Mixing
Properties of laplace type transform of $t^{alpha - 1}$
$begingroup$
Let $p>2, frac{1}{p} < alpha < 1- frac{1}{p}$ and define $g_alpha(t) := t^{alpha - 1} chi_{[1, infty]}$. Then $g_alpha in L^p(mathbb R)$. Define $$f(z) := int_1^infty g_alpha(t) exp(-izt) , dt.$$
for $zin mathbb C, operatorname{Im }z <0.$ Then:
$a)$ $f$ is holomorphic in the lower half plane,
$b)$ The limit $$f_0(x):= lim_{substack{yto 0 \ y < 0}} f(x+iy)$$ exists for all $xneq 0$,
$c)$ $f(z)$ does not satisfy an estimate of the form $$forall epsilon >0: quad lvert f(z) rvert leq C_epsilon exp((1+epsilon)lvert z rvert),$$
$d)$ $f_0$ is not $p$-integrable in any neighbourhood of $0$.
I managed to show $a)$ using Lebesgue's theorem on interchanging derivative and integral sign. However, I am unsure with the other assertions, especially $c)$ and $d)$: How does one estimate the integral from below? And how I can I show something about $f_0 $ if I don't explicitly know it? Any help appreciated!
real-analysis complex-analysis functional-analysis fourier-analysis laplace-transform
asked Jan 12 at 14:40
Staki42Staki42
1,154618
$endgroup$
add a comment |
$begingroup$
Let $p>2, frac{1}{p} < alpha < 1- frac{1}{p}$ and define $g_alpha(t) := t^{alpha - 1} chi_{[1, infty]}$. Then $g_alpha in L^p(mathbb R)$. Define $$f(z) := int_1^infty g_alpha(t) exp(-izt) , dt.$$
for $zin mathbb C, operatorname{Im }z <0.$ Then:
$a)$ $f$ is holomorphic in the lower half plane,
$b)$ The limit $$f_0(x):= lim_{substack{yto 0 \ y < 0}} f(x+iy)$$ exists for all $xneq 0$,
$c)$ $f(z)$ does not satisfy an estimate of the form $$forall epsilon >0: quad lvert f(z) rvert leq C_epsilon exp((1+epsilon)lvert z rvert),$$
$d)$ $f_0$ is not $p$-integrable in any neighbourhood of $0$.
I managed to show $a)$ using Lebesgue's theorem on interchanging derivative and integral sign. However, I am unsure with the other assertions, especially $c)$ and $d)$: How does one estimate the integral from below? And how I can I show something about $f_0 $ if I don't explicitly know it? Any help appreciated!
real-analysis complex-analysis functional-analysis fourier-analysis laplace-transform
asked Jan 12 at 14:40
Staki42Staki42
1,154618
$endgroup$
add a comment |
$begingroup$
Let $p>2, frac{1}{p} < alpha < 1- frac{1}{p}$ and define $g_alpha(t) := t^{alpha - 1} chi_{[1, infty]}$. Then $g_alpha in L^p(mathbb R)$. Define $$f(z) := int_1^infty g_alpha(t) exp(-izt) , dt.$$
for $zin mathbb C, operatorname{Im }z <0.$ Then:
$a)$ $f$ is holomorphic in the lower half plane,
$b)$ The limit $$f_0(x):= lim_{substack{yto 0 \ y < 0}} f(x+iy)$$ exists for all $xneq 0$,
$c)$ $f(z)$ does not satisfy an estimate of the form $$forall epsilon >0: quad lvert f(z) rvert leq C_epsilon exp((1+epsilon)lvert z rvert),$$
$d)$ $f_0$ is not $p$-integrable in any neighbourhood of $0$.
