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[SOLVED] continuous

dwsmith

Well-known member
Feb 1, 2012
1,673
How can I prove the below is continuous?

$$
\int_{-\pi}^{\pi}te^{xt}\cos(yt)g(t)dt \quad\text{and}\quad
-\int_{-\pi}^{\pi}te^{xt}\sin(yt)g(t)dt
$$

define the fourier transform of g as
$$
G(z) = \int_{-\pi}^{\pi}e^{zt}g(t)dt
$$

We know t, e^{xt}, sine, and cosine are continuous which means their products are continuous but we don't know about g(t).
 

chisigma

Well-known member
Feb 13, 2012
1,704
How can I prove the below is continuous?

$$
\int_{-\pi}^{\pi}te^{xt}\cos(yt)g(t)dt \quad\text{and}\quad
-\int_{-\pi}^{\pi}te^{xt}\sin(yt)g(t)dt
$$

define the fourier transform of g as
$$
G(z) = \int_{-\pi}^{\pi}e^{zt}g(t)dt
$$

We know t, e^{xt}, sine, and cosine are continuous which means their products are continuous but we don't know about g(t).
The definition of F.T. of g(*) You have given...

$\displaystyle \mathcal{F} \{g(t)\}= G(z)=\int_{- \pi}^{\pi} g(t)\ e^{z t}\ dt$ (1)

... implies that...

$\displaystyle G^{'}(z)= \int_{- \pi}^{\pi} t\ g(t)\ e^{z t}\ dt= \mathcal{F}\{t\ g(t)\}$ (2)

Now if $\displaystyle g(t)\ e^{z t}$ is integrable in $(-\pi\ ,\ \pi)$ the same is for $\displaystyle t\ g(t)\ e^{z t}$ so that G(z) is everywhere derivable and that means that G(z) is everywhere continuous...

Kind regards

$\chi$ $\sigma$
 

dwsmith

Well-known member
Feb 1, 2012
1,673
The definition of F.T. of g(*) You have given...

$\displaystyle \mathcal{F} \{g(t)\}= G(z)=\int_{- \pi}^{\pi} g(t)\ e^{z t}\ dt$ (1)

... implies that...

$\displaystyle G^{'}(z)= \int_{- \pi}^{\pi} t\ g(t)\ e^{z t}\ dt= \mathcal{F}\{t\ g(t)\}$ (2)

Now if $\displaystyle g(t)\ e^{z t}$ is integrable in $(-\pi\ ,\ \pi)$ the same is for $\displaystyle t\ g(t)\ e^{z t}$ so that G(z) is everywhere derivable and that means that G(z) is everywhere continuous...

Kind regards

$\chi$ $\sigma$
Can the derivative be slipped past the integral because of uniform continuous?
 

chisigma

Well-known member
Feb 13, 2012
1,704
Can the derivative be slipped past the integral because of uniform continuous?
The derivative of the function...

$\displaystyle g(z)= \int_{a}^{b} f(x,z)\ dx$ (1)

... exists if $f_{z} (x,z)$ exists and is continuous...

Kind regards

$\chi$ $\sigma$
 

dwsmith

Well-known member
Feb 1, 2012
1,673
Now if $\displaystyle g(t)\ e^{z t}$ is integrable in $(-\pi\ ,\ \pi)$ the same is for $\displaystyle t\ g(t)\ e^{z t}$ so that G(z) is everywhere derivable and that means that G(z) is everywhere continuous...
Don't we need the G' to be continuous because than the partials would be continuous?
 

chisigma

Well-known member
Feb 13, 2012
1,704
Don't we need the G' to be continuous because than the partials would be continuous?
We are interested only to the existence of G'(z), because if G'(z) exists then G(z) is continuous...

Kind regards

$\chi$ $\sigma$
 

dwsmith

Well-known member
Feb 1, 2012
1,673
We are interested only to the existence of G'(z), because if G'(z) exists then G(z) is continuous...

Kind regards

$\chi$ $\sigma$
I think you misunderstood the question. I want to show the first to integrals are continuous. I gave the Fourier Transform information because g(t) is defined by it.
 

chisigma

Well-known member
Feb 13, 2012
1,704
I think you misunderstood the question. I want to show the first to integrals are continuous. I gave the Fourier Transform information because g(t) is defined by it.
I'm sure to have misunderstood because continuous can be [if it is...] a function and not a definite integral...

Kind regards

$\chi$ $\sigma$
 

dwsmith

Well-known member
Feb 1, 2012
1,673
I'm sure to have misunderstood because continuous can be [if it is...] a function and not a definite integral...

Kind regards

$\chi$ $\sigma$
If the definite integrals are partial derivatives, then they can be continuous (or not) as well.