Why convergence of these two series is equivalent?
$begingroup$
In my notes I have the following:
For $sgeq 0$, define the Sobolev Space $H_s$ by:
$$H_s := left{fin L^2([0, 2pi], mathbb{C}),, : ,, sum_{k=-infty}^infty |k|^{2s}|hat{f_k}|^2 <inftyright}$$
And define the Inner Product as:
$$(f, g)_{(s)} := sum_{k=-infty}^infty hat{f_k}bar{hat{g_k}}(1 + k^2)^s$$
Note that $sum_{k=-infty}^infty (|k|^2)^s|hat{f_k}|^2$ converges iff $sum_{k=-infty}^infty (1+k^2)^s |hat{f_k}|^2$ converges.
How do we show the last statement is true? From how the lecturer put it down it seems VERY obvious as if you could do it in one line, but I couldn't show this is true.
My Attempt
I tried as follows: Since $sgeq 0$ then clearly:
$$(1 + k^2)^s geq (|k|^2)^s quad forall kinmathbb{Z}quad forall sgeq 0$$
since $k^2 = |k|^2$ for all $kinmathbb{Z}$. Therefore we must have:
$$(1+k^2)^s |hat{f_k}|^2 geq (|k|^2)^s |hat{f_k}|^2quad forall kinmathbb{Z}quad forall sgeq 0$$ so that we can show the direction "$Longleftarrow$" using the comparison test.
real-analysis calculus complex-analysis hilbert-spaces inner-product-space
asked Dec 18 '18 at 20:40
Euler_SalterEuler_Salter
2,0571335
$endgroup$
add a comment |
$begingroup$
In my notes I have the following:
For $sgeq 0$, define the Sobolev Space $H_s$ by:
$$H_s := left{fin L^2([0, 2pi], mathbb{C}),, : ,, sum_{k=-infty}^infty |k|^{2s}|hat{f_k}|^2 <inftyright}$$
And define the Inner Product as:
$$(f, g)_{(s)} := sum_{k=-infty}^infty hat{f_k}bar{hat{g_k}}(1 + k^2)^s$$
Note that $sum_{k=-infty}^infty (|k|^2)^s|hat{f_k}|^2$ converges iff $sum_{k=-infty}^infty (1+k^2)^s |hat{f_k}|^2$ converges.
How do we show the last statement is true? From how the lecturer put it down it seems VERY obvious as if you could do it in one line, but I couldn't show this is true.
My Attempt
I tried as follows: Since $sgeq 0$ then clearly:
$$(1 + k^2)^s geq (|k|^2)^s quad forall kinmathbb{Z}quad forall sgeq 0$$
since $k^2 = |k|^2$ for all $kinmathbb{Z}$. Therefore we must have:
$$(1+k^2)^s |hat{f_k}|^2 geq (|k|^2)^s |hat{f_k}|^2quad forall kinmathbb{Z}quad forall sgeq 0$$ so that we can show the direction "$Longleftarrow$" using the comparison test.
real-analysis calculus complex-analysis hilbert-spaces inner-product-space
asked Dec 18 '18 at 20:40
Euler_SalterEuler_Salter
2,0571335
$endgroup$
add a comment |
$begingroup$
In my notes I have the following:
For $sgeq 0$, define the Sobolev Space $H_s$ by:
$$H_s := left{fin L^2([0, 2pi], mathbb{C}),, : ,, sum_{k=-infty}^infty |k|^{2s}|hat{f_k}|^2 <inftyright}$$
And define the Inner Product as:
$$(f, g)_{(s)} := sum_{k=-infty}^infty hat{f_k}bar{hat{g_k}}(1 + k^2)^s$$
Note that $sum_{k=-infty}^infty (|k|^2)^s|hat{f_k}|^2$ converges iff $sum_{k=-infty}^infty (1+k^2)^s |hat{f_k}|^2$ converges.
How do we show the last statement is true? From how the lecturer put it down it seems VERY obvious as if you could do it in one line, but I couldn't show this is true.
