Show that $hat{f}$ is continuous and $hat{f(xi)}rightarrow 0 $ as $|xi|rightarrow 0$












1












$begingroup$


Suppose $f$ continuous and of moderate decrease. Denote the Fourier Transform of $f$ by $hat{f}$. I need to prove that $hat{f}$ is continuous and $hat{f(xi)}rightarrow 0 $ as $|xi|rightarrow 0$.



For the first part, I'm thinking to show that $|hat{f}(xi+h)-hat{f}(xi)|$ goes to zero when $h$ goes to zero. So, I have



$$ |hat{f}(xi+h)-hat{f}(xi)|=left|int_{mathbb{R}}f(x)e^{-2pi ix(xi+h)}dx-int_{mathbb{R}}f(x)e^{-2pi ixxi}dx right| $$
$$leqint_{mathbb{R}} |f(x)|left|e^{-2pi ixxi}right|left|e^{-2pi ixh}-1right|dx $$
If $hrightarrow 0$, so $e^{-2pi ixh}rightarrow 1$, and this implies $left|e^{-2pi ixh}-1right|rightarrow 0$. So, the integral goes to zero and $hat{f}$ is continuous.



Is this proof correct?



For the second part, the hint is to prove the following statement:
$$hat{f}(xi)=frac{1}{2}int_{mathbb{R}}left[f(x)-fleft(x-frac{1}{2xi}right)right]e^{-2pi i x xi}dx $$
If that is true, when $|xi|rightarrow 0$, $fleft(1-frac{1}{2xi}right)rightarrow f(x)$, and it implies $hat{f}(xi)rightarrow 0$. So, I just need to prove that statement.



All that I have is:



$$frac{1}{2}int_{mathbb{R}}left[f(x)-fleft(x-frac{1}{2xi}right)right]e^{-2pi i x xi}dx=frac{1}{2}hat{f}(xi)-frac{1}{2}int_{mathbb{R}}fleft(1-frac{1}{2xi}right)e^{-2pi i x xi}dx$$



So, prove the statement is equivalent to prove that



$$int_{mathbb{R}}fleft(1-frac{1}{2xi}right)e^{-2pi i x xi}dx=-frac{1}{2}hat{f}(xi)$$



How can I prove that?






fourier-analysis fourier-transform







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Please be careful about commuting limits and integrals, as the answer below notes, this is justified here by DCT
    $endgroup$
    – qbert
    Dec 5 '18 at 6:57
















1












$begingroup$


Suppose $f$ continuous and of moderate decrease. Denote the Fourier Transform of $f$ by $hat{f}$. I need to prove that $hat{f}$ is continuous and $hat{f(xi)}rightarrow 0 $ as $|xi|rightarrow 0$.



For the first part, I'm thinking to show that $|hat{f}(xi+h)-hat{f}(xi)|$ goes to zero when $h$ goes to zero. So, I have



$$ |hat{f}(xi+h)-hat{f}(xi)|=left|int_{mathbb{R}}f(x)e^{-2pi ix(xi+h)}dx-int_{mathbb{R}}f(x)e^{-2pi ixxi}dx right| $$
$$leqint_{mathbb{R}} |f(x)|left|e^{-2pi ixxi}right|left|e^{-2pi ixh}-1right|dx $$
If $hrightarrow 0$, so $e^{-2pi ixh}rightarrow 1$, and this implies $left|e^{-2pi ixh}-1right|rightarrow 0$. So, the integral goes to zero and $hat{f}$ is continuous.



Is this proof correct?



For the second part, the hint is to prove the following statement:
$$hat{f}(xi)=frac{1}{2}int_{mathbb{R}}left[f(x)-fleft(x-frac{1}{2xi}right)right]e^{-2pi i x xi}dx $$
If that is true, when $|xi|rightarrow 0$, $fleft(1-frac{1}{2xi}right)rightarrow f(x)$, and it implies $hat{f}(xi)rightarrow 0$. So, I just need to prove that statement.



