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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In PowerPoint, is there a keyboard shortcut for bulleted / numbered list?

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.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box; } 2 OneNote has two useful keyboard shortcuts for bulleted and numbered lists, namely Ctrl + . and Ctrl + - . Is there something similar for PowerPoint where the work with lists is also very common? Haven't found anything here: Use keyboard shortcuts in PowerPoint. keyboard-shortcuts microsoft-powerpoint share | improve this question edited Aug 14 '14 at 14:49 Linger 2,826 10 28 40
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How to put 3 figures in Latex with 2 figures side by side and 1 below these side by side images but in...

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1 0 I want to put 3 images in LateX such as 2 figures are side by side horizotally and 3rd figure below these 2 side by side figures but in middle. I have following code: begin{figure}[H] centering begin{minipage}{.5textwidth} centering includegraphics[width=.50linewidth]{Text/Images/Genelec_8010_AP.jpg} captionof{figure}{Genelec 8010 AP} label{fig:Genelec 8010} end{minipage}% begin{minipage}{.5textwidth} centering includegraphics[width=.69linewidth]{Text/Images/Genelec_8020_CPM.jpg} captionof{figure}{Genelec 8020 CPM} label{fig:Genelec 8020} end{minipage}% end{figure} %%Here is my 3rd picture code begin{figure} centering includegraphics[width=5cm]{Text/Images/Genelec_8030_BPM.jpg} caption{Genelec 8030 BPM} label{fig:Genelec 8030} end{figure} I want to ask ho
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