Mixing
Compute the Fourier transform of a $L^2$ function.
Clash Royale CLAN TAG#URR8PPP
up vote
2
down vote
favorite
Let $fin L^2(mathbb R)$. Is it possible to compute the Fourier transform of an $L^2(mathbb R)$ function ? I'm asking this question because in a book I'm reading, they take the Fourier transform of a function $fin L^2(mathbb R)$ but for me there is no reason to have that $$int_mathbb Rf(x)e^-2ipi xxi dx$$
is convergent. So, could it be a mistake from the author ?
fourier-transform
asked Jul 20 at 19:50
MSE
1,471315
add a comment |Â
up vote
2
down vote
favorite
Let $fin L^2(mathbb R)$. Is it possible to compute the Fourier transform of an $L^2(mathbb R)$ function ? I'm asking this question because in a book I'm reading, they take the Fourier transform of a function $fin L^2(mathbb R)$ but for me there is no reason to have that $$int_mathbb Rf(x)e^-2ipi xxi dx$$
is convergent. So, could it be a mistake from the author ?
fourier-transform
asked Jul 20 at 19:50
MSE
1,471315
See the riemann lebesgue lemma. en.wikipedia.org/wiki/Riemann%E2%80%93Lebesgue_lemma
– RHowe
Jul 20 at 19:55
In the text "Analysis" by Lieb & Loss it's explains very well: they construct the Fourier transform beginning by $L^1$ function then with Plancherel's theorem they construct it for $L^2$ and $L^p$
– Davide Morgante
Jul 20 at 20:01
@DavideMorgante: For $L^p$ it's unfortunately not possible in general. I guess that for $pin [1,2]$, we can extend Fourier transform $L^pto L^p'$ using Riesz-Thorin (with $frac1p+frac1p'=1$), but I don't think that it's possible for $p>2$ (but maybe I'm wrong...)
– Surb
Jul 20 at 20:09
@Surb I'm a little bit rusty when it comes to this specific mathematical construction. But from what I recall you're right, the authors used a sharper version of the Hausdorff-Young inequality which doesn't stand for $pgt 2$
– Davide Morgante
Jul 20 at 20:14
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Let $fin L^2(mathbb R)$. Is it possible to compute the Fourier transform of an $L^2(mathbb R)$ function ? I'm asking this question because in a book I'm reading, they take the Fourier transform of a function $fin L^2(mathbb R)$ but for me there is no reason to have that $$int_mathbb Rf(x)e^-2ipi xxi dx$$
is convergent. So, could it be a mistake from the author ?
fourier-transform
asked Jul 20 at 19:50
MSE
1,471315
Let $fin L^2(mathbb R)$. Is it possible to compute the Fourier transform of an $L^2(mathbb R)$ function ? I'm asking this question because in a book I'm reading, they take the Fourier transform of a function $fin L^2(mathbb R)$ but for me there is no reason to have that $$int_mathbb Rf(x)e^-2ipi xxi dx$$
is convergent. So, could it be a mistake from the author ?
fourier-transform
asked Jul 20 at 19:50
MSE
1,471315
asked Jul 20 at 19:50
MSE
1,471315
asked Jul 20 at 19:50
MSE
1,471315
asked Jul 20 at 19:50
MSE
1,471315
1,471315
See the riemann lebesgue lemma. en.wikipedia.org/wiki/Riemann%E2%80%93Lebesgue_lemma
– RHowe
Jul 20 at 19:55
In the text "Analysis" by Lieb & Loss it's explains very well: they construct the Fourier transform beginning by $L^1$ function then with Plancherel's theorem they construct it for $L^2$ and $L^p$
– Davide Morgante
Jul 20 at 20:01
@DavideMorgante: For $L^p$ it's unfortunately not possible in general. I guess that for $pin [1,2]$, we can extend Fourier transform $L^pto L^p'$ using Riesz-Thorin (with $frac1p+frac1p'=1$), but I don't think that it's possible for $p>2$ (but maybe I'm wrong...)
– Surb
Jul 20 at 20:09
@Surb I'm a little bit rusty when it comes to this specific mathematical construction. But from what I recall you're right, the authors used a sharper version of the Hausdorff-Young inequality which doesn't stand for $pgt 2$
– Davide Morgante
Jul 20 at 20:14
add a comment |Â
See the riemann lebesgue lemma. en.wikipedia.org/wiki/Riemann%E2%80%93Lebesgue_lemma
– RHowe
Jul 20 at 19:55
In the text "Analysis" by Lieb & Loss it's explains very well: they construct the Fourier transform beginning by $L^1$ function then with Plancherel's theorem they construct it for $L^2$ and $L^p$
– Davide Morgante
Jul 20 at 20:01
@DavideMorgante: For $L^p$ it's unfortunately not possible in general. I guess that for $pin [1,2]$, we can extend Fourier transform $L^pto L^p'$ using Riesz-Thorin (with $frac1p+frac1p'=1$), but I don't think that it's possible for $p>2$ (but maybe I'm wrong...)
– Surb
Jul 20 at 20:09
@Surb I'm a little bit rusty when it comes to this specific mathematical construction. But from what I recall you're right, the authors used a sharper version of the Hausdorff-Young inequality which doesn't stand for $pgt 2$
– Davide Morgante
Jul 20 at 20:14
See the riemann lebesgue lemma. en.wikipedia.org/wiki/Riemann%E2%80%93Lebesgue_lemma
– RHowe
Jul 20 at 19:55
See the riemann lebesgue lemma. en.wikipedia.org/wiki/Riemann%E2%80%93Lebesgue_lemma
– RHowe
Jul 20 at 19:55
In the text "Analysis" by Lieb & Loss it's explains very well: they construct the Fourier transform beginning by $L^1$ function then with Plancherel's theorem they construct it for $L^2$ and $L^p$
– Davide Morgante
Jul 20 at 20:01
In the text "Analysis" by Lieb & Loss it's explains very well: they construct the Fourier transform beginning by $L^1$ function then with Plancherel's theorem they construct it for $L^2$ and $L^p$
– Davide Morgante
Jul 20 at 20:01
@DavideMorgante: For $L^p$ it's unfortunately not possible in general. I guess that for $pin [1,2]$, we can extend Fourier transform $L^pto L^p'$ using Riesz-Thorin (with $frac1p+frac1p'=1$), but I don't think that it's possible for $p>2$ (but maybe I'm wrong...)
– Surb
Jul 20 at 20:09
@DavideMorgante: For $L^p$ it's unfortunately not possible in general. I guess that for $pin [1,2]$, we can extend Fourier transform $L^pto L^p'$ using Riesz-Thorin (with $frac1p+frac1p'=1$), but I don't think that it's possible for $p>2$ (but maybe I'm wrong...)