All that I have is:



$$frac{1}{2}int_{mathbb{R}}left[f(x)-fleft(x-frac{1}{2xi}right)right]e^{-2pi i x xi}dx=frac{1}{2}hat{f}(xi)-frac{1}{2}int_{mathbb{R}}fleft(1-frac{1}{2xi}right)e^{-2pi i x xi}dx$$



So, prove the statement is equivalent to prove that



$$int_{mathbb{R}}fleft(1-frac{1}{2xi}right)e^{-2pi i x xi}dx=-frac{1}{2}hat{f}(xi)$$



How can I prove that?






fourier-analysis fourier-transform







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Please be careful about commuting limits and integrals, as the answer below notes, this is justified here by DCT
    $endgroup$
    – qbert
    Dec 5 '18 at 6:57














1












1








1





$begingroup$


Suppose $f$ continuous and of moderate decrease. Denote the Fourier Transform of $f$ by $hat{f}$. I need to prove that $hat{f}$ is continuous and $hat{f(xi)}rightarrow 0 $ as $|xi|rightarrow 0$.



For the first part, I'm thinking to show that $|hat{f}(xi+h)-hat{f}(xi)|$ goes to zero when $h$ goes to zero. So, I have



$$ |hat{f}(xi+h)-hat{f}(xi)|=left|int_{mathbb{R}}f(x)e^{-2pi ix(xi+h)}dx-int_{mathbb{R}}f(x)e^{-2pi ixxi}dx right| $$
$$leqint_{mathbb{R}} |f(x)|left|e^{-2pi ixxi}right|left|e^{-2pi ixh}-1right|dx $$
If $hrightarrow 0$, so $e^{-2pi ixh}rightarrow 1$, and this implies $left|e^{-2pi ixh}-1right|rightarrow 0$. So, the integral goes to zero and $hat{f}$ is continuous.



Is this proof correct?



For the second part, the hint is to prove the following statement:
$$hat{f}(xi)=frac{1}{2}int_{mathbb{R}}left[f(x)-fleft(x-frac{1}{2xi}right)right]e^{-2pi i x xi}dx $$
If that is true, when $|xi|rightarrow 0$, $fleft(1-frac{1}{2xi}right)rightarrow f(x)$, and it implies $hat{f}(xi)rightarrow 0$. So, I just need to prove that statement.



All that I have is:



$$frac{1}{2}int_{mathbb{R}}left[f(x)-fleft(x-frac{1}{2xi}right)right]e^{-2pi i x xi}dx=frac{1}{2}hat{f}(xi)-frac{1}{2}int_{mathbb{R}}fleft(1-frac{1}{2xi}right)e^{-2pi i x xi}dx$$



So, prove the statement is equivalent to prove that



$$int_{mathbb{R}}fleft(1-frac{1}{2xi}right)e^{-2pi i x xi}dx=-frac{1}{2}hat{f}(xi)$$



How can I prove that?






fourier-analysis fourier-transform







share|cite|improve this question









$endgroup$




Suppose $f$ continuous and of moderate decrease. Denote the Fourier Transform of $f$ by $hat{f}$. I need to prove that $hat{f}$ is continuous and $hat{f(xi)}rightarrow 0 $ as $|xi|rightarrow 0$.



For the first part, I'm thinking to show that $|hat{f}(xi+h)-hat{f}(xi)|$ goes to zero when $h$ goes to zero. So, I have



$$ |hat{f}(xi+h)-hat{f}(xi)|=left|int_{mathbb{R}}f(x)e^{-2pi ix(xi+h)}dx-int_{mathbb{R}}f(x)e^{-2pi ixxi}dx right| $$
$$leqint_{mathbb{R}} |f(x)|left|e^{-2pi ixxi}right|left|e^{-2pi ixh}-1right|dx $$
If $hrightarrow 0$, so $e^{-2pi ixh}rightarrow 1$, and this implies $left|e^{-2pi ixh}-1right|rightarrow 0$. So, the integral goes to zero and $hat{f}$ is continuous.



