Mixing
A question on space of $l^p(mathbb{N})$
$begingroup$
1) give an example of ${x_n}$ such that $displaystylelim_{nto +infty}x_n=0 $ but
${x_n}notin l^p(mathbb{N}), p>1, forall p.$
2) show that $1le p<r$ then $l^p(mathbb{N})subsetneq l^r(mathbb{N}) $
for (1) we choose ${x_n}=frac{1}{n^frac{1}{p}}$ but I am not sure ..
some one can help ..thank you so much
real-analysis functional-analysis
edited Feb 3 at 9:58
Nosrati
26.5k62354
asked Feb 3 at 9:52
learnerlearner
1208
$endgroup$
add a comment |
$begingroup$
1) give an example of ${x_n}$ such that $displaystylelim_{nto +infty}x_n=0 $ but
${x_n}notin l^p(mathbb{N}), p>1, forall p.$
2) show that $1le p<r$ then $l^p(mathbb{N})subsetneq l^r(mathbb{N}) $
for (1) we choose ${x_n}=frac{1}{n^frac{1}{p}}$ but I am not sure ..
some one can help ..thank you so much
real-analysis functional-analysis
edited Feb 3 at 9:58
Nosrati
26.5k62354
asked Feb 3 at 9:52
learnerlearner
1208
$endgroup$
add a comment |
$begingroup$
1) give an example of ${x_n}$ such that $displaystylelim_{nto +infty}x_n=0 $ but
${x_n}notin l^p(mathbb{N}), p>1, forall p.$
2) show that $1le p<r$ then $l^p(mathbb{N})subsetneq l^r(mathbb{N}) $
for (1) we choose ${x_n}=frac{1}{n^frac{1}{p}}$ but I am not sure ..
some one can help ..thank you so much
real-analysis functional-analysis
edited Feb 3 at 9:58
Nosrati
26.5k62354
asked Feb 3 at 9:52
learnerlearner
1208
$endgroup$
1) give an example of ${x_n}$ such that $displaystylelim_{nto +infty}x_n=0 $ but
${x_n}notin l^p(mathbb{N}), p>1, forall p.$
2) show that $1le p<r$ then $l^p(mathbb{N})subsetneq l^r(mathbb{N}) $
for (1) we choose ${x_n}=frac{1}{n^frac{1}{p}}$ but I am not sure ..
some one can help ..thank you so much
real-analysis functional-analysis
real-analysis functional-analysis
edited Feb 3 at 9:58
Nosrati
26.5k62354
asked Feb 3 at 9:52
learnerlearner
1208
edited Feb 3 at 9:58
Nosrati
26.5k62354
asked Feb 3 at 9:52
learnerlearner
1208
edited Feb 3 at 9:58
Nosrati
26.5k62354
edited Feb 3 at 9:58
Nosrati
26.5k62354
edited Feb 3 at 9:58
Nosrati
26.5k62354
26.5k62354
asked Feb 3 at 9:52
learnerlearner
1208
asked Feb 3 at 9:52
learnerlearner
1208
asked Feb 3 at 9:52
learnerlearner
1208
1208
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Your example for part $(1)$ is fine since $sum |x_n|^p = sum frac1n = infty$ (as long as you are looking for one example for each value of $p$). If you want to find a single sequence that works for all values of $p$, one way to proceed is to combine a family of examples for single values of $p$. Let ${p_1, p_2, p_3, dots }$ be the prime numbers and define
$$x_n = begin{cases}
bigg(frac{1}{k+p_i} bigg)^{1/p_i} & text{ if } n = kp_i text{ for some } i \
0 & text{ otherwise}
end{cases}
$$
Then $x_n to 0$ as $n to infty$ (I'll leave this as an exercise) and for any $p > 1$ there is a $p_i > p$. As a result, the subsequence $(x_{kp_i})$ is not in $ell_p$, as in the first example, and hence neither is the sequence $(x_n)$.
For the second part, start by taking $(x_n) in ell^p$. We want to show $(x_n) in ell^r$. Note that since $|x_n|^p$ is summable, $x_n to 0$ as $n to infty$. In particular, for some $N$ and for all $n geq N$, $|x_n| < 1$ and so $|x_n|^r < |x_n|^p$ for $n geq N$.