I managed to show $a)$ using Lebesgue's theorem on interchanging derivative and integral sign. However, I am unsure with the other assertions, especially $c)$ and $d)$: How does one estimate the integral from below? And how I can I show something about $f_0 $ if I don't explicitly know it? Any help appreciated!
real-analysis complex-analysis functional-analysis fourier-analysis laplace-transform
asked Jan 12 at 14:40
Staki42Staki42
1,154618
$endgroup$
Let $p>2, frac{1}{p} < alpha < 1- frac{1}{p}$ and define $g_alpha(t) := t^{alpha - 1} chi_{[1, infty]}$. Then $g_alpha in L^p(mathbb R)$. Define $$f(z) := int_1^infty g_alpha(t) exp(-izt) , dt.$$
for $zin mathbb C, operatorname{Im }z <0.$ Then:
$a)$ $f$ is holomorphic in the lower half plane,
$b)$ The limit $$f_0(x):= lim_{substack{yto 0 \ y < 0}} f(x+iy)$$ exists for all $xneq 0$,
$c)$ $f(z)$ does not satisfy an estimate of the form $$forall epsilon >0: quad lvert f(z) rvert leq C_epsilon exp((1+epsilon)lvert z rvert),$$
$d)$ $f_0$ is not $p$-integrable in any neighbourhood of $0$.
I managed to show $a)$ using Lebesgue's theorem on interchanging derivative and integral sign. However, I am unsure with the other assertions, especially $c)$ and $d)$: How does one estimate the integral from below? And how I can I show something about $f_0 $ if I don't explicitly know it? Any help appreciated!
real-analysis complex-analysis functional-analysis fourier-analysis laplace-transform
real-analysis complex-analysis functional-analysis fourier-analysis laplace-transform
asked Jan 12 at 14:40
Staki42Staki42
1,154618
asked Jan 12 at 14:40
Staki42Staki42
1,154618
asked Jan 12 at 14:40
Staki42Staki42
1,154618
asked Jan 12 at 14:40
Staki42Staki42
1,154618
asked Jan 12 at 14:40
Staki42Staki42
1,154618
1,154618
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
First we prove that
$$tag{1}f(z) := int_1^infty t^{alpha-1} e^{-izt} , mathrm{d} t$$
is convergent (as an improper Riemann integral) for all $z = x-iy$ with $x ne 0$ and $y >0$. More presioulsy, the convergence is uniform if $|x| ge x_0$ for fixed $x_0>0$. Thus (1) defines a continuous function and we have for $x ne 0$ that
$$tag{2}f_0(x) = int_1^infty t^{alpha-1} e^{-ixt} , mathrm{d} t$$
Prove: For any $z= x-iy$ with $|x| ge x_0$ and $y ne 0$ we have with $1 le a < b$ that
begin{align}
tag{3}int_a^b t^{alpha-1} e^{-izt} dt = frac{1}{iz} ( a^{alpha-1} e^{-iza} - b^{alpha-1} e^{-izb}) - frac{alpha-1}{iz} int_a^b t^{alpha-2} e^{-izt} dt.
end{align}
The last line can be bounded by
$$frac{1}{x_0} (a^{alpha-1} + b^{alpha-1}) +frac{1-alpha}{x_0} int_a^b t^{alpha-2} dt le frac{2}{x_0} a^{alpha-1} $$
and thus (3) is a Cauchy sequence and thus convergent. In fact, it is (uniformly) convergent and thus continuous. (The whole argument is also known as Dirchlet's test, see for example here.)
Prove of (c) and (d): So we have shown (b). With the help of the explicit identity (2) we can also verify (c) and (d). In fact, we have (after coordinate of change) the identity
$$f_0(x) = x^{-alpha} int_x^infty s^{alpha-1} e^{-is} , mathrm{d} s.$$
Since the integrand is locally integrable (also in $s=0$) the function
$$g(x) := int_x^infty s^{alpha-1} e^{-is} , mathrm{d} s$$
is continuous in $x=0$ with
$$tag{4}g(0) = int_0^infty s^{alpha-1} e^{-is} , mathrm{d} s.$$
Thus, it remains to show that $g(0) ne 0$. Here we use the integral representation of (4). In fact, (4) is related to the Gamma function. One representation of the Gamma function (which was proven by Euler by using a contour shift argument) is
$$Gamma(w) = e^{i pi w /2} int_0^infty s^{w-1} e^{-is} , mathrm{d} s$$
if $0 < mathrm{Re}(w) < 1$. Hence
$$g(0) = e^{-i alpha pi/2} Gamma(alpha) neq 0.$$
All in all, we see that
$$tag{5} f_0(x) sim_alpha x^{-alpha} quad (text{for} x rightarrow 0) $$
up to an non-zero constant (depending on $alpha$).