Is this proof correct?



For the second part, the hint is to prove the following statement:
$$hat{f}(xi)=frac{1}{2}int_{mathbb{R}}left[f(x)-fleft(x-frac{1}{2xi}right)right]e^{-2pi i x xi}dx $$
If that is true, when $|xi|rightarrow 0$, $fleft(1-frac{1}{2xi}right)rightarrow f(x)$, and it implies $hat{f}(xi)rightarrow 0$. So, I just need to prove that statement.



All that I have is:



$$frac{1}{2}int_{mathbb{R}}left[f(x)-fleft(x-frac{1}{2xi}right)right]e^{-2pi i x xi}dx=frac{1}{2}hat{f}(xi)-frac{1}{2}int_{mathbb{R}}fleft(1-frac{1}{2xi}right)e^{-2pi i x xi}dx$$



So, prove the statement is equivalent to prove that



$$int_{mathbb{R}}fleft(1-frac{1}{2xi}right)e^{-2pi i x xi}dx=-frac{1}{2}hat{f}(xi)$$



How can I prove that?






fourier-analysis fourier-transform




fourier-analysis fourier-transform






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 5 '18 at 1:53









Mateus RochaMateus Rocha

817117




817117








  • 1




    $begingroup$
    Please be careful about commuting limits and integrals, as the answer below notes, this is justified here by DCT
    $endgroup$
    – qbert
    Dec 5 '18 at 6:57














  • 1




    $begingroup$
    Please be careful about commuting limits and integrals, as the answer below notes, this is justified here by DCT
    $endgroup$
    – qbert
    Dec 5 '18 at 6:57








1




1




$begingroup$
Please be careful about commuting limits and integrals, as the answer below notes, this is justified here by DCT
$endgroup$
– qbert
Dec 5 '18 at 6:57




$begingroup$
Please be careful about commuting limits and integrals, as the answer below notes, this is justified here by DCT
$endgroup$
– qbert
Dec 5 '18 at 6:57










1 Answer
1






active

oldest

votes


















2












$begingroup$

If by moderate decrease you mean that $int_{-infty}^infty |f(x)| , dx < infty$ then your proof of the first part is correct and the interchange of limit and integral is justified by the dominated convergence theorem.



For the second part, using the change of variables $x = u + dfrac{1}{2xi}$ we have, since $e^{-pi i} = -1$,



$$int_{-R}^R fleft(x- frac{1}{2xi}right) e^{-2pi i xi x} , dx = int_{-R-frac{1}{2xi}}^{R- frac{1}{2xi}} f(u) e^{-2pi i xi u}e^{-pi i} , du = -int_{-R-frac{1}{2xi}}^{R- frac{1}{2xi}} f(u) e^{-2pi i xi u} , du, $$



Thus,



$$tag{*}int_{-infty}^infty fleft(x- frac{1}{2xi}right) e^{-2pi i xi x} , dx = -lim_{Rto infty}int_{-R-frac{1}{2xi}}^{R- frac{1}{2xi}} f(u) e^{-2pi i xi u} , du = -int_{-infty}^{infty} f(u) e^{-2pi i xi u} , du \ - hat{f}(xi)$$



Subtracting (*) from $displaystylehat{f}(xi) = int_{-infty}^{infty} f(x) e^{-2pi i xi x} , dx $, we get



$$2hat{f}(xi) = int_{-infty}^infty fleft(xright) e^{-2pi i xi x} , dx - int_{-infty}^infty fleft(x- frac{1}{2xi}right) e^{-2pi i xi x} , dx, $$



which implies



$$hat{f}(xi) = frac{1}{2} int_{-infty}^infty left[f(x) - fleft(x- frac{1}{2xi}right)right] e^{-2pi i xi x} , dx$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    My definition to Moderate decrease is: Exists $A>0$ such that $|f(x)|leqfrac{A}{1+x^2}$ for all $xinmathbb{R}$. That definition implies yours?
    $endgroup$
    – Mateus Rocha
    Dec 5 '18 at 14:04










