Mixing
Is the set of sequences $(x_n)$ in $ell^1$ such that $|x_n|leqslantfrac1n$ for every $n$, compact in $ell^1$?
I was thinking about proving a set $S$ is compact in $l^{1}$ space.
$S = {f in l^{1} : |f(k)| leq frac{1}{k} forall k }$
where ${f_{n}}_{n=1}^{infty}$ be functions such that $f_{n} : Bbb{N} rightarrow Bbb{R}^{+}$
I thought of applying that a set $A subset l^{1}$ is compact if and only if $A$ is closed, bounded and $Sup_{x in A} sum_{j geq n}|x_{j}| rightarrow 0$ as $n rightarrow infty$.
Discussing a bit let me think of taking $f(n) = frac{1}{n}$, but it would not belong to $l^{1}$ as the harmonic series would not converge. How to proceed with this question.
EDIT -
Also, how the result changes if we would consider $l^{infty}$ instead of $l^{1}$ space. I think in this case we can take for a fixed integer $N$, $v_{k} = frac{1}{k}$ for $1 leq k leq N$ and $v_{k} =0$ otherwise.
then here $||v|| = 1$ and since $N$ is arbitrary so its bounded always, so the set is compact if we change the space from $l^{1}$ to $l^{infty}$. Is this correct?.
real-analysis sequences-and-series compactness lp-spaces
asked Dec 30 '18 at 21:10
BAYMAXBAYMAX
2,86121123
add a comment |
I was thinking about proving a set $S$ is compact in $l^{1}$ space.
$S = {f in l^{1} : |f(k)| leq frac{1}{k} forall k }$
where ${f_{n}}_{n=1}^{infty}$ be functions such that $f_{n} : Bbb{N} rightarrow Bbb{R}^{+}$
I thought of applying that a set $A subset l^{1}$ is compact if and only if $A$ is closed, bounded and $Sup_{x in A} sum_{j geq n}|x_{j}| rightarrow 0$ as $n rightarrow infty$.
Discussing a bit let me think of taking $f(n) = frac{1}{n}$, but it would not belong to $l^{1}$ as the harmonic series would not converge. How to proceed with this question.
EDIT -
Also, how the result changes if we would consider $l^{infty}$ instead of $l^{1}$ space. I think in this case we can take for a fixed integer $N$, $v_{k} = frac{1}{k}$ for $1 leq k leq N$ and $v_{k} =0$ otherwise.
then here $||v|| = 1$ and since $N$ is arbitrary so its bounded always, so the set is compact if we change the space from $l^{1}$ to $l^{infty}$. Is this correct?.
real-analysis sequences-and-series compactness lp-spaces
asked Dec 30 '18 at 21:10
BAYMAXBAYMAX
2,86121123
2
Consider every $f_N$ defined by $f_N(n)=frac1nmathbf 1_{nleqslant N}$, and proceed as it seems you wanted to, then the proof is direct.
– Did
Dec 30 '18 at 21:16
Nice!! the $1_{n leq N}$ is the characteristic function?
– BAYMAX
Dec 30 '18 at 21:27
add a comment |
I was thinking about proving a set $S$ is compact in $l^{1}$ space.
$S = {f in l^{1} : |f(k)| leq frac{1}{k} forall k }$
where ${f_{n}}_{n=1}^{infty}$ be functions such that $f_{n} : Bbb{N} rightarrow Bbb{R}^{+}$
I thought of applying that a set $A subset l^{1}$ is compact if and only if $A$ is closed, bounded and $Sup_{x in A} sum_{j geq n}|x_{j}| rightarrow 0$ as $n rightarrow infty$.
Discussing a bit let me think of taking $f(n) = frac{1}{n}$, but it would not belong to $l^{1}$ as the harmonic series would not converge. How to proceed with this question.