c) can not hold, because c) would imply that $f_0$ is bounded. (5) implies that $f_0$ is not $p$-integrable (note that $alpha p >1$) in any neighbourhood of $0$.
answered Jan 13 at 11:08
p4schp4sch
5,285217
$endgroup$
$begingroup$
Thanks for taking your time! What do you mean by "$(3)$ is a Cauchy sequence"? In which way do you interpret this as a sequence?
$endgroup$
– Staki42
Jan 13 at 12:29
$begingroup$
Never mind, I got it. Thanks!
$endgroup$
– Staki42
Jan 13 at 12:40
$begingroup$
More precisely, I should say 'Cauchy-net'. There is a generalization of sequences called nets, see Wikipedia. The bound shows that (3) can be made arbitary small if $a$ is large enough and that is exactly the property to be a Cauchy-net.
$endgroup$
– p4sch
Jan 13 at 12:47
$begingroup$
I see. But I also could just take a sequence, right?
$endgroup$
– Staki42
Jan 13 at 13:28
1
$begingroup$
No, for $y=0$ the integral is not absolute convergent, because $t^{alpha-1}$ is not in $L^1([1,infty)$. For $y >0$ it is not a problem, but we would like to determine the integral for $y=0$ and also see that it is continuous in $y$ in order to determine the limes.
$endgroup$
– p4sch
Jan 13 at 14:54
|
show 6 more comments
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1 Answer
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1 Answer
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oldest
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$begingroup$
First we prove that
$$tag{1}f(z) := int_1^infty t^{alpha-1} e^{-izt} , mathrm{d} t$$
is convergent (as an improper Riemann integral) for all $z = x-iy$ with $x ne 0$ and $y >0$. More presioulsy, the convergence is uniform if $|x| ge x_0$ for fixed $x_0>0$. Thus (1) defines a continuous function and we have for $x ne 0$ that
$$tag{2}f_0(x) = int_1^infty t^{alpha-1} e^{-ixt} , mathrm{d} t$$
Prove: For any $z= x-iy$ with $|x| ge x_0$ and $y ne 0$ we have with $1 le a < b$ that
begin{align}
tag{3}int_a^b t^{alpha-1} e^{-izt} dt = frac{1}{iz} ( a^{alpha-1} e^{-iza} - b^{alpha-1} e^{-izb}) - frac{alpha-1}{iz} int_a^b t^{alpha-2} e^{-izt} dt.
end{align}
The last line can be bounded by
$$frac{1}{x_0} (a^{alpha-1} + b^{alpha-1}) +frac{1-alpha}{x_0} int_a^b t^{alpha-2} dt le frac{2}{x_0} a^{alpha-1} $$
and thus (3) is a Cauchy sequence and thus convergent. In fact, it is (uniformly) convergent and thus continuous. (The whole argument is also known as Dirchlet's test, see for example here.)
Prove of (c) and (d): So we have shown (b). With the help of the explicit identity (2) we can also verify (c) and (d). In fact, we have (after coordinate of change) the identity
$$f_0(x) = x^{-alpha} int_x^infty s^{alpha-1} e^{-is} , mathrm{d} s.$$
Since the integrand is locally integrable (also in $s=0$) the function
$$g(x) := int_x^infty s^{alpha-1} e^{-is} , mathrm{d} s$$
is continuous in $x=0$ with
$$tag{4}g(0) = int_0^infty s^{alpha-1} e^{-is} , mathrm{d} s.$$
Thus, it remains to show that $g(0) ne 0$. Here we use the integral representation of (4). In fact, (4) is related to the Gamma function. One representation of the Gamma function (which was proven by Euler by using a contour shift argument) is
$$Gamma(w) = e^{i pi w /2} int_0^infty s^{w-1} e^{-is} , mathrm{d} s$$
if $0 < mathrm{Re}(w) < 1$. Hence
$$g(0) = e^{-i alpha pi/2} Gamma(alpha) neq 0.$$
All in all, we see that
$$tag{5} f_0(x) sim_alpha x^{-alpha} quad (text{for} x rightarrow 0) $$
up to an non-zero constant (depending on $alpha$).