  • $begingroup$
    @MateusRocha: Yes that implies that $f$ is absolutely integrable, a sufficient condition for the Fourier transform to exist and be uniformly continuous. Note that $int_{mathbb{R}} |f(x)| , dx leqslant int_{mathbb{R}} frac{A}{1 +x^2}, dx = Api < infty$. Then everything above works out well.
    $endgroup$
    – RRL
    Dec 5 '18 at 14:10













Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3026488%2fshow-that-hatf-is-continuous-and-hatf-xi-rightarrow-0-as-xi-rig%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

If by moderate decrease you mean that $int_{-infty}^infty |f(x)| , dx < infty$ then your proof of the first part is correct and the interchange of limit and integral is justified by the dominated convergence theorem.



For the second part, using the change of variables $x = u + dfrac{1}{2xi}$ we have, since $e^{-pi i} = -1$,



$$int_{-R}^R fleft(x- frac{1}{2xi}right) e^{-2pi i xi x} , dx = int_{-R-frac{1}{2xi}}^{R- frac{1}{2xi}} f(u) e^{-2pi i xi u}e^{-pi i} , du = -int_{-R-frac{1}{2xi}}^{R- frac{1}{2xi}} f(u) e^{-2pi i xi u} , du, $$



Thus,



$$tag{*}int_{-infty}^infty fleft(x- frac{1}{2xi}right) e^{-2pi i xi x} , dx = -lim_{Rto infty}int_{-R-frac{1}{2xi}}^{R- frac{1}{2xi}} f(u) e^{-2pi i xi u} , du = -int_{-infty}^{infty} f(u) e^{-2pi i xi u} , du \ - hat{f}(xi)$$



Subtracting (*) from $displaystylehat{f}(xi) = int_{-infty}^{infty} f(x) e^{-2pi i xi x} , dx $, we get



$$2hat{f}(xi) = int_{-infty}^infty fleft(xright) e^{-2pi i xi x} , dx - int_{-infty}^infty fleft(x- frac{1}{2xi}right) e^{-2pi i xi x} , dx, $$



which implies



$$hat{f}(xi) = frac{1}{2} int_{-infty}^infty left[f(x) - fleft(x- frac{1}{2xi}right)right] e^{-2pi i xi x} , dx$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    My definition to Moderate decrease is: Exists $A>0$ such that $|f(x)|leqfrac{A}{1+x^2}$ for all $xinmathbb{R}$. That definition implies yours?
    $endgroup$
    – Mateus Rocha
    Dec 5 '18 at 14:04










  • $begingroup$
    @MateusRocha: Yes that implies that $f$ is absolutely integrable, a sufficient condition for the Fourier transform to exist and be uniformly continuous. Note that $int_{mathbb{R}} |f(x)| , dx leqslant int_{mathbb{R}} frac{A}{1 +x^2}, dx = Api < infty$. Then everything above works out well.
    $endgroup$
    – RRL
    Dec 5 '18 at 14:10


















2












$begingroup$

If by moderate decrease you mean that $int_{-infty}^infty |f(x)| , dx < infty$ then your proof of the first part is correct and the interchange of limit and integral is justified by the dominated convergence theorem.



For the second part, using the change of variables $x = u + dfrac{1}{2xi}$ we have, since $e^{-pi i} = -1$,



$$int_{-R}^R fleft(x- frac{1}{2xi}right) e^{-2pi i xi x} , dx = int_{-R-frac{1}{2xi}}^{R- frac{1}{2xi}} f(u) e^{-2pi i xi u}e^{-pi i} , du = -int_{-R-frac{1}{2xi}}^{R- frac{1}{2xi}} f(u) e^{-2pi i xi u} , du, $$