EDIT -
Also, how the result changes if we would consider $l^{infty}$ instead of $l^{1}$ space. I think in this case we can take for a fixed integer $N$, $v_{k} = frac{1}{k}$ for $1 leq k leq N$ and $v_{k} =0$ otherwise.
then here $||v|| = 1$ and since $N$ is arbitrary so its bounded always, so the set is compact if we change the space from $l^{1}$ to $l^{infty}$. Is this correct?.
real-analysis sequences-and-series compactness lp-spaces
asked Dec 30 '18 at 21:10
BAYMAXBAYMAX
2,86121123
I was thinking about proving a set $S$ is compact in $l^{1}$ space.
$S = {f in l^{1} : |f(k)| leq frac{1}{k} forall k }$
where ${f_{n}}_{n=1}^{infty}$ be functions such that $f_{n} : Bbb{N} rightarrow Bbb{R}^{+}$
I thought of applying that a set $A subset l^{1}$ is compact if and only if $A$ is closed, bounded and $Sup_{x in A} sum_{j geq n}|x_{j}| rightarrow 0$ as $n rightarrow infty$.
Discussing a bit let me think of taking $f(n) = frac{1}{n}$, but it would not belong to $l^{1}$ as the harmonic series would not converge. How to proceed with this question.
EDIT -
Also, how the result changes if we would consider $l^{infty}$ instead of $l^{1}$ space. I think in this case we can take for a fixed integer $N$, $v_{k} = frac{1}{k}$ for $1 leq k leq N$ and $v_{k} =0$ otherwise.
then here $||v|| = 1$ and since $N$ is arbitrary so its bounded always, so the set is compact if we change the space from $l^{1}$ to $l^{infty}$. Is this correct?.
real-analysis sequences-and-series compactness lp-spaces
real-analysis sequences-and-series compactness lp-spaces
asked Dec 30 '18 at 21:10
BAYMAXBAYMAX
2,86121123
asked Dec 30 '18 at 21:10
BAYMAXBAYMAX
2,86121123
edited Dec 30 '18 at 22:16
BAYMAX
asked Dec 30 '18 at 21:10
BAYMAXBAYMAX
2,86121123
asked Dec 30 '18 at 21:10
BAYMAXBAYMAX
2,86121123
asked Dec 30 '18 at 21:10
BAYMAXBAYMAX
2,86121123
2,86121123
2
Consider every $f_N$ defined by $f_N(n)=frac1nmathbf 1_{nleqslant N}$, and proceed as it seems you wanted to, then the proof is direct.
– Did
Dec 30 '18 at 21:16
Nice!! the $1_{n leq N}$ is the characteristic function?
– BAYMAX
Dec 30 '18 at 21:27
add a comment |
2
Consider every $f_N$ defined by $f_N(n)=frac1nmathbf 1_{nleqslant N}$, and proceed as it seems you wanted to, then the proof is direct.
– Did
Dec 30 '18 at 21:16
Nice!! the $1_{n leq N}$ is the characteristic function?
– BAYMAX
Dec 30 '18 at 21:27
2
2
Consider every $f_N$ defined by $f_N(n)=frac1nmathbf 1_{nleqslant N}$, and proceed as it seems you wanted to, then the proof is direct.
– Did
Dec 30 '18 at 21:16
Consider every $f_N$ defined by $f_N(n)=frac1nmathbf 1_{nleqslant N}$, and proceed as it seems you wanted to, then the proof is direct.
– Did
Dec 30 '18 at 21:16
Nice!! the $1_{n leq N}$ is the characteristic function?
– BAYMAX
Dec 30 '18 at 21:27
Nice!! the $1_{n leq N}$ is the characteristic function?
– BAYMAX
Dec 30 '18 at 21:27
add a comment |
1 Answer
1
active
oldest
votes
The set $S$ is not bounded in $ell^1$: for a fixed integer $N$, let $v$ be the element of $S$ define as $v_k=1/k$ for $1leqslant kleqslant N$ and $v_k=0$ otherwise. Then
$leftlVert vrightrVert_1=sum_{k=1}^N1/k$ hence $sup_{vin S}leftlVert vrightrVert_1geqslant sum_{k=1}^N1/k$. As $N$ is arbitrary and the series $sum_{kgeqslant 1}1/k$ is infinite, we derive that $S$ is not bounded.