– Surb
Jul 20 at 20:09
@Surb I'm a little bit rusty when it comes to this specific mathematical construction. But from what I recall you're right, the authors used a sharper version of the Hausdorff-Young inequality which doesn't stand for $pgt 2$
– Davide Morgante
Jul 20 at 20:14
@Surb I'm a little bit rusty when it comes to this specific mathematical construction. But from what I recall you're right, the authors used a sharper version of the Hausdorff-Young inequality which doesn't stand for $pgt 2$
– Davide Morgante
Jul 20 at 20:14
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
2
down vote
The Fourier transform is a bijection from $mathcal S(mathbb R)longrightarrow mathcal S(mathbb R),$ where $mathcal S(mathbb R)$ denote the space of Schwartz function. It can be prolonged from $L^2(mathbb R)longrightarrow L^2(mathbb R)$ as follow : $mathcal S(mathbb R)$ is dense in $L^2$, therefore, if $fin L^2(mathbb R)$, there is $(f_n)subset mathcal S(mathbb R)$ s.t. $$f_nlongrightarrow fquad textin L^2(mathbb R),$$
and we define $$hat f:=lim_nto infty hat f_nquad textin L^2(mathbb R).$$
But if $fin L^2(mathbb R)$ and $fnotin L^1(mathbb R)$, then $$hat f(xi)neq int_mathbb Rf(x)e^-2ipi xxidx,$$
a priori.
This definition is motivated by the linearity of Fourier transform, Plancherel's equality and the completeness of $L^2(mathbb R)$.
answered Jul 20 at 20:01
Surb
36.3k84274
add a comment |Â
up vote
2
down vote
The pointwise convergence known as Carleson's theorem is a delicate problem in the sense that the proof is about 10 pages.The paper by Carleson focused on the Fourier series convergence while simpler proofs focused on convergence of Fourier transform , nevertheless the two are equivalent and the $ L^2$ convergence is not difficult to prove:
Given an $L^2$ function $f(t)$,
define $$hatf_n(s)=int_-n^nf(t)e^-istdt$$
It can be easily shown using Plancherel theorem that there is some $g(s) in L^2$ and $$lim_nrightarrowinftyint_-infty^infty(hatf_n(s)-g(s))^2ds=0$$
$$int_-infty^infty(g(s))^2ds=int_-infty^infty(f(t))^2dt$$
How to show $$lim_nrightarrowinftyint_-a^aleft(frac12piint_-n^ng(s)e^istds-f(t)right)^2dt=0$$ where $a in R$ :
First prove it for $f$ with compact support then generalise it to $L^2$ using inequalities.
Proof for $f(t) $ of compact support and $in L^2$ :
clearly $f(t) in L^1$ without loss of generality assume $f(t)=0$ outside $[-pi,pi]$ since we can always replace $f(t)$ with $f(t')$ such as $t=fracn2pi t'$ for $n in N$ and $t' in R$.
define $$g(s)=lim_ntoinftyint_-n^nf(t)e^-istdt$$
The relationship between fourier transform and fourier series for $f(t)$ above follows from
$$(1) lim_nrightarrowinftyint_-pi^pi(frac1piint_-pi^pif(t+u)fracsin(n+frac12)u2sin(u/2)du-frac1piint_-pi^pi f(t+u)fracsin(n+frac12)uudu)^2dt=0$$
which is simply consequence of Riemann-Lebesgue Lemma applied to the function $ f (t+u) k (u) $, where $k(u)=frac12sin(u/2)-frac1u$ , using remainder term for the Taylor's series of $sin(u/2)$, it easily seen $k(u) $ is bounded over $[-pi, pi]$
$$(2)frac12pi int_-(n+frac12)^(n+frac12)g(s)e^istds=frac1piint_-infty^inftyf(t+u)fracsin(n+frac12)uudu$$
This can be derived from fubini's theorem and this theorem : $mu$ is Lebesgue outer measure, $ A $ is a measurable set with finite measure. For all $n$ , $int_Af_n^2dmu le k $ where $ k in R $ , $ f_n $ is uniformly integrable
Norm convergence for Fourier series :
$$(3) lim_nrightarrowinftyint_-pi^pi(frac1piint_-pi^pif(t+u)fracsin(n+frac12)u2sin(u/2)du-f(t))^2dt=0$$
using Riemann Lebesgue Lemma on (2) we have $$frac12pi int_-(n+frac12)^(n+frac12)g(s)e^istds=frac1piint_-pi^pif(t+u)fracsin(n+frac12)uudu$$
Using Norm convergence of fourier series (3), Riemann-Lebesgue lemma, Cauchy-Schwartz Inequality,$a^2+b^2 ge 2ab$ and (1) and (2) we have:
$$lim_nrightarrowinftyint_-pi^pileft(frac12piint_-(n+frac12)^n+frac12g(s)e^istds-f(t)right)^2dt=0$$
Proof for $f(t) in L^2$:
define $f_m=f_[-m,m]$ where $m in N$ and its fourier transform is $hatf_m$
$$lim_mrightarrowinfty f_m=f$$
By Plancherel theorem : $lim_mrightarrowinfty hatf_m=lim_mrightarrowinftyint_-m^mf(t)e^-istdt=hatf$ in sense of $L^2$
our previous result for function with compact support : $$lim_nto inftyint_-pi^pi(frac12piint_-n^nhatf_m(s)e^istds-f_m(t))^2dt =0 $$
$$lim_nto inftyint_-pi^pi(frac12piint_-n^nhatf_m(s)e^istds-f_m(t))^2dt=lim_nto inftyint_-pi^pi(frac12piint_-n^nhatf_m(s)e^istds-f_m(t)_[-pi,pi])^2dt =lim_nto inftyint_-pi^pi(frac12piint_-n^nhatf_m(s)e^istds-f(t))^2dt=lim_nto inftyint_-pi^pi(frac12piint_-n^nhatf_n(s)e^istds-f(t))^2dt=0 $$
There is a subsequence $n$ such that $hatf_n to hatf $ ae. Let's work with such subsequence
$$lim_nto inftyint_-pi^pi(frac12piint_-n^nhatf(s)e^istds-frac12piint_-n^nhatf_n(s)e^istds)^2dt =$$$$lim_nto inftyint_-pi^pi(frac1piint_-infty^infty(f(t+u)_[-infty,-n]+f(t+u)_[n,infty])fracsin(n+frac12)uudu)^2dt $$ $$le 2(||f(u)_[-infty,-n]+f(u)_[n,infty]||_L^2+||fracsin(n+frac12)uu||_L^2)^-1/2$$ by Cauchy-Scwartz ineqaulity
because $a^2+b^2 ge 2ab$ , it follows from Cauchy-Schwartz inequality that:
$lim_nto inftyint_-pi^pi(frac12piint_-n^nhatf(s)e^istds-f(t))^2dtle$
$lim_nto inftyint_-pi^pi(|frac12piint_-n^nhatf(s)e^istds-f_n(t)|+|f_n(t)-f(t)|)^2dt=0 $
we proved the result for some sequence $n$ to see the result holds for all $n$ we observe that
$$lim_nto inftyint_-pi^pi(frac12piint_-(n+1)^(n+1)hatf(s)e^istds-int_-n^nhatf(s)e^istds)^2dt le$$$$lim_nto inftyint_-pi^pi(frac12piint_-(n+1)^(-n)|hatf(s)|ds+frac12piint_n^(n+1)|hatf(s)|ds)^2dt le lim_nto infty frac12pi(||hatf_[-n-1,-n]||_L^2+||hatf_[n,n+1]||_L^2)^2=0$$ by Hölder's Inequality
answered Jul 21 at 3:41
S.t.a.l.k.e.r
721320
add a comment |Â
up vote
2
down vote
Being $L^2(mathbbR)backslash L^1(mathbbR)neqemptyset$ and being $int_mathbbRf(x)e^-2pi ixxidx$ well defined as a Lebesgue integral iff $fin L^1(mathbbR)$, you are absolutely right to complain.