c) can not hold, because c) would imply that $f_0$ is bounded. (5) implies that $f_0$ is not $p$-integrable (note that $alpha p >1$) in any neighbourhood of $0$.
answered Jan 13 at 11:08
p4schp4sch
5,285217
$endgroup$
$begingroup$
Thanks for taking your time! What do you mean by "$(3)$ is a Cauchy sequence"? In which way do you interpret this as a sequence?
$endgroup$
– Staki42
Jan 13 at 12:29
$begingroup$
Never mind, I got it. Thanks!
$endgroup$
– Staki42
Jan 13 at 12:40
$begingroup$
More precisely, I should say 'Cauchy-net'. There is a generalization of sequences called nets, see Wikipedia. The bound shows that (3) can be made arbitary small if $a$ is large enough and that is exactly the property to be a Cauchy-net.
$endgroup$
– p4sch
Jan 13 at 12:47
$begingroup$
I see. But I also could just take a sequence, right?
$endgroup$
– Staki42
Jan 13 at 13:28
1
$begingroup$
No, for $y=0$ the integral is not absolute convergent, because $t^{alpha-1}$ is not in $L^1([1,infty)$. For $y >0$ it is not a problem, but we would like to determine the integral for $y=0$ and also see that it is continuous in $y$ in order to determine the limes.
$endgroup$
– p4sch
Jan 13 at 14:54
|
show 6 more comments
$begingroup$
First we prove that
$$tag{1}f(z) := int_1^infty t^{alpha-1} e^{-izt} , mathrm{d} t$$
is convergent (as an improper Riemann integral) for all $z = x-iy$ with $x ne 0$ and $y >0$. More presioulsy, the convergence is uniform if $|x| ge x_0$ for fixed $x_0>0$. Thus (1) defines a continuous function and we have for $x ne 0$ that
$$tag{2}f_0(x) = int_1^infty t^{alpha-1} e^{-ixt} , mathrm{d} t$$
Prove: For any $z= x-iy$ with $|x| ge x_0$ and $y ne 0$ we have with $1 le a < b$ that
begin{align}
tag{3}int_a^b t^{alpha-1} e^{-izt} dt = frac{1}{iz} ( a^{alpha-1} e^{-iza} - b^{alpha-1} e^{-izb}) - frac{alpha-1}{iz} int_a^b t^{alpha-2} e^{-izt} dt.
end{align}
The last line can be bounded by
$$frac{1}{x_0} (a^{alpha-1} + b^{alpha-1}) +frac{1-alpha}{x_0} int_a^b t^{alpha-2} dt le frac{2}{x_0} a^{alpha-1} $$
and thus (3) is a Cauchy sequence and thus convergent. In fact, it is (uniformly) convergent and thus continuous. (The whole argument is also known as Dirchlet's test, see for example here.)
Prove of (c) and (d): So we have shown (b). With the help of the explicit identity (2) we can also verify (c) and (d). In fact, we have (after coordinate of change) the identity
$$f_0(x) = x^{-alpha} int_x^infty s^{alpha-1} e^{-is} , mathrm{d} s.$$
Since the integrand is locally integrable (also in $s=0$) the function
$$g(x) := int_x^infty s^{alpha-1} e^{-is} , mathrm{d} s$$
is continuous in $x=0$ with
$$tag{4}g(0) = int_0^infty s^{alpha-1} e^{-is} , mathrm{d} s.$$
Thus, it remains to show that $g(0) ne 0$. Here we use the integral representation of (4). In fact, (4) is related to the Gamma function. One representation of the Gamma function (which was proven by Euler by using a contour shift argument) is
$$Gamma(w) = e^{i pi w /2} int_0^infty s^{w-1} e^{-is} , mathrm{d} s$$
if $0 < mathrm{Re}(w) < 1$. Hence
$$g(0) = e^{-i alpha pi/2} Gamma(alpha) neq 0.$$
All in all, we see that
$$tag{5} f_0(x) sim_alpha x^{-alpha} quad (text{for} x rightarrow 0) $$
up to an non-zero constant (depending on $alpha$).
