Mixing
Techniques for proving non-uniform-convergence of improper integrals
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I have been reviewing uniform convergence of series and improper integrals with parameters. Working through examples I found it easier to prove uniform convergence (using M-test, Dirichlet-Abel test, etc.) than to refute it. I would like to get a better idea of techniques to prove that uniform convergence fails.
I could show that the improper integral $I(r) =int_0^infty frac{sin(rx)}{x} , dx$ converges uniformly for $r geq c > 0$. The Dirichlet test works because $1/x to 0$ and we have uniformly bounded
$$left|int_0^t sin(rx), dxright| = frac{|1 - cos(rt)|}{r} leq frac{2}{c}$$
Since $I(0) = 0$ and $I(r) > 0$ for $r > 0$ we know that $I(r)$ is not continuous and the improper integral cannot converge uniformly for $r in [0,infty)$. In this case we have the information about the value to apply the continuity theorem.
But how can you show the convergence is not uniform directly from the definition of uniform convergence?
real-analysis uniform-convergence
asked Jan 18 at 22:59
scobacoscobaco
20739
$endgroup$
add a comment |
$begingroup$
I have been reviewing uniform convergence of series and improper integrals with parameters. Working through examples I found it easier to prove uniform convergence (using M-test, Dirichlet-Abel test, etc.) than to refute it. I would like to get a better idea of techniques to prove that uniform convergence fails.
I could show that the improper integral $I(r) =int_0^infty frac{sin(rx)}{x} , dx$ converges uniformly for $r geq c > 0$. The Dirichlet test works because $1/x to 0$ and we have uniformly bounded
$$left|int_0^t sin(rx), dxright| = frac{|1 - cos(rt)|}{r} leq frac{2}{c}$$
Since $I(0) = 0$ and $I(r) > 0$ for $r > 0$ we know that $I(r)$ is not continuous and the improper integral cannot converge uniformly for $r in [0,infty)$. In this case we have the information about the value to apply the continuity theorem.
But how can you show the convergence is not uniform directly from the definition of uniform convergence?
real-analysis uniform-convergence
asked Jan 18 at 22:59
scobacoscobaco
20739
$endgroup$
$begingroup$
"Since $I(0) = 0$ and $I(r) > 0$ for $r > 0$ we know that $I(r)$ is not continuous" That's not the reason. The reason is that $I(r)$ is a positive constant for $r>0.$
$endgroup$
– zhw.
Jan 19 at 18:00
add a comment |
$begingroup$
I have been reviewing uniform convergence of series and improper integrals with parameters. Working through examples I found it easier to prove uniform convergence (using M-test, Dirichlet-Abel test, etc.) than to refute it. I would like to get a better idea of techniques to prove that uniform convergence fails.
I could show that the improper integral $I(r) =int_0^infty frac{sin(rx)}{x} , dx$ converges uniformly for $r geq c > 0$. The Dirichlet test works because $1/x to 0$ and we have uniformly bounded
$$left|int_0^t sin(rx), dxright| = frac{|1 - cos(rt)|}{r} leq frac{2}{c}$$
Since $I(0) = 0$ and $I(r) > 0$ for $r > 0$ we know that $I(r)$ is not continuous and the improper integral cannot converge uniformly for $r in [0,infty)$. In this case we have the information about the value to apply the continuity theorem.
But how can you show the convergence is not uniform directly from the definition of uniform convergence?
real-analysis uniform-convergence
asked Jan 18 at 22:59
scobacoscobaco
20739
$endgroup$
I have been reviewing uniform convergence of series and improper integrals with parameters. Working through examples I found it easier to prove uniform convergence (using M-test, Dirichlet-Abel test, etc.) than to refute it. I would like to get a better idea of techniques to prove that uniform convergence fails.
I could show that the improper integral $I(r) =int_0^infty frac{sin(rx)}{x} , dx$ converges uniformly for $r geq c > 0$. The Dirichlet test works because $1/x to 0$ and we have uniformly bounded
$$left|int_0^t sin(rx), dxright| = frac{|1 - cos(rt)|}{r} leq frac{2}{c}$$
Since $I(0) = 0$ and $I(r) > 0$ for $r > 0$ we know that $I(r)$ is not continuous and the improper integral cannot converge uniformly for $r in [0,infty)$. In this case we have the information about the value to apply the continuity theorem.
But how can you show the convergence is not uniform directly from the definition of uniform convergence?
real-analysis uniform-convergence
real-analysis uniform-convergence
asked Jan 18 at 22:59
scobacoscobaco
20739
asked Jan 18 at 22:59
scobacoscobaco
20739
asked Jan 18 at 22:59
scobacoscobaco
20739
asked Jan 18 at 22:59
scobacoscobaco
20739
asked Jan 18 at 22:59
scobacoscobaco
20739
20739
$begingroup$
"Since $I(0) = 0$ and $I(r) > 0$ for $r > 0$ we know that $I(r)$ is not continuous" That's not the reason. The reason is that $I(r)$ is a positive constant for $r>0.$
$endgroup$
– zhw.
Jan 19 at 18:00
add a comment |
$begingroup$
"Since $I(0) = 0$ and $I(r) > 0$ for $r > 0$ we know that $I(r)$ is not continuous" That's not the reason. The reason is that $I(r)$ is a positive constant for $r>0.$
$endgroup$
– zhw.
Jan 19 at 18:00
$begingroup$
"Since $I(0) = 0$ and $I(r) > 0$ for $r > 0$ we know that $I(r)$ is not continuous" That's not the reason. The reason is that $I(r)$ is a positive constant for $r>0.$
$endgroup$
– zhw.