By the way, you can define the Fourier transform $mathcalF$ for $L^2(mathbbR)$ by a little detour, first defining it by that formula for $L^1(mathbbR)$ functions, then proving that $$forall fin L^1(mathbbR)cap L^2(mathbbR), |mathcalF(f)|_2=|f|_2,$$
so the Fourier transform is an isometry and then can be extended uniquely to an isometry on the whole $L^2(mathbbR)$ by continuity and density.
Now, having defined the Fourier transform in $L^2(mathbbR)$ via this indirect route, you might ask yourself if there is a more practical procedure to compute it. The first idea that comes to mind is to try to give meaning to the $L^1(mathbbR)$ formula: $$ mathcalF(f)(xi)=int_mathbbRf(x)e^-2pi ixxidx,$$
also in the $L^2(mathbbR)$ context.
As initially said, this can't be, in general, interpreted as a Lebesgue integral. However, what about interpreting it as a improper integral? After all, for subsets $KsubsetmathbbR$ of finite Lebesgue measure, by Holder inequality, we have that $L^2 (K)subset L^1 (K)$ and so, if $fin L^2 (mathbbR)$ and $M>0$, it is well defined the integral: $$int_-M^Mf(x)e^-2pi i xxidx,$$ and we can try to get the limit of this quantity as $Mrightarrowinfty$. So the question: does it hold true for $fin L^2 (mathbbR)$ that $$mathcalF(f)(xi)=lim_Mrightarrowinfty int_-M^Mf(x)e^-2pi i xxidx?$$
Well, by Carleson's theorem, this limit exists for a.e. $xiinmathbbR$ and that formula holds true if interpreted as an a.e. $xiinmathbbR$ equality. So, at the end, we have recovered a meaning for the original formula via an improper integral, obtaining in this manner an explicit way to calculate the Fourier transform of a $L^2(mathbbR)$ function.
answered Jul 20 at 20:46
Bob
1,467522
Just for notation, $L^2backslash L^1$ looks weird since $L^1notsubset L^2$. I would use more $L^2cap (L^1)^c$.
– Surb
Jul 21 at 8:01
1
@surb en.wikipedia.org/wiki/…
– Bob
Jul 21 at 8:12
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
The Fourier transform is a bijection from $mathcal S(mathbb R)longrightarrow mathcal S(mathbb R),$ where $mathcal S(mathbb R)$ denote the space of Schwartz function. It can be prolonged from $L^2(mathbb R)longrightarrow L^2(mathbb R)$ as follow : $mathcal S(mathbb R)$ is dense in $L^2$, therefore, if $fin L^2(mathbb R)$, there is $(f_n)subset mathcal S(mathbb R)$ s.t. $$f_nlongrightarrow fquad textin L^2(mathbb R),$$
and we define $$hat f:=lim_nto infty hat f_nquad textin L^2(mathbb R).$$
But if $fin L^2(mathbb R)$ and $fnotin L^1(mathbb R)$, then $$hat f(xi)neq int_mathbb Rf(x)e^-2ipi xxidx,$$
a priori.
This definition is motivated by the linearity of Fourier transform, Plancherel's equality and the completeness of $L^2(mathbb R)$.
answered Jul 20 at 20:01
Surb
36.3k84274
add a comment |Â
up vote
2
down vote
The Fourier transform is a bijection from $mathcal S(mathbb R)longrightarrow mathcal S(mathbb R),$ where $mathcal S(mathbb R)$ denote the space of Schwartz function. It can be prolonged from $L^2(mathbb R)longrightarrow L^2(mathbb R)$ as follow : $mathcal S(mathbb R)$ is dense in $L^2$, therefore, if $fin L^2(mathbb R)$, there is $(f_n)subset mathcal S(mathbb R)$ s.t. $$f_nlongrightarrow fquad textin L^2(mathbb R),$$
and we define $$hat f:=lim_nto infty hat f_nquad textin L^2(mathbb R).$$
But if $fin L^2(mathbb R)$ and $fnotin L^1(mathbb R)$, then $$hat f(xi)neq int_mathbb Rf(x)e^-2ipi xxidx,$$
a priori.
This definition is motivated by the linearity of Fourier transform, Plancherel's equality and the completeness of $L^2(mathbb R)$.
answered Jul 20 at 20:01
Surb
36.3k84274
add a comment |Â
up vote
2
down vote
up vote
2
down vote
The Fourier transform is a bijection from $mathcal S(mathbb R)longrightarrow mathcal S(mathbb R),$ where $mathcal S(mathbb R)$ denote the space of Schwartz function. It can be prolonged from $L^2(mathbb R)longrightarrow L^2(mathbb R)$ as follow : $mathcal S(mathbb R)$ is dense in $L^2$, therefore, if $fin L^2(mathbb R)$, there is $(f_n)subset mathcal S(mathbb R)$ s.t. $$f_nlongrightarrow fquad textin L^2(mathbb R),$$
and we define $$hat f:=lim_nto infty hat f_nquad textin L^2(mathbb R).$$
But if $fin L^2(mathbb R)$ and $fnotin L^1(mathbb R)$, then $$hat f(xi)neq int_mathbb Rf(x)e^-2ipi xxidx,$$
a priori.
This definition is motivated by the linearity of Fourier transform, Plancherel's equality and the completeness of $L^2(mathbb R)$.
answered Jul 20 at 20:01
Surb
36.3k84274
The Fourier transform is a bijection from $mathcal S(mathbb R)longrightarrow mathcal S(mathbb R),$ where $mathcal S(mathbb R)$ denote the space of Schwartz function. It can be prolonged from $L^2(mathbb R)longrightarrow L^2(mathbb R)$ as follow : $mathcal S(mathbb R)$ is dense in $L^2$, therefore, if $fin L^2(mathbb R)$, there is $(f_n)subset mathcal S(mathbb R)$ s.t. $$f_nlongrightarrow fquad textin L^2(mathbb R),$$
and we define $$hat f:=lim_nto infty hat f_nquad textin L^2(mathbb R).$$
But if $fin L^2(mathbb R)$ and $fnotin L^1(mathbb R)$, then $$hat f(xi)neq int_mathbb Rf(x)e^-2ipi xxidx,$$
a priori.
This definition is motivated by the linearity of Fourier transform, Plancherel's equality and the completeness of $L^2(mathbb R)$.
answered Jul 20 at 20:01
Surb
36.3k84274
answered Jul 20 at 20:01
Surb
36.3k84274
answered Jul 20 at 20:01
Surb
36.3k84274
answered Jul 20 at 20:01
Surb
36.3k84274
36.3k84274
add a comment |Â
add a comment |Â
up vote
2
down vote
The pointwise convergence known as Carleson's theorem is a delicate problem in the sense that the proof is about 10 pages.The paper by Carleson focused on the Fourier series convergence while simpler proofs focused on convergence of Fourier transform , nevertheless the two are equivalent and the $ L^2$ convergence is not difficult to prove:
Given an $L^2$ function $f(t)$,
define $$hatf_n(s)=int_-n^nf(t)e^-istdt$$
It can be easily shown using Plancherel theorem that there is some $g(s) in L^2$ and $$lim_nrightarrowinftyint_-infty^infty(hatf_n(s)-g(s))^2ds=0$$
$$int_-infty^infty(g(s))^2ds=int_-infty^infty(f(t))^2dt$$
How to show $$lim_nrightarrowinftyint_-a^aleft(frac12piint_-n^ng(s)e^istds-f(t)right)^2dt=0$$ where $a in R$ :
First prove it for $f$ with compact support then generalise it to $L^2$ using inequalities.