Thus,



$$tag{*}int_{-infty}^infty fleft(x- frac{1}{2xi}right) e^{-2pi i xi x} , dx = -lim_{Rto infty}int_{-R-frac{1}{2xi}}^{R- frac{1}{2xi}} f(u) e^{-2pi i xi u} , du = -int_{-infty}^{infty} f(u) e^{-2pi i xi u} , du \ - hat{f}(xi)$$



Subtracting (*) from $displaystylehat{f}(xi) = int_{-infty}^{infty} f(x) e^{-2pi i xi x} , dx $, we get



$$2hat{f}(xi) = int_{-infty}^infty fleft(xright) e^{-2pi i xi x} , dx - int_{-infty}^infty fleft(x- frac{1}{2xi}right) e^{-2pi i xi x} , dx, $$



which implies



$$hat{f}(xi) = frac{1}{2} int_{-infty}^infty left[f(x) - fleft(x- frac{1}{2xi}right)right] e^{-2pi i xi x} , dx$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    My definition to Moderate decrease is: Exists $A>0$ such that $|f(x)|leqfrac{A}{1+x^2}$ for all $xinmathbb{R}$. That definition implies yours?
    $endgroup$
    – Mateus Rocha
    Dec 5 '18 at 14:04










  • $begingroup$
    @MateusRocha: Yes that implies that $f$ is absolutely integrable, a sufficient condition for the Fourier transform to exist and be uniformly continuous. Note that $int_{mathbb{R}} |f(x)| , dx leqslant int_{mathbb{R}} frac{A}{1 +x^2}, dx = Api < infty$. Then everything above works out well.
    $endgroup$
    – RRL
    Dec 5 '18 at 14:10
















2












2








2





$begingroup$

If by moderate decrease you mean that $int_{-infty}^infty |f(x)| , dx < infty$ then your proof of the first part is correct and the interchange of limit and integral is justified by the dominated convergence theorem.



For the second part, using the change of variables $x = u + dfrac{1}{2xi}$ we have, since $e^{-pi i} = -1$,



$$int_{-R}^R fleft(x- frac{1}{2xi}right) e^{-2pi i xi x} , dx = int_{-R-frac{1}{2xi}}^{R- frac{1}{2xi}} f(u) e^{-2pi i xi u}e^{-pi i} , du = -int_{-R-frac{1}{2xi}}^{R- frac{1}{2xi}} f(u) e^{-2pi i xi u} , du, $$



Thus,



$$tag{*}int_{-infty}^infty fleft(x- frac{1}{2xi}right) e^{-2pi i xi x} , dx = -lim_{Rto infty}int_{-R-frac{1}{2xi}}^{R- frac{1}{2xi}} f(u) e^{-2pi i xi u} , du = -int_{-infty}^{infty} f(u) e^{-2pi i xi u} , du \ - hat{f}(xi)$$



Subtracting (*) from $displaystylehat{f}(xi) = int_{-infty}^{infty} f(x) e^{-2pi i xi x} , dx $, we get



$$2hat{f}(xi) = int_{-infty}^infty fleft(xright) e^{-2pi i xi x} , dx - int_{-infty}^infty fleft(x- frac{1}{2xi}right) e^{-2pi i xi x} , dx, $$



which implies



$$hat{f}(xi) = frac{1}{2} int_{-infty}^infty left[f(x) - fleft(x- frac{1}{2xi}right)right] e^{-2pi i xi x} , dx$$






share|cite|improve this answer









$endgroup$



If by moderate decrease you mean that $int_{-infty}^infty |f(x)| , dx < infty$ then your proof of the first part is correct and the interchange of limit and integral is justified by the dominated convergence theorem.