Jan 19 at 18:00
$begingroup$
"Since $I(0) = 0$ and $I(r) > 0$ for $r > 0$ we know that $I(r)$ is not continuous" That's not the reason. The reason is that $I(r)$ is a positive constant for $r>0.$
$endgroup$
– zhw.
Jan 19 at 18:00
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
We know
$$I=int_0^infty frac{sin x}{x}, dx >0.$$
Suppose there exists $M>0$ such that $b>M$ implies
$$|int_b^infty frac{sin (rx)}{x}, dx |<frac{I}{2}$$
for all $rge 0.$ Letting $x=y/r$ then gives
$$|int_{rb}^infty frac{sin y}{y}), dy |<frac{I}{2}$$
for all $rge 0.$ Now let $rto 0^+$ to arrive at $Ile I/2,$ contradiction.
answered Jan 19 at 18:12
zhw.zhw.
73.6k43175
$endgroup$
$begingroup$
Thanks. I realize now this is a bad example for my since the integral doesn't even depend on $r > 0$, but seeing the approach is helpful.
$endgroup$
– scobaco
Jan 25 at 20:54
add a comment |
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1 Answer
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active
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1 Answer
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active
oldest
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$begingroup$
We know
$$I=int_0^infty frac{sin x}{x}, dx >0.$$
Suppose there exists $M>0$ such that $b>M$ implies
$$|int_b^infty frac{sin (rx)}{x}, dx |<frac{I}{2}$$
for all $rge 0.$ Letting $x=y/r$ then gives
$$|int_{rb}^infty frac{sin y}{y}), dy |<frac{I}{2}$$
for all $rge 0.$ Now let $rto 0^+$ to arrive at $Ile I/2,$ contradiction.
answered Jan 19 at 18:12
zhw.zhw.
73.6k43175
$endgroup$
$begingroup$
Thanks. I realize now this is a bad example for my since the integral doesn't even depend on $r > 0$, but seeing the approach is helpful.
$endgroup$
– scobaco
Jan 25 at 20:54
add a comment |
$begingroup$
We know
$$I=int_0^infty frac{sin x}{x}, dx >0.$$
Suppose there exists $M>0$ such that $b>M$ implies
$$|int_b^infty frac{sin (rx)}{x}, dx |<frac{I}{2}$$
for all $rge 0.$ Letting $x=y/r$ then gives
$$|int_{rb}^infty frac{sin y}{y}), dy |<frac{I}{2}$$
for all $rge 0.$ Now let $rto 0^+$ to arrive at $Ile I/2,$ contradiction.
answered Jan 19 at 18:12
zhw.zhw.
73.6k43175
$endgroup$
$begingroup$
Thanks. I realize now this is a bad example for my since the integral doesn't even depend on $r > 0$, but seeing the approach is helpful.
$endgroup$
– scobaco
Jan 25 at 20:54
add a comment |
$begingroup$
We know
$$I=int_0^infty frac{sin x}{x}, dx >0.$$
Suppose there exists $M>0$ such that $b>M$ implies
$$|int_b^infty frac{sin (rx)}{x}, dx |<frac{I}{2}$$
for all $rge 0.$ Letting $x=y/r$ then gives
$$|int_{rb}^infty frac{sin y}{y}), dy |<frac{I}{2}$$
for all $rge 0.$ Now let $rto 0^+$ to arrive at $Ile I/2,$ contradiction.
answered Jan 19 at 18:12
zhw.zhw.
73.6k43175
$endgroup$
We know
$$I=int_0^infty frac{sin x}{x}, dx >0.$$
Suppose there exists $M>0$ such that $b>M$ implies
$$|int_b^infty frac{sin (rx)}{x}, dx |<frac{I}{2}$$
for all $rge 0.$ Letting $x=y/r$ then gives
$$|int_{rb}^infty frac{sin y}{y}), dy |<frac{I}{2}$$
for all $rge 0.$ Now let $rto 0^+$ to arrive at $Ile I/2,$ contradiction.
answered Jan 19 at 18:12
zhw.zhw.
73.6k43175
answered Jan 19 at 18:12
zhw.zhw.
73.6k43175
answered Jan 19 at 18:12
zhw.zhw.
73.6k43175
answered Jan 19 at 18:12
zhw.zhw.
73.6k43175
73.6k43175
$begingroup$
Thanks. I realize now this is a bad example for my since the integral doesn't even depend on $r > 0$, but seeing the approach is helpful.
$endgroup$
– scobaco
Jan 25 at 20:54
add a comment |
$begingroup$
Thanks. I realize now this is a bad example for my since the integral doesn't even depend on $r > 0$, but seeing the approach is helpful.
$endgroup$
– scobaco
Jan 25 at 20:54
$begingroup$
Thanks. I realize now this is a bad example for my since the integral doesn't even depend on $r > 0$, but seeing the approach is helpful.
$endgroup$
– scobaco
Jan 25 at 20:54
$begingroup$
Thanks. I realize now this is a bad example for my since the integral doesn't even depend on $r > 0$, but seeing the approach is helpful.
$endgroup$
– scobaco
Jan 25 at 20:54
add a comment |
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$begingroup$
"Since $I(0) = 0$ and $I(r) > 0$ for $r > 0$ we know that $I(r)$ is not continuous" That's not the reason. The reason is that $I(r)$ is a positive constant for $r>0.$
$endgroup$
– zhw.
Jan 19 at 18:00