Proof for $f(t) $ of compact support and $in L^2$ :
clearly $f(t) in L^1$ without loss of generality assume $f(t)=0$ outside $[-pi,pi]$ since we can always replace $f(t)$ with $f(t')$ such as $t=fracn2pi t'$ for $n in N$ and $t' in R$.
define $$g(s)=lim_ntoinftyint_-n^nf(t)e^-istdt$$
The relationship between fourier transform and fourier series for $f(t)$ above follows from
$$(1) lim_nrightarrowinftyint_-pi^pi(frac1piint_-pi^pif(t+u)fracsin(n+frac12)u2sin(u/2)du-frac1piint_-pi^pi f(t+u)fracsin(n+frac12)uudu)^2dt=0$$
which is simply consequence of Riemann-Lebesgue Lemma applied to the function $ f (t+u) k (u) $, where $k(u)=frac12sin(u/2)-frac1u$ , using remainder term for the Taylor's series of $sin(u/2)$, it easily seen $k(u) $ is bounded over $[-pi, pi]$
$$(2)frac12pi int_-(n+frac12)^(n+frac12)g(s)e^istds=frac1piint_-infty^inftyf(t+u)fracsin(n+frac12)uudu$$
This can be derived from fubini's theorem and this theorem : $mu$ is Lebesgue outer measure, $ A $ is a measurable set with finite measure. For all $n$ , $int_Af_n^2dmu le k $ where $ k in R $ , $ f_n $ is uniformly integrable
Norm convergence for Fourier series :
$$(3) lim_nrightarrowinftyint_-pi^pi(frac1piint_-pi^pif(t+u)fracsin(n+frac12)u2sin(u/2)du-f(t))^2dt=0$$
using Riemann Lebesgue Lemma on (2) we have $$frac12pi int_-(n+frac12)^(n+frac12)g(s)e^istds=frac1piint_-pi^pif(t+u)fracsin(n+frac12)uudu$$
Using Norm convergence of fourier series (3), Riemann-Lebesgue lemma, Cauchy-Schwartz Inequality,$a^2+b^2 ge 2ab$ and (1) and (2) we have:
$$lim_nrightarrowinftyint_-pi^pileft(frac12piint_-(n+frac12)^n+frac12g(s)e^istds-f(t)right)^2dt=0$$
Proof for $f(t) in L^2$:
define $f_m=f_[-m,m]$ where $m in N$ and its fourier transform is $hatf_m$
$$lim_mrightarrowinfty f_m=f$$
By Plancherel theorem : $lim_mrightarrowinfty hatf_m=lim_mrightarrowinftyint_-m^mf(t)e^-istdt=hatf$ in sense of $L^2$
our previous result for function with compact support : $$lim_nto inftyint_-pi^pi(frac12piint_-n^nhatf_m(s)e^istds-f_m(t))^2dt =0 $$
$$lim_nto inftyint_-pi^pi(frac12piint_-n^nhatf_m(s)e^istds-f_m(t))^2dt=lim_nto inftyint_-pi^pi(frac12piint_-n^nhatf_m(s)e^istds-f_m(t)_[-pi,pi])^2dt =lim_nto inftyint_-pi^pi(frac12piint_-n^nhatf_m(s)e^istds-f(t))^2dt=lim_nto inftyint_-pi^pi(frac12piint_-n^nhatf_n(s)e^istds-f(t))^2dt=0 $$
There is a subsequence $n$ such that $hatf_n to hatf $ ae. Let's work with such subsequence
$$lim_nto inftyint_-pi^pi(frac12piint_-n^nhatf(s)e^istds-frac12piint_-n^nhatf_n(s)e^istds)^2dt =$$$$lim_nto inftyint_-pi^pi(frac1piint_-infty^infty(f(t+u)_[-infty,-n]+f(t+u)_[n,infty])fracsin(n+frac12)uudu)^2dt $$ $$le 2(||f(u)_[-infty,-n]+f(u)_[n,infty]||_L^2+||fracsin(n+frac12)uu||_L^2)^-1/2$$ by Cauchy-Scwartz ineqaulity
because $a^2+b^2 ge 2ab$ , it follows from Cauchy-Schwartz inequality that:
$lim_nto inftyint_-pi^pi(frac12piint_-n^nhatf(s)e^istds-f(t))^2dtle$
$lim_nto inftyint_-pi^pi(|frac12piint_-n^nhatf(s)e^istds-f_n(t)|+|f_n(t)-f(t)|)^2dt=0 $
we proved the result for some sequence $n$ to see the result holds for all $n$ we observe that
$$lim_nto inftyint_-pi^pi(frac12piint_-(n+1)^(n+1)hatf(s)e^istds-int_-n^nhatf(s)e^istds)^2dt le$$$$lim_nto inftyint_-pi^pi(frac12piint_-(n+1)^(-n)|hatf(s)|ds+frac12piint_n^(n+1)|hatf(s)|ds)^2dt le lim_nto infty frac12pi(||hatf_[-n-1,-n]||_L^2+||hatf_[n,n+1]||_L^2)^2=0$$ by Hölder's Inequality
answered Jul 21 at 3:41
S.t.a.l.k.e.r
721320
add a comment |Â
up vote
2
down vote
The pointwise convergence known as Carleson's theorem is a delicate problem in the sense that the proof is about 10 pages.The paper by Carleson focused on the Fourier series convergence while simpler proofs focused on convergence of Fourier transform , nevertheless the two are equivalent and the $ L^2$ convergence is not difficult to prove:
Given an $L^2$ function $f(t)$,
define $$hatf_n(s)=int_-n^nf(t)e^-istdt$$
It can be easily shown using Plancherel theorem that there is some $g(s) in L^2$ and $$lim_nrightarrowinftyint_-infty^infty(hatf_n(s)-g(s))^2ds=0$$
$$int_-infty^infty(g(s))^2ds=int_-infty^infty(f(t))^2dt$$
How to show $$lim_nrightarrowinftyint_-a^aleft(frac12piint_-n^ng(s)e^istds-f(t)right)^2dt=0$$ where $a in R$ :
First prove it for $f$ with compact support then generalise it to $L^2$ using inequalities.