Hence $$sum |x_n|^r leq sum_{n=1}^{N-1} |x_n|^r + sum_N^infty |x_n|^p < infty.$$
To finish, you want to find $(y_n)$ such that $sum |y_n|^r < infty$ but $sum |y_n|^p = infty$. For this, you should try to adapt your example from part $(1)$ by replacing $frac{1}{p}$ by another suitable power of $frac1n$.
answered Feb 3 at 10:10
Rhys SteeleRhys Steele
7,9301931
$endgroup$
$begingroup$
Correct me if I'm wrong, but $frac1{n^frac1p}$ as an example does not make sense as it will not converge for all $p>1$.
$endgroup$
– Anthony Ter
Feb 3 at 10:14
$begingroup$
@AnthonyTer……. I think this example not wrong
$endgroup$
– learner
Feb 3 at 10:16
$begingroup$
@AnthonyTer This depends on how you parse the exercise I think. I'd read it as requiring you to, for each $p$, give an example rather than give a single example that works for every $p$. With the latter reading you are of course correct.
$endgroup$
– Rhys Steele
Feb 3 at 10:29
$begingroup$
@RhysSteele Oh ok, I can see how it can be interpreted both ways. Could you think of an example that satisfies the condition the way I read it? I couldn't.
$endgroup$
– Anthony Ter
Feb 3 at 10:39
$begingroup$
@RhysSteele..sorry I m not understand last two steps of your proof
$endgroup$
– learner
Feb 3 at 10:52
|
show 4 more comments
$begingroup$
Another example for part 1 of the exercise would be $$x_n = frac{1}{ln n}$$
In fact, by observing that $frac{1/{(ln n)^p}}{1/n} to infty$ and that the harmonic series diverges, you can conclude that:
$$sum_{n=2}^{infty} frac{1}{(ln n)^p} = infty$$ for every $p$.
answered Feb 3 at 11:20
HarnakHarnak
1,3691512
$endgroup$
add a comment |
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2 Answers
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2 Answers
2
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$begingroup$
Your example for part $(1)$ is fine since $sum |x_n|^p = sum frac1n = infty$ (as long as you are looking for one example for each value of $p$). If you want to find a single sequence that works for all values of $p$, one way to proceed is to combine a family of examples for single values of $p$. Let ${p_1, p_2, p_3, dots }$ be the prime numbers and define
$$x_n = begin{cases}
bigg(frac{1}{k+p_i} bigg)^{1/p_i} & text{ if } n = kp_i text{ for some } i \
0 & text{ otherwise}
end{cases}
$$
Then $x_n to 0$ as $n to infty$ (I'll leave this as an exercise) and for any $p > 1$ there is a $p_i > p$. As a result, the subsequence $(x_{kp_i})$ is not in $ell_p$, as in the first example, and hence neither is the sequence $(x_n)$.
For the second part, start by taking $(x_n) in ell^p$. We want to show $(x_n) in ell^r$. Note that since $|x_n|^p$ is summable, $x_n to 0$ as $n to infty$. In particular, for some $N$ and for all $n geq N$, $|x_n| < 1$ and so $|x_n|^r < |x_n|^p$ for $n geq N$.
Hence $$sum |x_n|^r leq sum_{n=1}^{N-1} |x_n|^r + sum_N^infty |x_n|^p < infty.$$
To finish, you want to find $(y_n)$ such that $sum |y_n|^r < infty$ but $sum |y_n|^p = infty$. For this, you should try to adapt your example from part $(1)$ by replacing $frac{1}{p}$ by another suitable power of $frac1n$.
answered Feb 3 at 10:10
Rhys SteeleRhys Steele
7,9301931
$endgroup$
$begingroup$
Correct me if I'm wrong, but $frac1{n^frac1p}$ as an example does not make sense as it will not converge for all $p>1$.
$endgroup$
– Anthony Ter
Feb 3 at 10:14
$begingroup$
@AnthonyTer……. I think this example not wrong
$endgroup$
– learner
Feb 3 at 10:16
$begingroup$
@AnthonyTer This depends on how you parse the exercise I think. I'd read it as requiring you to, for each $p$, give an example rather than give a single example that works for every $p$. With the latter reading you are of course correct.