If we consider instead $ell^infty$, that is,
$$S = left{f in l^{infty} : |f(k)| leq frac{1}{k} forall k right},
$$
then $S$ is closed (since the functional $fmapsto f(k)$ is continuous for all $k$) and relatively compact: for a fixed $varepsilon$, choose $K$ such that $1/Kltvarepsilon$ and use relative compactness of $left[-1,1right]^k$.
answered Dec 30 '18 at 21:16
Davide GiraudoDavide Giraudo
125k16150261
nice solution! I was thinking how would the answer change if we consider $l^{infty}$ instead of $l^{1}$ ? I htink it is still unbounded in $l^{infty}$ too, so its not compact in both $l^{1}$ and $l^{infty}$?
– BAYMAX
Dec 30 '18 at 21:28
I think for the case of $l^{infty}$, for a fixed integer $N$, we could take $v_{k} = k$ for $1 leq k leq N$ then $||v||_{infty} = N$ but since $N$ is arbitrary so its unbounded.
– BAYMAX
Dec 30 '18 at 21:34
@BAYMAX Is $kle 1/k$?
– David C. Ullrich
Dec 31 '18 at 16:44
Nope, I realised that, so please see my edit in the question
– BAYMAX
Dec 31 '18 at 21:28
Usually, it is better to ask a separate question. For this time I addressed a sketch of prove in the edit.
– Davide Giraudo
Dec 31 '18 at 21:56
add a comment |
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1 Answer
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The set $S$ is not bounded in $ell^1$: for a fixed integer $N$, let $v$ be the element of $S$ define as $v_k=1/k$ for $1leqslant kleqslant N$ and $v_k=0$ otherwise. Then
$leftlVert vrightrVert_1=sum_{k=1}^N1/k$ hence $sup_{vin S}leftlVert vrightrVert_1geqslant sum_{k=1}^N1/k$. As $N$ is arbitrary and the series $sum_{kgeqslant 1}1/k$ is infinite, we derive that $S$ is not bounded.
If we consider instead $ell^infty$, that is,
$$S = left{f in l^{infty} : |f(k)| leq frac{1}{k} forall k right},
$$
then $S$ is closed (since the functional $fmapsto f(k)$ is continuous for all $k$) and relatively compact: for a fixed $varepsilon$, choose $K$ such that $1/Kltvarepsilon$ and use relative compactness of $left[-1,1right]^k$.
answered Dec 30 '18 at 21:16
Davide GiraudoDavide Giraudo
125k16150261
nice solution! I was thinking how would the answer change if we consider $l^{infty}$ instead of $l^{1}$ ? I htink it is still unbounded in $l^{infty}$ too, so its not compact in both $l^{1}$ and $l^{infty}$?
– BAYMAX
Dec 30 '18 at 21:28
I think for the case of $l^{infty}$, for a fixed integer $N$, we could take $v_{k} = k$ for $1 leq k leq N$ then $||v||_{infty} = N$ but since $N$ is arbitrary so its unbounded.
– BAYMAX
Dec 30 '18 at 21:34
@BAYMAX Is $kle 1/k$?
– David C. Ullrich
Dec 31 '18 at 16:44
Nope, I realised that, so please see my edit in the question
– BAYMAX
Dec 31 '18 at 21:28
Usually, it is better to ask a separate question. For this time I addressed a sketch of prove in the edit.
– Davide Giraudo
Dec 31 '18 at 21:56
add a comment |
The set $S$ is not bounded in $ell^1$: for a fixed integer $N$, let $v$ be the element of $S$ define as $v_k=1/k$ for $1leqslant kleqslant N$ and $v_k=0$ otherwise. Then
$leftlVert vrightrVert_1=sum_{k=1}^N1/k$ hence $sup_{vin S}leftlVert vrightrVert_1geqslant sum_{k=1}^N1/k$. As $N$ is arbitrary and the series $sum_{kgeqslant 1}1/k$ is infinite, we derive that $S$ is not bounded.