Proof for $f(t) $ of compact support and $in L^2$ :
clearly $f(t) in L^1$ without loss of generality assume $f(t)=0$ outside $[-pi,pi]$ since we can always replace $f(t)$ with $f(t')$ such as $t=fracn2pi t'$ for $n in N$ and $t' in R$.
define $$g(s)=lim_ntoinftyint_-n^nf(t)e^-istdt$$
The relationship between fourier transform and fourier series for $f(t)$ above follows from
$$(1) lim_nrightarrowinftyint_-pi^pi(frac1piint_-pi^pif(t+u)fracsin(n+frac12)u2sin(u/2)du-frac1piint_-pi^pi f(t+u)fracsin(n+frac12)uudu)^2dt=0$$
which is simply consequence of Riemann-Lebesgue Lemma applied to the function $ f (t+u) k (u) $, where $k(u)=frac12sin(u/2)-frac1u$ , using remainder term for the Taylor's series of $sin(u/2)$, it easily seen $k(u) $ is bounded over $[-pi, pi]$
$$(2)frac12pi int_-(n+frac12)^(n+frac12)g(s)e^istds=frac1piint_-infty^inftyf(t+u)fracsin(n+frac12)uudu$$
This can be derived from fubini's theorem and this theorem : $mu$ is Lebesgue outer measure, $ A $ is a measurable set with finite measure. For all $n$ , $int_Af_n^2dmu le k $ where $ k in R $ , $ f_n $ is uniformly integrable
Norm convergence for Fourier series :
$$(3) lim_nrightarrowinftyint_-pi^pi(frac1piint_-pi^pif(t+u)fracsin(n+frac12)u2sin(u/2)du-f(t))^2dt=0$$
using Riemann Lebesgue Lemma on (2) we have $$frac12pi int_-(n+frac12)^(n+frac12)g(s)e^istds=frac1piint_-pi^pif(t+u)fracsin(n+frac12)uudu$$
Using Norm convergence of fourier series (3), Riemann-Lebesgue lemma, Cauchy-Schwartz Inequality,$a^2+b^2 ge 2ab$ and (1) and (2) we have:
$$lim_nrightarrowinftyint_-pi^pileft(frac12piint_-(n+frac12)^n+frac12g(s)e^istds-f(t)right)^2dt=0$$
Proof for $f(t) in L^2$:
define $f_m=f_[-m,m]$ where $m in N$ and its fourier transform is $hatf_m$
$$lim_mrightarrowinfty f_m=f$$
By Plancherel theorem : $lim_mrightarrowinfty hatf_m=lim_mrightarrowinftyint_-m^mf(t)e^-istdt=hatf$ in sense of $L^2$
our previous result for function with compact support : $$lim_nto inftyint_-pi^pi(frac12piint_-n^nhatf_m(s)e^istds-f_m(t))^2dt =0 $$
$$lim_nto inftyint_-pi^pi(frac12piint_-n^nhatf_m(s)e^istds-f_m(t))^2dt=lim_nto inftyint_-pi^pi(frac12piint_-n^nhatf_m(s)e^istds-f_m(t)_[-pi,pi])^2dt =lim_nto inftyint_-pi^pi(frac12piint_-n^nhatf_m(s)e^istds-f(t))^2dt=lim_nto inftyint_-pi^pi(frac12piint_-n^nhatf_n(s)e^istds-f(t))^2dt=0 $$
There is a subsequence $n$ such that $hatf_n to hatf $ ae. Let's work with such subsequence
$$lim_nto inftyint_-pi^pi(frac12piint_-n^nhatf(s)e^istds-frac12piint_-n^nhatf_n(s)e^istds)^2dt =$$$$lim_nto inftyint_-pi^pi(frac1piint_-infty^infty(f(t+u)_[-infty,-n]+f(t+u)_[n,infty])fracsin(n+frac12)uudu)^2dt $$ $$le 2(||f(u)_[-infty,-n]+f(u)_[n,infty]||_L^2+||fracsin(n+frac12)uu||_L^2)^-1/2$$ by Cauchy-Scwartz ineqaulity
because $a^2+b^2 ge 2ab$ , it follows from Cauchy-Schwartz inequality that:
$lim_nto inftyint_-pi^pi(frac12piint_-n^nhatf(s)e^istds-f(t))^2dtle$
$lim_nto inftyint_-pi^pi(|frac12piint_-n^nhatf(s)e^istds-f_n(t)|+|f_n(t)-f(t)|)^2dt=0 $
we proved the result for some sequence $n$ to see the result holds for all $n$ we observe that
$$lim_nto inftyint_-pi^pi(frac12piint_-(n+1)^(n+1)hatf(s)e^istds-int_-n^nhatf(s)e^istds)^2dt le$$$$lim_nto inftyint_-pi^pi(frac12piint_-(n+1)^(-n)|hatf(s)|ds+frac12piint_n^(n+1)|hatf(s)|ds)^2dt le lim_nto infty frac12pi(||hatf_[-n-1,-n]||_L^2+||hatf_[n,n+1]||_L^2)^2=0$$ by Hölder's Inequality
answered Jul 21 at 3:41
S.t.a.l.k.e.r
721320
add a comment |Â
up vote
2
down vote
up vote
2
down vote
The pointwise convergence known as Carleson's theorem is a delicate problem in the sense that the proof is about 10 pages.The paper by Carleson focused on the Fourier series convergence while simpler proofs focused on convergence of Fourier transform , nevertheless the two are equivalent and the $ L^2$ convergence is not difficult to prove:
Given an $L^2$ function $f(t)$,
define $$hatf_n(s)=int_-n^nf(t)e^-istdt$$
It can be easily shown using Plancherel theorem that there is some $g(s) in L^2$ and $$lim_nrightarrowinftyint_-infty^infty(hatf_n(s)-g(s))^2ds=0$$
$$int_-infty^infty(g(s))^2ds=int_-infty^infty(f(t))^2dt$$
How to show $$lim_nrightarrowinftyint_-a^aleft(frac12piint_-n^ng(s)e^istds-f(t)right)^2dt=0$$ where $a in R$ :
First prove it for $f$ with compact support then generalise it to $L^2$ using inequalities.
Proof for $f(t) $ of compact support and $in L^2$ :
clearly $f(t) in L^1$ without loss of generality assume $f(t)=0$ outside $[-pi,pi]$ since we can always replace $f(t)$ with $f(t')$ such as $t=fracn2pi t'$ for $n in N$ and $t' in R$.