For the second part, using the change of variables $x = u + dfrac{1}{2xi}$ we have, since $e^{-pi i} = -1$,



$$int_{-R}^R fleft(x- frac{1}{2xi}right) e^{-2pi i xi x} , dx = int_{-R-frac{1}{2xi}}^{R- frac{1}{2xi}} f(u) e^{-2pi i xi u}e^{-pi i} , du = -int_{-R-frac{1}{2xi}}^{R- frac{1}{2xi}} f(u) e^{-2pi i xi u} , du, $$



Thus,



$$tag{*}int_{-infty}^infty fleft(x- frac{1}{2xi}right) e^{-2pi i xi x} , dx = -lim_{Rto infty}int_{-R-frac{1}{2xi}}^{R- frac{1}{2xi}} f(u) e^{-2pi i xi u} , du = -int_{-infty}^{infty} f(u) e^{-2pi i xi u} , du \ - hat{f}(xi)$$



Subtracting (*) from $displaystylehat{f}(xi) = int_{-infty}^{infty} f(x) e^{-2pi i xi x} , dx $, we get



$$2hat{f}(xi) = int_{-infty}^infty fleft(xright) e^{-2pi i xi x} , dx - int_{-infty}^infty fleft(x- frac{1}{2xi}right) e^{-2pi i xi x} , dx, $$



which implies



$$hat{f}(xi) = frac{1}{2} int_{-infty}^infty left[f(x) - fleft(x- frac{1}{2xi}right)right] e^{-2pi i xi x} , dx$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 5 '18 at 6:53









RRLRRL

50.5k42573




50.5k42573












  • $begingroup$
    My definition to Moderate decrease is: Exists $A>0$ such that $|f(x)|leqfrac{A}{1+x^2}$ for all $xinmathbb{R}$. That definition implies yours?
    $endgroup$
    – Mateus Rocha
    Dec 5 '18 at 14:04










  • $begingroup$
    @MateusRocha: Yes that implies that $f$ is absolutely integrable, a sufficient condition for the Fourier transform to exist and be uniformly continuous. Note that $int_{mathbb{R}} |f(x)| , dx leqslant int_{mathbb{R}} frac{A}{1 +x^2}, dx = Api < infty$. Then everything above works out well.
    $endgroup$
    – RRL
    Dec 5 '18 at 14:10




















  • $begingroup$
    My definition to Moderate decrease is: Exists $A>0$ such that $|f(x)|leqfrac{A}{1+x^2}$ for all $xinmathbb{R}$. That definition implies yours?
    $endgroup$
    – Mateus Rocha
    Dec 5 '18 at 14:04










  • $begingroup$
    @MateusRocha: Yes that implies that $f$ is absolutely integrable, a sufficient condition for the Fourier transform to exist and be uniformly continuous. Note that $int_{mathbb{R}} |f(x)| , dx leqslant int_{mathbb{R}} frac{A}{1 +x^2}, dx = Api < infty$. Then everything above works out well.
    $endgroup$
    – RRL
    Dec 5 '18 at 14:10


















$begingroup$
My definition to Moderate decrease is: Exists $A>0$ such that $|f(x)|leqfrac{A}{1+x^2}$ for all $xinmathbb{R}$. That definition implies yours?
$endgroup$
– Mateus Rocha
Dec 5 '18 at 14:04




$begingroup$
My definition to Moderate decrease is: Exists $A>0$ such that $|f(x)|leqfrac{A}{1+x^2}$ for all $xinmathbb{R}$. That definition implies yours?
$endgroup$
– Mateus Rocha
Dec 5 '18 at 14:04












$begingroup$
@MateusRocha: Yes that implies that $f$ is absolutely integrable, a sufficient condition for the Fourier transform to exist and be uniformly continuous. Note that $int_{mathbb{R}} |f(x)| , dx leqslant int_{mathbb{R}} frac{A}{1 +x^2}, dx = Api < infty$. Then everything above works out well.
$endgroup$
– RRL
Dec 5 '18 at 14:10






$begingroup$
@MateusRocha: Yes that implies that $f$ is absolutely integrable, a sufficient condition for the Fourier transform to exist and be uniformly continuous. Note that $int_{mathbb{R}} |f(x)| , dx leqslant int_{mathbb{R}} frac{A}{1 +x^2}, dx = Api < infty$. Then everything above works out well.
$endgroup$
– RRL
Dec 5 '18 at 14:10




















draft saved

draft discarded




















































Thanks for contributing an answer to Mathematics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid …



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3026488%2fshow-that-hatf-is-continuous-and-hatf-xi-rightarrow-0-as-xi-rig%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







-

Popular posts from this blog

Bundesstraße 106

Image
Dieser Artikel beschreibt die Bundesstraße 106 in Deutschland. Zur gleichnamigen Straße in Österreich siehe Mölltal Straße. Vorlage:Infobox hochrangige Straße/Wartung/DE-B Bundesstraße 106 in Deutschland Karte Basisdaten Betreiber: Deutschland  Bundesrepublik Deutschland Straßenbeginn: Wismar ( 53° 54′  N , 11° 24′  O 53.900536 11.396875 ) Straßenende: Schwerin ( 53° 36′  N , 11° 24′  O 53.592785 11.404892 ) Gesamtlänge: 40,7 km Bundesland: Mecklenburg-Vorpommern Ausbauzustand: zweistreifig [1] Bundesstraße 106 in Schwerin (Umgehungsstraße) Straßenverlauf Land Mecklenburg-Vorpommern Landkreis Nordwestmecklenburg Ortsumgehung Wismar Wallensteingraben über   Dorf Mecklenburg Wallensteingraben Groß Stieten Bad Kleinen OT Niendorf Bahnstrecke Lübeck–Bad Kleinen Zickhusen Lübstorf Klein Trebbow OT Kirch Stück
Read more

Verónica Boquete

Image
Verónica Boquete im Trikot der Nationalmannschaft (2013) Personalia Name Verónica Boquete Giadans Geburtstag 9. April 1987 Geburtsort Santiago de Compostela, Spanien Größe 159 cm Position Mittelfeld / Sturm Juniorinnen Jahre Station 2001–2005 Xuventú Aguiño Frauen Jahre Station Spiele (Tore) 1 2005–2008 Prainsa Zaragoza 83 (40) 2008–2010 Espanyol Barcelona 53 (30) 2010 Buffalo Flash 9 0 (9) 2010 Chicago Red Stars 3 0 (1) 2010–2011 Espanyol Barcelona 26 (41) 2011 Philadelphia Independence 13 0 (5) 2011 Energija Woronesch 5 0 (1) 2012–2014 Tyresö FF 39 (14) 2014 Portland Thorns FC 15 0 (4) 2014–2015 1. FFC Frankfurt 23 0 (7) 2015–2016 FC Bayern München 9 0 (0) 2016–2018 Paris Saint-Germain 31 0 (8) 2018 Peking BG Phoenix 2019– Utah Royals FC 0 0 (0) Nationalmannschaft Jahre Auswahl
Read more

Ida-Boy-Ed-Garten

Image
Die Lage des Ida-Boy-Ed-Gartens, rot markiert Der Ida-Boy-Ed-Garten Der Ida-Boy-Ed-Garten, Blick von der Wakenitzmauer durch das Tor in der Stadtmauer Der Ida-Boy-Ed-Garten ist eine Straße der Lübecker Altstadt. Inhaltsverzeichnis 1 Lage 2 Geschichte 3 Literatur 4 Weblinks Lage | Der etwa 90 Meter lange Ida-Boy-Ed-Garten befindet sich an der nördlichen Spitze der Altstadtinsel, vor dem Burgtor. Die in einem Bogen durch eine Grünanlage hangaufwärts führende Straße verbindet die Wakenitzmauer mit der Burgtorbrücke. Geschichte | Der heutige Ida-Boy-Ed-Garten wurde ursprünglich als Verlängerung der Wakenitzmauer angelegt. 1903 wurde in den erhaltenen Teil der mittelalterlichen Stadtmauer, an dem die Wakenitzmauer bei der Einmündung der Kaiserstraße begann, ein Tor mit einem großen Bogen für die Durchführung der Straße und zwei kleineren seitlichen Öffnungen für Fußgänger gebrochen, um für die neu errichteten Wohnhäuser in diesem Bereich
Read more