$endgroup$
– Rhys Steele
Feb 3 at 10:29
$begingroup$
@RhysSteele Oh ok, I can see how it can be interpreted both ways. Could you think of an example that satisfies the condition the way I read it? I couldn't.
$endgroup$
– Anthony Ter
Feb 3 at 10:39
$begingroup$
@RhysSteele..sorry I m not understand last two steps of your proof
$endgroup$
– learner
Feb 3 at 10:52
|
show 4 more comments
$begingroup$
Your example for part $(1)$ is fine since $sum |x_n|^p = sum frac1n = infty$ (as long as you are looking for one example for each value of $p$). If you want to find a single sequence that works for all values of $p$, one way to proceed is to combine a family of examples for single values of $p$. Let ${p_1, p_2, p_3, dots }$ be the prime numbers and define
$$x_n = begin{cases}
bigg(frac{1}{k+p_i} bigg)^{1/p_i} & text{ if } n = kp_i text{ for some } i \
0 & text{ otherwise}
end{cases}
$$
Then $x_n to 0$ as $n to infty$ (I'll leave this as an exercise) and for any $p > 1$ there is a $p_i > p$. As a result, the subsequence $(x_{kp_i})$ is not in $ell_p$, as in the first example, and hence neither is the sequence $(x_n)$.
For the second part, start by taking $(x_n) in ell^p$. We want to show $(x_n) in ell^r$. Note that since $|x_n|^p$ is summable, $x_n to 0$ as $n to infty$. In particular, for some $N$ and for all $n geq N$, $|x_n| < 1$ and so $|x_n|^r < |x_n|^p$ for $n geq N$.
Hence $$sum |x_n|^r leq sum_{n=1}^{N-1} |x_n|^r + sum_N^infty |x_n|^p < infty.$$
To finish, you want to find $(y_n)$ such that $sum |y_n|^r < infty$ but $sum |y_n|^p = infty$. For this, you should try to adapt your example from part $(1)$ by replacing $frac{1}{p}$ by another suitable power of $frac1n$.
answered Feb 3 at 10:10
Rhys SteeleRhys Steele
7,9301931
$endgroup$
$begingroup$
Correct me if I'm wrong, but $frac1{n^frac1p}$ as an example does not make sense as it will not converge for all $p>1$.
$endgroup$
– Anthony Ter
Feb 3 at 10:14
$begingroup$
@AnthonyTer……. I think this example not wrong
$endgroup$
– learner
Feb 3 at 10:16
$begingroup$
@AnthonyTer This depends on how you parse the exercise I think. I'd read it as requiring you to, for each $p$, give an example rather than give a single example that works for every $p$. With the latter reading you are of course correct.
$endgroup$
– Rhys Steele
Feb 3 at 10:29
$begingroup$
@RhysSteele Oh ok, I can see how it can be interpreted both ways. Could you think of an example that satisfies the condition the way I read it? I couldn't.
$endgroup$
– Anthony Ter
Feb 3 at 10:39
$begingroup$
@RhysSteele..sorry I m not understand last two steps of your proof
$endgroup$
– learner
Feb 3 at 10:52
|
show 4 more comments
$begingroup$
Your example for part $(1)$ is fine since $sum |x_n|^p = sum frac1n = infty$ (as long as you are looking for one example for each value of $p$). If you want to find a single sequence that works for all values of $p$, one way to proceed is to combine a family of examples for single values of $p$. Let ${p_1, p_2, p_3, dots }$ be the prime numbers and define
$$x_n = begin{cases}
bigg(frac{1}{k+p_i} bigg)^{1/p_i} & text{ if } n = kp_i text{ for some } i \
0 & text{ otherwise}
end{cases}
$$
Then $x_n to 0$ as $n to infty$ (I'll leave this as an exercise) and for any $p > 1$ there is a $p_i > p$. As a result, the subsequence $(x_{kp_i})$ is not in $ell_p$, as in the first example, and hence neither is the sequence $(x_n)$.
For the second part, start by taking $(x_n) in ell^p$. We want to show $(x_n) in ell^r$. Note that since $|x_n|^p$ is summable, $x_n to 0$ as $n to infty$. In particular, for some $N$ and for all $n geq N$, $|x_n| < 1$ and so $|x_n|^r < |x_n|^p$ for $n geq N$.