My Attempt
I tried as follows: Since $sgeq 0$ then clearly:
$$(1 + k^2)^s geq (|k|^2)^s quad forall kinmathbb{Z}quad forall sgeq 0$$
since $k^2 = |k|^2$ for all $kinmathbb{Z}$. Therefore we must have:
$$(1+k^2)^s |hat{f_k}|^2 geq (|k|^2)^s |hat{f_k}|^2quad forall kinmathbb{Z}quad forall sgeq 0$$ so that we can show the direction "$Longleftarrow$" using the comparison test.
real-analysis calculus complex-analysis hilbert-spaces inner-product-space
asked Dec 18 '18 at 20:40
Euler_SalterEuler_Salter
2,0571335
$endgroup$
In my notes I have the following:
For $sgeq 0$, define the Sobolev Space $H_s$ by:
$$H_s := left{fin L^2([0, 2pi], mathbb{C}),, : ,, sum_{k=-infty}^infty |k|^{2s}|hat{f_k}|^2 <inftyright}$$
And define the Inner Product as:
$$(f, g)_{(s)} := sum_{k=-infty}^infty hat{f_k}bar{hat{g_k}}(1 + k^2)^s$$
Note that $sum_{k=-infty}^infty (|k|^2)^s|hat{f_k}|^2$ converges iff $sum_{k=-infty}^infty (1+k^2)^s |hat{f_k}|^2$ converges.
How do we show the last statement is true? From how the lecturer put it down it seems VERY obvious as if you could do it in one line, but I couldn't show this is true.
My Attempt
I tried as follows: Since $sgeq 0$ then clearly:
$$(1 + k^2)^s geq (|k|^2)^s quad forall kinmathbb{Z}quad forall sgeq 0$$
since $k^2 = |k|^2$ for all $kinmathbb{Z}$. Therefore we must have:
$$(1+k^2)^s |hat{f_k}|^2 geq (|k|^2)^s |hat{f_k}|^2quad forall kinmathbb{Z}quad forall sgeq 0$$ so that we can show the direction "$Longleftarrow$" using the comparison test.
real-analysis calculus complex-analysis hilbert-spaces inner-product-space
real-analysis calculus complex-analysis hilbert-spaces inner-product-space
asked Dec 18 '18 at 20:40
Euler_SalterEuler_Salter
2,0571335
asked Dec 18 '18 at 20:40
Euler_SalterEuler_Salter
2,0571335
edited Dec 18 '18 at 20:46
Euler_Salter
asked Dec 18 '18 at 20:40
Euler_SalterEuler_Salter
2,0571335
asked Dec 18 '18 at 20:40
Euler_SalterEuler_Salter
2,0571335
asked Dec 18 '18 at 20:40
Euler_SalterEuler_Salter
2,0571335
2,0571335
add a comment |
add a comment |
2 Answers
2
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$begingroup$
Hint:
Note the general terms of both series (which are positive) are equivalent sice
$$lim_{ktoinfty}frac{(1+k^2)^s|hat f_k|^2}{k^{2s}|hat f_k|^2}=lim_{kto infty}Bigl(1+frac1{k^2} Bigr)^s=1.$$
Hence both series converge or both diverge.
answered Dec 18 '18 at 20:56
BernardBernard
119k740113
$endgroup$
$begingroup$
Oh I see what you did there. That's a pretty nice spot. How did you realize you had to use this?
$endgroup$
– Euler_Salter
Dec 18 '18 at 20:58
add a comment |
$begingroup$
Hint: Use that $|k|ge 1$ for $kinmathbb{Z}setminus{0}$, so $2k^2 ge k^2+1$.
answered Dec 18 '18 at 20:56
AugSBAugSB
3,33421733
$endgroup$
$begingroup$
Okay thanks, I'll work on that!
$endgroup$
– Euler_Salter
Dec 18 '18 at 21:01
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
Hint:
Note the general terms of both series (which are positive) are equivalent sice
$$lim_{ktoinfty}frac{(1+k^2)^s|hat f_k|^2}{k^{2s}|hat f_k|^2}=lim_{kto infty}Bigl(1+frac1{k^2} Bigr)^s=1.$$
Hence both series converge or both diverge.
answered Dec 18 '18 at 20:56
BernardBernard
119k740113
$endgroup$
$begingroup$
Oh I see what you did there. That's a pretty nice spot. How did you realize you had to use this?