c) can not hold, because c) would imply that $f_0$ is bounded. (5) implies that $f_0$ is not $p$-integrable (note that $alpha p >1$) in any neighbourhood of $0$.
answered Jan 13 at 11:08
p4schp4sch
5,285217
$endgroup$
$begingroup$
Thanks for taking your time! What do you mean by "$(3)$ is a Cauchy sequence"? In which way do you interpret this as a sequence?
$endgroup$
– Staki42
Jan 13 at 12:29
$begingroup$
Never mind, I got it. Thanks!
$endgroup$
– Staki42
Jan 13 at 12:40
$begingroup$
More precisely, I should say 'Cauchy-net'. There is a generalization of sequences called nets, see Wikipedia. The bound shows that (3) can be made arbitary small if $a$ is large enough and that is exactly the property to be a Cauchy-net.
$endgroup$
– p4sch
Jan 13 at 12:47
$begingroup$
I see. But I also could just take a sequence, right?
$endgroup$
– Staki42
Jan 13 at 13:28
1
$begingroup$
No, for $y=0$ the integral is not absolute convergent, because $t^{alpha-1}$ is not in $L^1([1,infty)$. For $y >0$ it is not a problem, but we would like to determine the integral for $y=0$ and also see that it is continuous in $y$ in order to determine the limes.
$endgroup$
– p4sch
Jan 13 at 14:54
|
show 6 more comments
$begingroup$
First we prove that
$$tag{1}f(z) := int_1^infty t^{alpha-1} e^{-izt} , mathrm{d} t$$
is convergent (as an improper Riemann integral) for all $z = x-iy$ with $x ne 0$ and $y >0$. More presioulsy, the convergence is uniform if $|x| ge x_0$ for fixed $x_0>0$. Thus (1) defines a continuous function and we have for $x ne 0$ that
$$tag{2}f_0(x) = int_1^infty t^{alpha-1} e^{-ixt} , mathrm{d} t$$
Prove: For any $z= x-iy$ with $|x| ge x_0$ and $y ne 0$ we have with $1 le a < b$ that
begin{align}
tag{3}int_a^b t^{alpha-1} e^{-izt} dt = frac{1}{iz} ( a^{alpha-1} e^{-iza} - b^{alpha-1} e^{-izb}) - frac{alpha-1}{iz} int_a^b t^{alpha-2} e^{-izt} dt.
end{align}
The last line can be bounded by
$$frac{1}{x_0} (a^{alpha-1} + b^{alpha-1}) +frac{1-alpha}{x_0} int_a^b t^{alpha-2} dt le frac{2}{x_0} a^{alpha-1} $$
and thus (3) is a Cauchy sequence and thus convergent. In fact, it is (uniformly) convergent and thus continuous. (The whole argument is also known as Dirchlet's test, see for example here.)
Prove of (c) and (d): So we have shown (b). With the help of the explicit identity (2) we can also verify (c) and (d). In fact, we have (after coordinate of change) the identity
$$f_0(x) = x^{-alpha} int_x^infty s^{alpha-1} e^{-is} , mathrm{d} s.$$
Since the integrand is locally integrable (also in $s=0$) the function
$$g(x) := int_x^infty s^{alpha-1} e^{-is} , mathrm{d} s$$
is continuous in $x=0$ with
$$tag{4}g(0) = int_0^infty s^{alpha-1} e^{-is} , mathrm{d} s.$$
Thus, it remains to show that $g(0) ne 0$. Here we use the integral representation of (4). In fact, (4) is related to the Gamma function. One representation of the Gamma function (which was proven by Euler by using a contour shift argument) is
$$Gamma(w) = e^{i pi w /2} int_0^infty s^{w-1} e^{-is} , mathrm{d} s$$
if $0 < mathrm{Re}(w) < 1$. Hence
$$g(0) = e^{-i alpha pi/2} Gamma(alpha) neq 0.$$
All in all, we see that
$$tag{5} f_0(x) sim_alpha x^{-alpha} quad (text{for} x rightarrow 0) $$
up to an non-zero constant (depending on $alpha$).