If we consider instead $ell^infty$, that is,
$$S = left{f in l^{infty} : |f(k)| leq frac{1}{k} forall k right},
$$
then $S$ is closed (since the functional $fmapsto f(k)$ is continuous for all $k$) and relatively compact: for a fixed $varepsilon$, choose $K$ such that $1/Kltvarepsilon$ and use relative compactness of $left[-1,1right]^k$.
answered Dec 30 '18 at 21:16
Davide GiraudoDavide Giraudo
125k16150261
nice solution! I was thinking how would the answer change if we consider $l^{infty}$ instead of $l^{1}$ ? I htink it is still unbounded in $l^{infty}$ too, so its not compact in both $l^{1}$ and $l^{infty}$?
– BAYMAX
Dec 30 '18 at 21:28
I think for the case of $l^{infty}$, for a fixed integer $N$, we could take $v_{k} = k$ for $1 leq k leq N$ then $||v||_{infty} = N$ but since $N$ is arbitrary so its unbounded.
– BAYMAX
Dec 30 '18 at 21:34
@BAYMAX Is $kle 1/k$?
– David C. Ullrich
Dec 31 '18 at 16:44
Nope, I realised that, so please see my edit in the question
– BAYMAX
Dec 31 '18 at 21:28
Usually, it is better to ask a separate question. For this time I addressed a sketch of prove in the edit.
– Davide Giraudo
Dec 31 '18 at 21:56
add a comment |
The set $S$ is not bounded in $ell^1$: for a fixed integer $N$, let $v$ be the element of $S$ define as $v_k=1/k$ for $1leqslant kleqslant N$ and $v_k=0$ otherwise. Then
$leftlVert vrightrVert_1=sum_{k=1}^N1/k$ hence $sup_{vin S}leftlVert vrightrVert_1geqslant sum_{k=1}^N1/k$. As $N$ is arbitrary and the series $sum_{kgeqslant 1}1/k$ is infinite, we derive that $S$ is not bounded.
If we consider instead $ell^infty$, that is,
$$S = left{f in l^{infty} : |f(k)| leq frac{1}{k} forall k right},
$$
then $S$ is closed (since the functional $fmapsto f(k)$ is continuous for all $k$) and relatively compact: for a fixed $varepsilon$, choose $K$ such that $1/Kltvarepsilon$ and use relative compactness of $left[-1,1right]^k$.
answered Dec 30 '18 at 21:16
Davide GiraudoDavide Giraudo
125k16150261
The set $S$ is not bounded in $ell^1$: for a fixed integer $N$, let $v$ be the element of $S$ define as $v_k=1/k$ for $1leqslant kleqslant N$ and $v_k=0$ otherwise. Then
$leftlVert vrightrVert_1=sum_{k=1}^N1/k$ hence $sup_{vin S}leftlVert vrightrVert_1geqslant sum_{k=1}^N1/k$. As $N$ is arbitrary and the series $sum_{kgeqslant 1}1/k$ is infinite, we derive that $S$ is not bounded.
If we consider instead $ell^infty$, that is,
$$S = left{f in l^{infty} : |f(k)| leq frac{1}{k} forall k right},
$$
then $S$ is closed (since the functional $fmapsto f(k)$ is continuous for all $k$) and relatively compact: for a fixed $varepsilon$, choose $K$ such that $1/Kltvarepsilon$ and use relative compactness of $left[-1,1right]^k$.
answered Dec 30 '18 at 21:16
Davide GiraudoDavide Giraudo
125k16150261
edited Dec 31 '18 at 21:55
answered Dec 30 '18 at 21:16
Davide GiraudoDavide Giraudo
125k16150261
answered Dec 30 '18 at 21:16
Davide GiraudoDavide Giraudo
125k16150261
answered Dec 30 '18 at 21:16
Davide GiraudoDavide Giraudo
125k16150261
125k16150261
nice solution! I was thinking how would the answer change if we consider $l^{infty}$ instead of $l^{1}$ ? I htink it is still unbounded in $l^{infty}$ too, so its not compact in both $l^{1}$ and $l^{infty}$?
