_n-f|_L^1+|hat f_n(alpha )|undersetsubstackalpha longrightarrow |f_n-f|_L^1undersetnto infty longrightarrow 0.$$
answered Jul 20 at 20:47
Surb
36.3k84274
edited Jul 20 at 21:18
answered Jul 20 at 20:47
Surb
36.3k84274
answered Jul 20 at 20:47
Surb
36.3k84274
answered Jul 20 at 20:47
Surb
36.3k84274
36.3k84274
Of course, this requires one to know that $hat f$ is a Schwartz function when $f$ is a Schwartz function, which is a much more advanced fact.
– User8128
Jul 20 at 20:52
@User8128: Thank you for your comment. In fact I just answer to the OP to an other question here where I just mentioned the fact that Fourier transform was a bijection Schwartz to Schwartz. I'm not sure it's an advanced fact, but indeed it require some background in functional analysis :-)
– Surb
Jul 20 at 20:55
add a comment |Â
Of course, this requires one to know that $hat f$ is a Schwartz function when $f$ is a Schwartz function, which is a much more advanced fact.
– User8128
Jul 20 at 20:52
@User8128: Thank you for your comment. In fact I just answer to the OP to an other question here where I just mentioned the fact that Fourier transform was a bijection Schwartz to Schwartz. I'm not sure it's an advanced fact, but indeed it require some background in functional analysis :-)
– Surb
Jul 20 at 20:55
Of course, this requires one to know that $hat f$ is a Schwartz function when $f$ is a Schwartz function, which is a much more advanced fact.
– User8128
Jul 20 at 20:52
Of course, this requires one to know that $hat f$ is a Schwartz function when $f$ is a Schwartz function, which is a much more advanced fact.
– User8128
Jul 20 at 20:52
@User8128: Thank you for your comment. In fact I just answer to the OP to an other question here where I just mentioned the fact that Fourier transform was a bijection Schwartz to Schwartz. I'm not sure it's an advanced fact, but indeed it require some background in functional analysis :-)
– Surb
Jul 20 at 20:55
@User8128: Thank you for your comment. In fact I just answer to the OP to an other question here where I just mentioned the fact that Fourier transform was a bijection Schwartz to Schwartz. I'm not sure it's an advanced fact, but indeed it require some background in functional analysis :-)
– Surb
Jul 20 at 20:55
add a comment |Â
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1
This is the Riemann-Lebesgue lemma. The typical thing to do is to prove it for $C^infty_c(mathbb R)$ functions (where you can just integrate by parts and explicitly take the limit) and pass to $L^1(mathbb R)$ functions by density.
– User8128
Jul 20 at 20:43
This is the Riemann–Lebesgue lemma. Remember that step functions are dense in $L^1$.
– Chappers
Jul 20 at 20:44