$endgroup$
– Euler_Salter
Dec 18 '18 at 20:58
add a comment |
$begingroup$
Hint:
Note the general terms of both series (which are positive) are equivalent sice
$$lim_{ktoinfty}frac{(1+k^2)^s|hat f_k|^2}{k^{2s}|hat f_k|^2}=lim_{kto infty}Bigl(1+frac1{k^2} Bigr)^s=1.$$
Hence both series converge or both diverge.
answered Dec 18 '18 at 20:56
BernardBernard
119k740113
$endgroup$
$begingroup$
Oh I see what you did there. That's a pretty nice spot. How did you realize you had to use this?
$endgroup$
– Euler_Salter
Dec 18 '18 at 20:58
add a comment |
$begingroup$
Hint:
Note the general terms of both series (which are positive) are equivalent sice
$$lim_{ktoinfty}frac{(1+k^2)^s|hat f_k|^2}{k^{2s}|hat f_k|^2}=lim_{kto infty}Bigl(1+frac1{k^2} Bigr)^s=1.$$
Hence both series converge or both diverge.
answered Dec 18 '18 at 20:56
BernardBernard
119k740113
$endgroup$
Hint:
Note the general terms of both series (which are positive) are equivalent sice
$$lim_{ktoinfty}frac{(1+k^2)^s|hat f_k|^2}{k^{2s}|hat f_k|^2}=lim_{kto infty}Bigl(1+frac1{k^2} Bigr)^s=1.$$
Hence both series converge or both diverge.
answered Dec 18 '18 at 20:56
BernardBernard
119k740113
answered Dec 18 '18 at 20:56
BernardBernard
119k740113
answered Dec 18 '18 at 20:56
BernardBernard
119k740113
answered Dec 18 '18 at 20:56
BernardBernard
119k740113
119k740113
$begingroup$
Oh I see what you did there. That's a pretty nice spot. How did you realize you had to use this?
$endgroup$
– Euler_Salter
Dec 18 '18 at 20:58
add a comment |
$begingroup$
Oh I see what you did there. That's a pretty nice spot. How did you realize you had to use this?
$endgroup$
– Euler_Salter
Dec 18 '18 at 20:58
$begingroup$
Oh I see what you did there. That's a pretty nice spot. How did you realize you had to use this?
$endgroup$
– Euler_Salter
Dec 18 '18 at 20:58
$begingroup$
Oh I see what you did there. That's a pretty nice spot. How did you realize you had to use this?
$endgroup$
– Euler_Salter
Dec 18 '18 at 20:58
add a comment |
$begingroup$
Hint: Use that $|k|ge 1$ for $kinmathbb{Z}setminus{0}$, so $2k^2 ge k^2+1$.
answered Dec 18 '18 at 20:56
AugSBAugSB
3,33421733
$endgroup$
$begingroup$
Okay thanks, I'll work on that!
$endgroup$
– Euler_Salter
Dec 18 '18 at 21:01
add a comment |
$begingroup$
Hint: Use that $|k|ge 1$ for $kinmathbb{Z}setminus{0}$, so $2k^2 ge k^2+1$.
answered Dec 18 '18 at 20:56
AugSBAugSB
3,33421733
$endgroup$
$begingroup$
Okay thanks, I'll work on that!
$endgroup$
– Euler_Salter
Dec 18 '18 at 21:01
add a comment |
$begingroup$
Hint: Use that $|k|ge 1$ for $kinmathbb{Z}setminus{0}$, so $2k^2 ge k^2+1$.
answered Dec 18 '18 at 20:56
AugSBAugSB
3,33421733
$endgroup$
Hint: Use that $|k|ge 1$ for $kinmathbb{Z}setminus{0}$, so $2k^2 ge k^2+1$.
answered Dec 18 '18 at 20:56
AugSBAugSB
3,33421733
answered Dec 18 '18 at 20:56
AugSBAugSB
3,33421733
answered Dec 18 '18 at 20:56
AugSBAugSB
3,33421733
answered Dec 18 '18 at 20:56
AugSBAugSB
3,33421733
3,33421733
$begingroup$
Okay thanks, I'll work on that!
$endgroup$
– Euler_Salter
Dec 18 '18 at 21:01
add a comment |
$begingroup$
Okay thanks, I'll work on that!
$endgroup$
– Euler_Salter
Dec 18 '18 at 21:01
$begingroup$
Okay thanks, I'll work on that!
$endgroup$
– Euler_Salter
Dec 18 '18 at 21:01
$begingroup$
Okay thanks, I'll work on that!
$endgroup$
– Euler_Salter
Dec 18 '18 at 21:01
add a comment |
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