define $$g(s)=lim_ntoinftyint_-n^nf(t)e^-istdt$$
The relationship between fourier transform and fourier series for $f(t)$ above follows from
$$(1) lim_nrightarrowinftyint_-pi^pi(frac1piint_-pi^pif(t+u)fracsin(n+frac12)u2sin(u/2)du-frac1piint_-pi^pi f(t+u)fracsin(n+frac12)uudu)^2dt=0$$
which is simply consequence of Riemann-Lebesgue Lemma applied to the function $ f (t+u) k (u) $, where $k(u)=frac12sin(u/2)-frac1u$ , using remainder term for the Taylor's series of $sin(u/2)$, it easily seen $k(u) $ is bounded over $[-pi, pi]$
$$(2)frac12pi int_-(n+frac12)^(n+frac12)g(s)e^istds=frac1piint_-infty^inftyf(t+u)fracsin(n+frac12)uudu$$
This can be derived from fubini's theorem and this theorem : $mu$ is Lebesgue outer measure, $ A $ is a measurable set with finite measure. For all $n$ , $int_Af_n^2dmu le k $ where $ k in R $ , $ f_n $ is uniformly integrable
Norm convergence for Fourier series :
$$(3) lim_nrightarrowinftyint_-pi^pi(frac1piint_-pi^pif(t+u)fracsin(n+frac12)u2sin(u/2)du-f(t))^2dt=0$$
using Riemann Lebesgue Lemma on (2) we have $$frac12pi int_-(n+frac12)^(n+frac12)g(s)e^istds=frac1piint_-pi^pif(t+u)fracsin(n+frac12)uudu$$
Using Norm convergence of fourier series (3), Riemann-Lebesgue lemma, Cauchy-Schwartz Inequality,$a^2+b^2 ge 2ab$ and (1) and (2) we have:
$$lim_nrightarrowinftyint_-pi^pileft(frac12piint_-(n+frac12)^n+frac12g(s)e^istds-f(t)right)^2dt=0$$
Proof for $f(t) in L^2$:
define $f_m=f_[-m,m]$ where $m in N$ and its fourier transform is $hatf_m$
$$lim_mrightarrowinfty f_m=f$$
By Plancherel theorem : $lim_mrightarrowinfty hatf_m=lim_mrightarrowinftyint_-m^mf(t)e^-istdt=hatf$ in sense of $L^2$
our previous result for function with compact support : $$lim_nto inftyint_-pi^pi(frac12piint_-n^nhatf_m(s)e^istds-f_m(t))^2dt =0 $$
$$lim_nto inftyint_-pi^pi(frac12piint_-n^nhatf_m(s)e^istds-f_m(t))^2dt=lim_nto inftyint_-pi^pi(frac12piint_-n^nhatf_m(s)e^istds-f_m(t)_[-pi,pi])^2dt =lim_nto inftyint_-pi^pi(frac12piint_-n^nhatf_m(s)e^istds-f(t))^2dt=lim_nto inftyint_-pi^pi(frac12piint_-n^nhatf_n(s)e^istds-f(t))^2dt=0 $$
There is a subsequence $n$ such that $hatf_n to hatf $ ae. Let's work with such subsequence
$$lim_nto inftyint_-pi^pi(frac12piint_-n^nhatf(s)e^istds-frac12piint_-n^nhatf_n(s)e^istds)^2dt =$$$$lim_nto inftyint_-pi^pi(frac1piint_-infty^infty(f(t+u)_[-infty,-n]+f(t+u)_[n,infty])fracsin(n+frac12)uudu)^2dt $$ $$le 2(||f(u)_[-infty,-n]+f(u)_[n,infty]||_L^2+||fracsin(n+frac12)uu||_L^2)^-1/2$$ by Cauchy-Scwartz ineqaulity
because $a^2+b^2 ge 2ab$ , it follows from Cauchy-Schwartz inequality that:
$lim_nto inftyint_-pi^pi(frac12piint_-n^nhatf(s)e^istds-f(t))^2dtle$
$lim_nto inftyint_-pi^pi(|frac12piint_-n^nhatf(s)e^istds-f_n(t)|+|f_n(t)-f(t)|)^2dt=0 $
we proved the result for some sequence $n$ to see the result holds for all $n$ we observe that
$$lim_nto inftyint_-pi^pi(frac12piint_-(n+1)^(n+1)hatf(s)e^istds-int_-n^nhatf(s)e^istds)^2dt le$$$$lim_nto inftyint_-pi^pi(frac12piint_-(n+1)^(-n)|hatf(s)|ds+frac12piint_n^(n+1)|hatf(s)|ds)^2dt le lim_nto infty frac12pi(||hatf_[-n-1,-n]||_L^2+||hatf_[n,n+1]||_L^2)^2=0$$ by Hölder's Inequality
answered Jul 21 at 3:41
S.t.a.l.k.e.r
721320
The pointwise convergence known as Carleson's theorem is a delicate problem in the sense that the proof is about 10 pages.The paper by Carleson focused on the Fourier series convergence while simpler proofs focused on convergence of Fourier transform , nevertheless the two are equivalent and the $ L^2$ convergence is not difficult to prove:
Given an $L^2$ function $f(t)$,
define $$hatf_n(s)=int_-n^nf(t)e^-istdt$$
It can be easily shown using Plancherel theorem that there is some $g(s) in L^2$ and $$lim_nrightarrowinftyint_-infty^infty(hatf_n(s)-g(s))^2ds=0$$
$$int_-infty^infty(g(s))^2ds=int_-infty^infty(f(t))^2dt$$
How to show $$lim_nrightarrowinftyint_-a^aleft(frac12piint_-n^ng(s)e^istds-f(t)right)^2dt=0$$ where $a in R$ :
First prove it for $f$ with compact support then generalise it to $L^2$ using inequalities.
Proof for $f(t) $ of compact support and $in L^2$ :
clearly $f(t) in L^1$ without loss of generality assume $f(t)=0$ outside $[-pi,pi]$ since we can always replace $f(t)$ with $f(t')$ such as $t=fracn2pi t'$ for $n in N$ and $t' in R$.
define $$g(s)=lim_ntoinftyint_-n^nf(t)e^-istdt$$
The relationship between fourier transform and fourier series for $f(t)$ above follows from
$$(1) lim_nrightarrowinftyint_-pi^pi(frac1piint_-pi^pif(t+u)fracsin(n+frac12)u2sin(u/2)du-frac1piint_-pi^pi f(t+u)fracsin(n+frac12)uudu)^2dt=0$$
which is simply consequence of Riemann-Lebesgue Lemma applied to the function $ f (t+u) k (u) $, where $k(u)=frac12sin(u/2)-frac1u$ , using remainder term for the Taylor's series of $sin(u/2)$, it easily seen $k(u) $ is bounded over $[-pi, pi]$
$$(2)frac12pi int_-(n+frac12)^(n+frac12)g(s)e^istds=frac1piint_-infty^inftyf(t+u)fracsin(n+frac12)uudu$$
This can be derived from fubini's theorem and this theorem : $mu$ is Lebesgue outer measure, $ A $ is a measurable set with finite measure. For all $n$ , $int_Af_n^2dmu le k $ where $ k in R $ , $ f_n $ is uniformly integrable
Norm convergence for Fourier series :
$$(3) lim_nrightarrowinftyint_-pi^pi(frac1piint_-pi^pif(t+u)fracsin(n+frac12)u2sin(u/2)du-f(t))^2dt=0$$
using Riemann Lebesgue Lemma on (2) we have $$frac12pi int_-(n+frac12)^(n+frac12)g(s)e^istds=frac1piint_-pi^pif(t+u)fracsin(n+frac12)uudu$$
Using Norm convergence of fourier series (3), Riemann-Lebesgue lemma, Cauchy-Schwartz Inequality,$a^2+b^2 ge 2ab$ and (1) and (2) we have:
$$lim_nrightarrowinftyint_-pi^pileft(frac12piint_-(n+frac12)^n+frac12g(s)e^istds-f(t)right)^2dt=0$$
Proof for $f(t) in L^2$:
define $f_m=f_[-m,m]$ where $m in N$ and its fourier transform is $hatf_m$
$$lim_mrightarrowinfty f_m=f$$
By Plancherel theorem : $lim_mrightarrowinfty hatf_m=lim_mrightarrowinftyint_-m^mf(t)e^-istdt=hatf$ in sense of $L^2$