Hence $$sum |x_n|^r leq sum_{n=1}^{N-1} |x_n|^r + sum_N^infty |x_n|^p < infty.$$
To finish, you want to find $(y_n)$ such that $sum |y_n|^r < infty$ but $sum |y_n|^p = infty$. For this, you should try to adapt your example from part $(1)$ by replacing $frac{1}{p}$ by another suitable power of $frac1n$.
answered Feb 3 at 10:10
Rhys SteeleRhys Steele
7,9301931
$endgroup$
Your example for part $(1)$ is fine since $sum |x_n|^p = sum frac1n = infty$ (as long as you are looking for one example for each value of $p$). If you want to find a single sequence that works for all values of $p$, one way to proceed is to combine a family of examples for single values of $p$. Let ${p_1, p_2, p_3, dots }$ be the prime numbers and define
$$x_n = begin{cases}
bigg(frac{1}{k+p_i} bigg)^{1/p_i} & text{ if } n = kp_i text{ for some } i \
0 & text{ otherwise}
end{cases}
$$
Then $x_n to 0$ as $n to infty$ (I'll leave this as an exercise) and for any $p > 1$ there is a $p_i > p$. As a result, the subsequence $(x_{kp_i})$ is not in $ell_p$, as in the first example, and hence neither is the sequence $(x_n)$.
For the second part, start by taking $(x_n) in ell^p$. We want to show $(x_n) in ell^r$. Note that since $|x_n|^p$ is summable, $x_n to 0$ as $n to infty$. In particular, for some $N$ and for all $n geq N$, $|x_n| < 1$ and so $|x_n|^r < |x_n|^p$ for $n geq N$.
Hence $$sum |x_n|^r leq sum_{n=1}^{N-1} |x_n|^r + sum_N^infty |x_n|^p < infty.$$
To finish, you want to find $(y_n)$ such that $sum |y_n|^r < infty$ but $sum |y_n|^p = infty$. For this, you should try to adapt your example from part $(1)$ by replacing $frac{1}{p}$ by another suitable power of $frac1n$.
answered Feb 3 at 10:10
Rhys SteeleRhys Steele
7,9301931
edited Feb 3 at 11:09
answered Feb 3 at 10:10
Rhys SteeleRhys Steele
7,9301931
answered Feb 3 at 10:10
Rhys SteeleRhys Steele
7,9301931
answered Feb 3 at 10:10
Rhys SteeleRhys Steele
7,9301931
7,9301931
$begingroup$
Correct me if I'm wrong, but $frac1{n^frac1p}$ as an example does not make sense as it will not converge for all $p>1$.
$endgroup$
– Anthony Ter
Feb 3 at 10:14
$begingroup$
@AnthonyTer……. I think this example not wrong
$endgroup$
– learner
Feb 3 at 10:16
$begingroup$
@AnthonyTer This depends on how you parse the exercise I think. I'd read it as requiring you to, for each $p$, give an example rather than give a single example that works for every $p$. With the latter reading you are of course correct.
$endgroup$
– Rhys Steele
Feb 3 at 10:29
$begingroup$
@RhysSteele Oh ok, I can see how it can be interpreted both ways. Could you think of an example that satisfies the condition the way I read it? I couldn't.
$endgroup$
– Anthony Ter
Feb 3 at 10:39
$begingroup$
@RhysSteele..sorry I m not understand last two steps of your proof
$endgroup$
– learner
Feb 3 at 10:52
|
show 4 more comments
$begingroup$
Correct me if I'm wrong, but $frac1{n^frac1p}$ as an example does not make sense as it will not converge for all $p>1$.
$endgroup$
– Anthony Ter
Feb 3 at 10:14
$begingroup$
@AnthonyTer……. I think this example not wrong
$endgroup$
– learner
Feb 3 at 10:16
$begingroup$
@AnthonyTer This depends on how you parse the exercise I think. I'd read it as requiring you to, for each $p$, give an example rather than give a single example that works for every $p$. With the latter reading you are of course correct.
$endgroup$
– Rhys Steele
Feb 3 at 10:29
$begingroup$
@RhysSteele Oh ok, I can see how it can be interpreted both ways. Could you think of an example that satisfies the condition the way I read it? I couldn't.
