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











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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













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1 Answer
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1 Answer
1






active

oldest

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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

50.7k42573




50.7k42573












  • $begingroup$
    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




















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Plaza Victoria

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Plaza Victoria Plaza Victoria en 2017 Otros nombres Plaza de La Victoria Localización El Almendral, Valparaíso, Chile Vías adyacentes Chacabuco, Molina, Plaza Victoria y Edwards Creación 1841 Metro Bellavista Ubicación 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 Coordenadas: 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 [editar datos en Wikidata] La Plaza Victoria , también llamada Plaza de la Victoria , [ 1 ] ​ es una plaza ubicada en el sector El Almendral del plan de la ciudad de Valparaíso, Chile, de importancia patrimonial, que incluye valiosas estatuas y especies vegetales. [ 2 ] ​ La plaza se creó como tal en la primera mitad del siglo XIX, tomando previamente los nombres de Plaza Nueva y Plaza de Orrego . Si bien desde sus inicios tuvo relevancia histórica, no fue sino a partir de fines de los años 1860 cuando comenzó a cobrar mayor importancia y se convirtió en uno de los principales centros sociales de l
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Puebla de Zaragoza

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Para otros usos de este término, véase Puebla (desambiguación). Heroica Puebla de Zaragoza Localidad De arriba a abajo de izquierda a derecha: Estrella de Puebla, Volcán Popocatépetl, Catedral de Puebla, Monumento a Ignacio Zaragoza, Estadio Cuauhtémoc, Angelópolis y Vista panorámica de la ciudad. Escudo Otros nombres: El relicario de América/ La ciudad de los Ángeles/ La Angelópolis Lema: Muy noble y muy leal ciudad de Puebla de los Ángeles Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en México Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en Puebla Mapa interactivo Coordenadas 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Coordenadas: 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Idioma oficial Español Entidad Localidad  • País   México  • Estado Puebla  • Municipio Puebla Presidente municipal Claudia Rivera Vivanc
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Musa

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Para el género de plantas, véase Musa (planta). Apolo y la Nueve Musas En la mitología griega, las musas (en griego antiguo μοῦσαι « mousai» ) son, según los escritores más antiguos, las divinidades inspiradoras de las Artes: cada una de ellas está relacionada con ramas artísticas y del conocimiento. Son hijas de Zeus y de Mnemósine, compañeras del séquito de Apolo, dios olímpico de la música y patrón de las bellas artes, quien tuvo romances con cada una de ellas, dejando descendientes. Bajaban a la tierra a susurrar ideas e inspirar a aquellos mortales que las invocaran. En la época más arcaica eran las ninfas inspiradoras de las fuentes, en las cuales eran adoradas. Finalmente, alrededor de los siglos VIII-VII a. C. [ 1 ] ​ prevaleció en todo el territorio de la Hélade la adoración de las nueve Musas, que son Calíope, Clío, Erato, Euterpe, Melpómene, Polimnia, Talía, Terpsícore y Urania. El culto a las musas era originalmente de Tracia y Beocia, y fueron de vital impo
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