our previous result for function with compact support : $$lim_nto inftyint_-pi^pi(frac12piint_-n^nhatf_m(s)e^istds-f_m(t))^2dt =0 $$
$$lim_nto inftyint_-pi^pi(frac12piint_-n^nhatf_m(s)e^istds-f_m(t))^2dt=lim_nto inftyint_-pi^pi(frac12piint_-n^nhatf_m(s)e^istds-f_m(t)_[-pi,pi])^2dt =lim_nto inftyint_-pi^pi(frac12piint_-n^nhatf_m(s)e^istds-f(t))^2dt=lim_nto inftyint_-pi^pi(frac12piint_-n^nhatf_n(s)e^istds-f(t))^2dt=0 $$
There is a subsequence $n$ such that $hatf_n to hatf $ ae. Let's work with such subsequence
$$lim_nto inftyint_-pi^pi(frac12piint_-n^nhatf(s)e^istds-frac12piint_-n^nhatf_n(s)e^istds)^2dt =$$$$lim_nto inftyint_-pi^pi(frac1piint_-infty^infty(f(t+u)_[-infty,-n]+f(t+u)_[n,infty])fracsin(n+frac12)uudu)^2dt $$ $$le 2(||f(u)_[-infty,-n]+f(u)_[n,infty]||_L^2+||fracsin(n+frac12)uu||_L^2)^-1/2$$ by Cauchy-Scwartz ineqaulity
because $a^2+b^2 ge 2ab$ , it follows from Cauchy-Schwartz inequality that:
$lim_nto inftyint_-pi^pi(frac12piint_-n^nhatf(s)e^istds-f(t))^2dtle$
$lim_nto inftyint_-pi^pi(|frac12piint_-n^nhatf(s)e^istds-f_n(t)|+|f_n(t)-f(t)|)^2dt=0 $
we proved the result for some sequence $n$ to see the result holds for all $n$ we observe that
$$lim_nto inftyint_-pi^pi(frac12piint_-(n+1)^(n+1)hatf(s)e^istds-int_-n^nhatf(s)e^istds)^2dt le$$$$lim_nto inftyint_-pi^pi(frac12piint_-(n+1)^(-n)|hatf(s)|ds+frac12piint_n^(n+1)|hatf(s)|ds)^2dt le lim_nto infty frac12pi(||hatf_[-n-1,-n]||_L^2+||hatf_[n,n+1]||_L^2)^2=0$$ by Hölder's Inequality
answered Jul 21 at 3:41
S.t.a.l.k.e.r
721320
edited Jul 21 at 3:59
answered Jul 21 at 3:41
S.t.a.l.k.e.r
721320
answered Jul 21 at 3:41
S.t.a.l.k.e.r
721320
answered Jul 21 at 3:41
S.t.a.l.k.e.r
721320
721320
add a comment |Â
add a comment |Â
up vote
2
down vote
Being $L^2(mathbbR)backslash L^1(mathbbR)neqemptyset$ and being $int_mathbbRf(x)e^-2pi ixxidx$ well defined as a Lebesgue integral iff $fin L^1(mathbbR)$, you are absolutely right to complain.
By the way, you can define the Fourier transform $mathcalF$ for $L^2(mathbbR)$ by a little detour, first defining it by that formula for $L^1(mathbbR)$ functions, then proving that $$forall fin L^1(mathbbR)cap L^2(mathbbR), |mathcalF(f)|_2=|f|_2,$$
so the Fourier transform is an isometry and then can be extended uniquely to an isometry on the whole $L^2(mathbbR)$ by continuity and density.
Now, having defined the Fourier transform in $L^2(mathbbR)$ via this indirect route, you might ask yourself if there is a more practical procedure to compute it. The first idea that comes to mind is to try to give meaning to the $L^1(mathbbR)$ formula: $$ mathcalF(f)(xi)=int_mathbbRf(x)e^-2pi ixxidx,$$
also in the $L^2(mathbbR)$ context.
As initially said, this can't be, in general, interpreted as a Lebesgue integral. However, what about interpreting it as a improper integral? After all, for subsets $KsubsetmathbbR$ of finite Lebesgue measure, by Holder inequality, we have that $L^2 (K)subset L^1 (K)$ and so, if $fin L^2 (mathbbR)$ and $M>0$, it is well defined the integral: $$int_-M^Mf(x)e^-2pi i xxidx,$$ and we can try to get the limit of this quantity as $Mrightarrowinfty$. So the question: does it hold true for $fin L^2 (mathbbR)$ that $$mathcalF(f)(xi)=lim_Mrightarrowinfty int_-M^Mf(x)e^-2pi i xxidx?$$
Well, by Carleson's theorem, this limit exists for a.e. $xiinmathbbR$ and that formula holds true if interpreted as an a.e. $xiinmathbbR$ equality. So, at the end, we have recovered a meaning for the original formula via an improper integral, obtaining in this manner an explicit way to calculate the Fourier transform of a $L^2(mathbbR)$ function.
answered Jul 20 at 20:46
Bob
1,467522
Just for notation, $L^2backslash L^1$ looks weird since $L^1notsubset L^2$. I would use more $L^2cap (L^1)^c$.
– Surb
Jul 21 at 8:01
1
@surb en.wikipedia.org/wiki/…
– Bob
Jul 21 at 8:12
add a comment |Â
up vote
2
down vote
Being $L^2(mathbbR)backslash L^1(mathbbR)neqemptyset$ and being $int_mathbbRf(x)e^-2pi ixxidx$ well defined as a Lebesgue integral iff $fin L^1(mathbbR)$, you are absolutely right to complain.
By the way, you can define the Fourier transform $mathcalF$ for $L^2(mathbbR)$ by a little detour, first defining it by that formula for $L^1(mathbbR)$ functions, then proving that $$forall fin L^1(mathbbR)cap L^2(mathbbR), |mathcalF(f)|_2=|f|_2,$$
so the Fourier transform is an isometry and then can be extended uniquely to an isometry on the whole $L^2(mathbbR)$ by continuity and density.
Now, having defined the Fourier transform in $L^2(mathbbR)$ via this indirect route, you might ask yourself if there is a more practical procedure to compute it. The first idea that comes to mind is to try to give meaning to the $L^1(mathbbR)$ formula: $$ mathcalF(f)(xi)=int_mathbbRf(x)e^-2pi ixxidx,$$
also in the $L^2(mathbbR)$ context.
As initially said, this can't be, in general, interpreted as a Lebesgue integral. However, what about interpreting it as a improper integral? After all, for subsets $KsubsetmathbbR$ of finite Lebesgue measure, by Holder inequality, we have that $L^2 (K)subset L^1 (K)$ and so, if $fin L^2 (mathbbR)$ and $M>0$, it is well defined the integral: $$int_-M^Mf(x)e^-2pi i xxidx,$$ and we can try to get the limit of this quantity as $Mrightarrowinfty$. So the question: does it hold true for $fin L^2 (mathbbR)$ that $$mathcalF(f)(xi)=lim_Mrightarrowinfty int_-M^Mf(x)e^-2pi i xxidx?$$
Well, by Carleson's theorem, this limit exists for a.e. $xiinmathbbR$ and that formula holds true if interpreted as an a.e. $xiinmathbbR$ equality. So, at the end, we have recovered a meaning for the original formula via an improper integral, obtaining in this manner an explicit way to calculate the Fourier transform of a $L^2(mathbbR)$ function.
answered Jul 20 at 20:46
Bob
1,467522
Just for notation, $L^2backslash L^1$ looks weird since $L^1notsubset L^2$. I would use more $L^2cap (L^1)^c$.