$endgroup$
– Anthony Ter
Feb 3 at 10:39
$begingroup$
@RhysSteele..sorry I m not understand last two steps of your proof
$endgroup$
– learner
Feb 3 at 10:52
$begingroup$
Correct me if I'm wrong, but $frac1{n^frac1p}$ as an example does not make sense as it will not converge for all $p>1$.
$endgroup$
– Anthony Ter
Feb 3 at 10:14
$begingroup$
Correct me if I'm wrong, but $frac1{n^frac1p}$ as an example does not make sense as it will not converge for all $p>1$.
$endgroup$
– Anthony Ter
Feb 3 at 10:14
$begingroup$
@AnthonyTer……. I think this example not wrong
$endgroup$
– learner
Feb 3 at 10:16
$begingroup$
@AnthonyTer……. I think this example not wrong
$endgroup$
– learner
Feb 3 at 10:16
$begingroup$
@AnthonyTer This depends on how you parse the exercise I think. I'd read it as requiring you to, for each $p$, give an example rather than give a single example that works for every $p$. With the latter reading you are of course correct.
$endgroup$
– Rhys Steele
Feb 3 at 10:29
$begingroup$
@AnthonyTer This depends on how you parse the exercise I think. I'd read it as requiring you to, for each $p$, give an example rather than give a single example that works for every $p$. With the latter reading you are of course correct.
$endgroup$
– Rhys Steele
Feb 3 at 10:29
$begingroup$
@RhysSteele Oh ok, I can see how it can be interpreted both ways. Could you think of an example that satisfies the condition the way I read it? I couldn't.
$endgroup$
– Anthony Ter
Feb 3 at 10:39
$begingroup$
@RhysSteele Oh ok, I can see how it can be interpreted both ways. Could you think of an example that satisfies the condition the way I read it? I couldn't.
$endgroup$
– Anthony Ter
Feb 3 at 10:39
$begingroup$
@RhysSteele..sorry I m not understand last two steps of your proof
$endgroup$
– learner
Feb 3 at 10:52
$begingroup$
@RhysSteele..sorry I m not understand last two steps of your proof
$endgroup$
– learner
Feb 3 at 10:52
|
show 4 more comments
$begingroup$
Another example for part 1 of the exercise would be $$x_n = frac{1}{ln n}$$
In fact, by observing that $frac{1/{(ln n)^p}}{1/n} to infty$ and that the harmonic series diverges, you can conclude that:
$$sum_{n=2}^{infty} frac{1}{(ln n)^p} = infty$$ for every $p$.
answered Feb 3 at 11:20
HarnakHarnak
1,3691512
$endgroup$
add a comment |
$begingroup$
Another example for part 1 of the exercise would be $$x_n = frac{1}{ln n}$$
In fact, by observing that $frac{1/{(ln n)^p}}{1/n} to infty$ and that the harmonic series diverges, you can conclude that:
$$sum_{n=2}^{infty} frac{1}{(ln n)^p} = infty$$ for every $p$.
answered Feb 3 at 11:20
HarnakHarnak
1,3691512
$endgroup$
add a comment |
$begingroup$
Another example for part 1 of the exercise would be $$x_n = frac{1}{ln n}$$
In fact, by observing that $frac{1/{(ln n)^p}}{1/n} to infty$ and that the harmonic series diverges, you can conclude that:
$$sum_{n=2}^{infty} frac{1}{(ln n)^p} = infty$$ for every $p$.
answered Feb 3 at 11:20
HarnakHarnak
1,3691512
$endgroup$
Another example for part 1 of the exercise would be $$x_n = frac{1}{ln n}$$
In fact, by observing that $frac{1/{(ln n)^p}}{1/n} to infty$ and that the harmonic series diverges, you can conclude that:
$$sum_{n=2}^{infty} frac{1}{(ln n)^p} = infty$$ for every $p$.
answered Feb 3 at 11:20
HarnakHarnak
1,3691512
answered Feb 3 at 11:20
HarnakHarnak
1,3691512
answered Feb 3 at 11:20
HarnakHarnak
1,3691512
answered Feb 3 at 11:20
HarnakHarnak
1,3691512
1,3691512
add a comment |
add a comment |
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