c) can not hold, because c) would imply that $f_0$ is bounded. (5) implies that $f_0$ is not $p$-integrable (note that $alpha p >1$) in any neighbourhood of $0$.
answered Jan 13 at 11:08
p4schp4sch
5,285217
$endgroup$
First we prove that
$$tag{1}f(z) := int_1^infty t^{alpha-1} e^{-izt} , mathrm{d} t$$
is convergent (as an improper Riemann integral) for all $z = x-iy$ with $x ne 0$ and $y >0$. More presioulsy, the convergence is uniform if $|x| ge x_0$ for fixed $x_0>0$. Thus (1) defines a continuous function and we have for $x ne 0$ that
$$tag{2}f_0(x) = int_1^infty t^{alpha-1} e^{-ixt} , mathrm{d} t$$
Prove: For any $z= x-iy$ with $|x| ge x_0$ and $y ne 0$ we have with $1 le a < b$ that
begin{align}
tag{3}int_a^b t^{alpha-1} e^{-izt} dt = frac{1}{iz} ( a^{alpha-1} e^{-iza} - b^{alpha-1} e^{-izb}) - frac{alpha-1}{iz} int_a^b t^{alpha-2} e^{-izt} dt.
end{align}
The last line can be bounded by
$$frac{1}{x_0} (a^{alpha-1} + b^{alpha-1}) +frac{1-alpha}{x_0} int_a^b t^{alpha-2} dt le frac{2}{x_0} a^{alpha-1} $$
and thus (3) is a Cauchy sequence and thus convergent. In fact, it is (uniformly) convergent and thus continuous. (The whole argument is also known as Dirchlet's test, see for example here.)
Prove of (c) and (d): So we have shown (b). With the help of the explicit identity (2) we can also verify (c) and (d). In fact, we have (after coordinate of change) the identity
$$f_0(x) = x^{-alpha} int_x^infty s^{alpha-1} e^{-is} , mathrm{d} s.$$
Since the integrand is locally integrable (also in $s=0$) the function
$$g(x) := int_x^infty s^{alpha-1} e^{-is} , mathrm{d} s$$
is continuous in $x=0$ with
$$tag{4}g(0) = int_0^infty s^{alpha-1} e^{-is} , mathrm{d} s.$$
Thus, it remains to show that $g(0) ne 0$. Here we use the integral representation of (4). In fact, (4) is related to the Gamma function. One representation of the Gamma function (which was proven by Euler by using a contour shift argument) is
$$Gamma(w) = e^{i pi w /2} int_0^infty s^{w-1} e^{-is} , mathrm{d} s$$
if $0 < mathrm{Re}(w) < 1$. Hence
$$g(0) = e^{-i alpha pi/2} Gamma(alpha) neq 0.$$
All in all, we see that
$$tag{5} f_0(x) sim_alpha x^{-alpha} quad (text{for} x rightarrow 0) $$
up to an non-zero constant (depending on $alpha$).
c) can not hold, because c) would imply that $f_0$ is bounded. (5) implies that $f_0$ is not $p$-integrable (note that $alpha p >1$) in any neighbourhood of $0$.
answered Jan 13 at 11:08
p4schp4sch
5,285217
edited Jan 13 at 15:57
answered Jan 13 at 11:08
p4schp4sch
5,285217
answered Jan 13 at 11:08
p4schp4sch
5,285217
answered Jan 13 at 11:08
p4schp4sch
5,285217
5,285217
$begingroup$
Thanks for taking your time! What do you mean by "$(3)$ is a Cauchy sequence"? In which way do you interpret this as a sequence?
$endgroup$
– Staki42
Jan 13 at 12:29
$begingroup$
Never mind, I got it. Thanks!