– Surb
Jul 21 at 8:01
1
@surb en.wikipedia.org/wiki/…
– Bob
Jul 21 at 8:12
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Being $L^2(mathbbR)backslash L^1(mathbbR)neqemptyset$ and being $int_mathbbRf(x)e^-2pi ixxidx$ well defined as a Lebesgue integral iff $fin L^1(mathbbR)$, you are absolutely right to complain.
By the way, you can define the Fourier transform $mathcalF$ for $L^2(mathbbR)$ by a little detour, first defining it by that formula for $L^1(mathbbR)$ functions, then proving that $$forall fin L^1(mathbbR)cap L^2(mathbbR), |mathcalF(f)|_2=|f|_2,$$
so the Fourier transform is an isometry and then can be extended uniquely to an isometry on the whole $L^2(mathbbR)$ by continuity and density.
Now, having defined the Fourier transform in $L^2(mathbbR)$ via this indirect route, you might ask yourself if there is a more practical procedure to compute it. The first idea that comes to mind is to try to give meaning to the $L^1(mathbbR)$ formula: $$ mathcalF(f)(xi)=int_mathbbRf(x)e^-2pi ixxidx,$$
also in the $L^2(mathbbR)$ context.
As initially said, this can't be, in general, interpreted as a Lebesgue integral. However, what about interpreting it as a improper integral? After all, for subsets $KsubsetmathbbR$ of finite Lebesgue measure, by Holder inequality, we have that $L^2 (K)subset L^1 (K)$ and so, if $fin L^2 (mathbbR)$ and $M>0$, it is well defined the integral: $$int_-M^Mf(x)e^-2pi i xxidx,$$ and we can try to get the limit of this quantity as $Mrightarrowinfty$. So the question: does it hold true for $fin L^2 (mathbbR)$ that $$mathcalF(f)(xi)=lim_Mrightarrowinfty int_-M^Mf(x)e^-2pi i xxidx?$$
Well, by Carleson's theorem, this limit exists for a.e. $xiinmathbbR$ and that formula holds true if interpreted as an a.e. $xiinmathbbR$ equality. So, at the end, we have recovered a meaning for the original formula via an improper integral, obtaining in this manner an explicit way to calculate the Fourier transform of a $L^2(mathbbR)$ function.
answered Jul 20 at 20:46
Bob
1,467522
Being $L^2(mathbbR)backslash L^1(mathbbR)neqemptyset$ and being $int_mathbbRf(x)e^-2pi ixxidx$ well defined as a Lebesgue integral iff $fin L^1(mathbbR)$, you are absolutely right to complain.
By the way, you can define the Fourier transform $mathcalF$ for $L^2(mathbbR)$ by a little detour, first defining it by that formula for $L^1(mathbbR)$ functions, then proving that $$forall fin L^1(mathbbR)cap L^2(mathbbR), |mathcalF(f)|_2=|f|_2,$$
so the Fourier transform is an isometry and then can be extended uniquely to an isometry on the whole $L^2(mathbbR)$ by continuity and density.
Now, having defined the Fourier transform in $L^2(mathbbR)$ via this indirect route, you might ask yourself if there is a more practical procedure to compute it. The first idea that comes to mind is to try to give meaning to the $L^1(mathbbR)$ formula: $$ mathcalF(f)(xi)=int_mathbbRf(x)e^-2pi ixxidx,$$
also in the $L^2(mathbbR)$ context.
As initially said, this can't be, in general, interpreted as a Lebesgue integral. However, what about interpreting it as a improper integral? After all, for subsets $KsubsetmathbbR$ of finite Lebesgue measure, by Holder inequality, we have that $L^2 (K)subset L^1 (K)$ and so, if $fin L^2 (mathbbR)$ and $M>0$, it is well defined the integral: $$int_-M^Mf(x)e^-2pi i xxidx,$$ and we can try to get the limit of this quantity as $Mrightarrowinfty$. So the question: does it hold true for $fin L^2 (mathbbR)$ that $$mathcalF(f)(xi)=lim_Mrightarrowinfty int_-M^Mf(x)e^-2pi i xxidx?$$
Well, by Carleson's theorem, this limit exists for a.e. $xiinmathbbR$ and that formula holds true if interpreted as an a.e. $xiinmathbbR$ equality. So, at the end, we have recovered a meaning for the original formula via an improper integral, obtaining in this manner an explicit way to calculate the Fourier transform of a $L^2(mathbbR)$ function.
answered Jul 20 at 20:46
Bob
1,467522
edited Jul 21 at 9:13
answered Jul 20 at 20:46
Bob
1,467522
answered Jul 20 at 20:46
Bob
1,467522
answered Jul 20 at 20:46
Bob
1,467522
1,467522
Just for notation, $L^2backslash L^1$ looks weird since $L^1notsubset L^2$. I would use more $L^2cap (L^1)^c$.
– Surb
Jul 21 at 8:01
1
@surb en.wikipedia.org/wiki/…
– Bob
Jul 21 at 8:12
add a comment |Â
Just for notation, $L^2backslash L^1$ looks weird since $L^1notsubset L^2$. I would use more $L^2cap (L^1)^c$.
– Surb
Jul 21 at 8:01
1
@surb en.wikipedia.org/wiki/…
– Bob
Jul 21 at 8:12
Just for notation, $L^2backslash L^1$ looks weird since $L^1notsubset L^2$. I would use more $L^2cap (L^1)^c$.
– Surb
Jul 21 at 8:01
Just for notation, $L^2backslash L^1$ looks weird since $L^1notsubset L^2$. I would use more $L^2cap (L^1)^c$.
– Surb
Jul 21 at 8:01
1
1
@surb en.wikipedia.org/wiki/…
– Bob
Jul 21 at 8:12
@surb en.wikipedia.org/wiki/…
– Bob
Jul 21 at 8:12
add a comment |Â
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
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2857971%2fcompute-the-fourier-transform-of-a-l2-function%23new-answer', 'question_page');
);
Post as a guest
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
Post as a guest
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
Post as a guest
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
See the riemann lebesgue lemma. en.wikipedia.org/wiki/Riemann%E2%80%93Lebesgue_lemma
– RHowe
Jul 20 at 19:55
In the text "Analysis" by Lieb & Loss it's explains very well: they construct the Fourier transform beginning by $L^1$ function then with Plancherel's theorem they construct it for $L^2$ and $L^p$
– Davide Morgante
Jul 20 at 20:01
@DavideMorgante: For $L^p$ it's unfortunately not possible in general. I guess that for $pin [1,2]$, we can extend Fourier transform $L^pto L^p'$ using Riesz-Thorin (with $frac1p+frac1p'=1$), but I don't think that it's possible for $p>2$ (but maybe I'm wrong...)
– Surb
Jul 20 at 20:09
@Surb I'm a little bit rusty when it comes to this specific mathematical construction. But from what I recall you're right, the authors used a sharper version of the Hausdorff-Young inequality which doesn't stand for $pgt 2$
– Davide Morgante
Jul 20 at 20:14