– BAYMAX
Dec 30 '18 at 21:28
I think for the case of $l^{infty}$, for a fixed integer $N$, we could take $v_{k} = k$ for $1 leq k leq N$ then $||v||_{infty} = N$ but since $N$ is arbitrary so its unbounded.
– BAYMAX
Dec 30 '18 at 21:34
@BAYMAX Is $kle 1/k$?
– David C. Ullrich
Dec 31 '18 at 16:44
Nope, I realised that, so please see my edit in the question
– BAYMAX
Dec 31 '18 at 21:28
Usually, it is better to ask a separate question. For this time I addressed a sketch of prove in the edit.
– Davide Giraudo
Dec 31 '18 at 21:56
add a comment |
nice solution! I was thinking how would the answer change if we consider $l^{infty}$ instead of $l^{1}$ ? I htink it is still unbounded in $l^{infty}$ too, so its not compact in both $l^{1}$ and $l^{infty}$?
– BAYMAX
Dec 30 '18 at 21:28
I think for the case of $l^{infty}$, for a fixed integer $N$, we could take $v_{k} = k$ for $1 leq k leq N$ then $||v||_{infty} = N$ but since $N$ is arbitrary so its unbounded.
– BAYMAX
Dec 30 '18 at 21:34
@BAYMAX Is $kle 1/k$?
– David C. Ullrich
Dec 31 '18 at 16:44
Nope, I realised that, so please see my edit in the question
– BAYMAX
Dec 31 '18 at 21:28
Usually, it is better to ask a separate question. For this time I addressed a sketch of prove in the edit.
– Davide Giraudo
Dec 31 '18 at 21:56
nice solution! I was thinking how would the answer change if we consider $l^{infty}$ instead of $l^{1}$ ? I htink it is still unbounded in $l^{infty}$ too, so its not compact in both $l^{1}$ and $l^{infty}$?
– BAYMAX
Dec 30 '18 at 21:28
nice solution! I was thinking how would the answer change if we consider $l^{infty}$ instead of $l^{1}$ ? I htink it is still unbounded in $l^{infty}$ too, so its not compact in both $l^{1}$ and $l^{infty}$?
– BAYMAX
Dec 30 '18 at 21:28
I think for the case of $l^{infty}$, for a fixed integer $N$, we could take $v_{k} = k$ for $1 leq k leq N$ then $||v||_{infty} = N$ but since $N$ is arbitrary so its unbounded.
– BAYMAX
Dec 30 '18 at 21:34
I think for the case of $l^{infty}$, for a fixed integer $N$, we could take $v_{k} = k$ for $1 leq k leq N$ then $||v||_{infty} = N$ but since $N$ is arbitrary so its unbounded.
– BAYMAX
Dec 30 '18 at 21:34
@BAYMAX Is $kle 1/k$?
– David C. Ullrich
Dec 31 '18 at 16:44
@BAYMAX Is $kle 1/k$?
– David C. Ullrich
Dec 31 '18 at 16:44
Nope, I realised that, so please see my edit in the question
– BAYMAX
Dec 31 '18 at 21:28
Nope, I realised that, so please see my edit in the question
– BAYMAX
Dec 31 '18 at 21:28
Usually, it is better to ask a separate question. For this time I addressed a sketch of prove in the edit.
– Davide Giraudo
Dec 31 '18 at 21:56
Usually, it is better to ask a separate question. For this time I addressed a sketch of prove in the edit.
– Davide Giraudo
Dec 31 '18 at 21:56
add a comment |
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2
Consider every $f_N$ defined by $f_N(n)=frac1nmathbf 1_{nleqslant N}$, and proceed as it seems you wanted to, then the proof is direct.
– Did
Dec 30 '18 at 21:16
Nice!! the $1_{n leq N}$ is the characteristic function?
– BAYMAX
Dec 30 '18 at 21:27