$endgroup$
– Staki42
Jan 13 at 12:40
$begingroup$
More precisely, I should say 'Cauchy-net'. There is a generalization of sequences called nets, see Wikipedia. The bound shows that (3) can be made arbitary small if $a$ is large enough and that is exactly the property to be a Cauchy-net.
$endgroup$
– p4sch
Jan 13 at 12:47
$begingroup$
I see. But I also could just take a sequence, right?
$endgroup$
– Staki42
Jan 13 at 13:28
1
$begingroup$
No, for $y=0$ the integral is not absolute convergent, because $t^{alpha-1}$ is not in $L^1([1,infty)$. For $y >0$ it is not a problem, but we would like to determine the integral for $y=0$ and also see that it is continuous in $y$ in order to determine the limes.
$endgroup$
– p4sch
Jan 13 at 14:54
|
show 6 more comments
$begingroup$
Thanks for taking your time! What do you mean by "$(3)$ is a Cauchy sequence"? In which way do you interpret this as a sequence?
$endgroup$
– Staki42
Jan 13 at 12:29
$begingroup$
Never mind, I got it. Thanks!
$endgroup$
– Staki42
Jan 13 at 12:40
$begingroup$
More precisely, I should say 'Cauchy-net'. There is a generalization of sequences called nets, see Wikipedia. The bound shows that (3) can be made arbitary small if $a$ is large enough and that is exactly the property to be a Cauchy-net.
$endgroup$
– p4sch
Jan 13 at 12:47
$begingroup$
I see. But I also could just take a sequence, right?
$endgroup$
– Staki42
Jan 13 at 13:28
1
$begingroup$
No, for $y=0$ the integral is not absolute convergent, because $t^{alpha-1}$ is not in $L^1([1,infty)$. For $y >0$ it is not a problem, but we would like to determine the integral for $y=0$ and also see that it is continuous in $y$ in order to determine the limes.
$endgroup$
– p4sch
Jan 13 at 14:54
$begingroup$
Thanks for taking your time! What do you mean by "$(3)$ is a Cauchy sequence"? In which way do you interpret this as a sequence?
$endgroup$
– Staki42
Jan 13 at 12:29
$begingroup$
Thanks for taking your time! What do you mean by "$(3)$ is a Cauchy sequence"? In which way do you interpret this as a sequence?
$endgroup$
– Staki42
Jan 13 at 12:29
$begingroup$
Never mind, I got it. Thanks!
$endgroup$
– Staki42
Jan 13 at 12:40
$begingroup$
Never mind, I got it. Thanks!
$endgroup$
– Staki42
Jan 13 at 12:40
$begingroup$
More precisely, I should say 'Cauchy-net'. There is a generalization of sequences called nets, see Wikipedia. The bound shows that (3) can be made arbitary small if $a$ is large enough and that is exactly the property to be a Cauchy-net.
$endgroup$
– p4sch
Jan 13 at 12:47
$begingroup$
More precisely, I should say 'Cauchy-net'. There is a generalization of sequences called nets, see Wikipedia. The bound shows that (3) can be made arbitary small if $a$ is large enough and that is exactly the property to be a Cauchy-net.
$endgroup$
– p4sch
Jan 13 at 12:47
$begingroup$
I see. But I also could just take a sequence, right?
$endgroup$
– Staki42
Jan 13 at 13:28
$begingroup$
I see. But I also could just take a sequence, right?
$endgroup$
– Staki42
Jan 13 at 13:28
1
1
$begingroup$
No, for $y=0$ the integral is not absolute convergent, because $t^{alpha-1}$ is not in $L^1([1,infty)$. For $y >0$ it is not a problem, but we would like to determine the integral for $y=0$ and also see that it is continuous in $y$ in order to determine the limes.
$endgroup$
– p4sch
Jan 13 at 14:54
$begingroup$
No, for $y=0$ the integral is not absolute convergent, because $t^{alpha-1}$ is not in $L^1([1,infty)$. For $y >0$ it is not a problem, but we would like to determine the integral for $y=0$ and also see that it is continuous in $y$ in order to determine the limes.
$endgroup$
– p4sch
Jan 13 at 14:54
|
show 6 more comments
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