Is the set of sequences $(x_n)$ in $ell^1$ such that $|x_n|leqslantfrac1n$ for every $n$, compact in $ell^1$?












2












$begingroup$


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







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    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.
    $endgroup$
    – Did
    Dec 30 '18 at 21:16












  • $begingroup$
    Nice!! the $1_{n leq N}$ is the characteristic function?
    $endgroup$
    – BAYMAX
    Dec 30 '18 at 21:27
















2












$begingroup$


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







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    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.
    $endgroup$
    – Did
    Dec 30 '18 at 21:16












  • $begingroup$
    Nice!! the $1_{n leq N}$ is the characteristic function?
    $endgroup$
    – BAYMAX
    Dec 30 '18 at 21:27














2












2








2


1



$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 30 '18 at 22:16







BAYMAX

















asked Dec 30 '18 at 21:10









BAYMAXBAYMAX

3,04421326




3,04421326








  • 2




    $begingroup$
    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.
    $endgroup$
    – Did
    Dec 30 '18 at 21:16












  • $begingroup$
    Nice!! the $1_{n leq N}$ is the characteristic function?
    $endgroup$
    – BAYMAX
    Dec 30 '18 at 21:27














  • 2




    $begingroup$
    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.
    $endgroup$
    – Did
    Dec 30 '18 at 21:16












  • $begingroup$
    Nice!! the $1_{n leq N}$ is the characteristic function?
    $endgroup$
    – BAYMAX
    Dec 30 '18 at 21:27








2




2




$begingroup$
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.
$endgroup$
– Did
Dec 30 '18 at 21:16






$begingroup$
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.
$endgroup$
– Did
Dec 30 '18 at 21:16














$begingroup$
Nice!! the $1_{n leq N}$ is the characteristic function?
$endgroup$
– BAYMAX
Dec 30 '18 at 21:27




$begingroup$
Nice!! the $1_{n leq N}$ is the characteristic function?
$endgroup$
– BAYMAX
Dec 30 '18 at 21:27










1 Answer
1






active

oldest

votes


















3












$begingroup$

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 $left[-1,1right]^k$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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 $l^{infty}$?
    $endgroup$
    – BAYMAX
    Dec 30 '18 at 21:28












  • $begingroup$
    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.
    $endgroup$
    – BAYMAX
    Dec 30 '18 at 21:34










  • $begingroup$
    @BAYMAX Is $kle 1/k$?
    $endgroup$
    – David C. Ullrich
    Dec 31 '18 at 16:44










  • $begingroup$
    Nope, I realised that, so please see my edit in the question
    $endgroup$
    – BAYMAX
    Dec 31 '18 at 21:28










  • $begingroup$
    Usually, it is better to ask a separate question. For this time I addressed a sketch of prove in the edit.
    $endgroup$
    – Davide Giraudo
    Dec 31 '18 at 21:56












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1 Answer
1






active

oldest

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









3












$begingroup$

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 $left[-1,1right]^k$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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 $l^{infty}$?
    $endgroup$
    – BAYMAX
    Dec 30 '18 at 21:28












  • $begingroup$
    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.
    $endgroup$
    – BAYMAX
    Dec 30 '18 at 21:34










  • $begingroup$
    @BAYMAX Is $kle 1/k$?
    $endgroup$
    – David C. Ullrich
    Dec 31 '18 at 16:44










  • $begingroup$
    Nope, I realised that, so please see my edit in the question
    $endgroup$
    – BAYMAX
    Dec 31 '18 at 21:28










  • $begingroup$
    Usually, it is better to ask a separate question. For this time I addressed a sketch of prove in the edit.
    $endgroup$
    – Davide Giraudo
    Dec 31 '18 at 21:56
















3












$begingroup$

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 $left[-1,1right]^k$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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 $l^{infty}$?
    $endgroup$
    – BAYMAX
    Dec 30 '18 at 21:28












  • $begingroup$
    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.
    $endgroup$
    – BAYMAX
    Dec 30 '18 at 21:34










  • $begingroup$
    @BAYMAX Is $kle 1/k$?
    $endgroup$
    – David C. Ullrich
    Dec 31 '18 at 16:44










  • $begingroup$
    Nope, I realised that, so please see my edit in the question
    $endgroup$
    – BAYMAX
    Dec 31 '18 at 21:28










  • $begingroup$
    Usually, it is better to ask a separate question. For this time I addressed a sketch of prove in the edit.
    $endgroup$
    – Davide Giraudo
    Dec 31 '18 at 21:56














3












3








3





$begingroup$

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 $left[-1,1right]^k$.






share|cite|improve this answer











$endgroup$



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 $left[-1,1right]^k$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 31 '18 at 21:55

























answered Dec 30 '18 at 21:16









Davide GiraudoDavide Giraudo

128k17157269




128k17157269












  • $begingroup$
    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 $l^{infty}$?
    $endgroup$
    – BAYMAX
    Dec 30 '18 at 21:28












  • $begingroup$
    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.
    $endgroup$
    – BAYMAX
    Dec 30 '18 at 21:34










  • $begingroup$
    @BAYMAX Is $kle 1/k$?
    $endgroup$
    – David C. Ullrich
    Dec 31 '18 at 16:44










  • $begingroup$
    Nope, I realised that, so please see my edit in the question
    $endgroup$
    – BAYMAX
    Dec 31 '18 at 21:28










  • $begingroup$
    Usually, it is better to ask a separate question. For this time I addressed a sketch of prove in the edit.
    $endgroup$
    – Davide Giraudo
    Dec 31 '18 at 21:56


















  • $begingroup$
    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 $l^{infty}$?
    $endgroup$
    – BAYMAX
    Dec 30 '18 at 21:28












  • $begingroup$
    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.
    $endgroup$
    – BAYMAX
    Dec 30 '18 at 21:34










  • $begingroup$
    @BAYMAX Is $kle 1/k$?
    $endgroup$
    – David C. Ullrich
    Dec 31 '18 at 16:44










  • $begingroup$
    Nope, I realised that, so please see my edit in the question
    $endgroup$
    – BAYMAX
    Dec 31 '18 at 21:28










  • $begingroup$
    Usually, it is better to ask a separate question. For this time I addressed a sketch of prove in the edit.
    $endgroup$
    – Davide Giraudo
    Dec 31 '18 at 21:56
















$begingroup$
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 $l^{infty}$?
$endgroup$
– BAYMAX
Dec 30 '18 at 21:28






$begingroup$
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 $l^{infty}$?
$endgroup$
– BAYMAX
Dec 30 '18 at 21:28














$begingroup$
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.
$endgroup$
– BAYMAX
Dec 30 '18 at 21:34




$begingroup$
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.
$endgroup$
– BAYMAX
Dec 30 '18 at 21:34












$begingroup$
@BAYMAX Is $kle 1/k$?
$endgroup$
– David C. Ullrich
Dec 31 '18 at 16:44




$begingroup$
@BAYMAX Is $kle 1/k$?
$endgroup$
– David C. Ullrich
Dec 31 '18 at 16:44












$begingroup$
Nope, I realised that, so please see my edit in the question
$endgroup$
– BAYMAX
Dec 31 '18 at 21:28




$begingroup$
Nope, I realised that, so please see my edit in the question
$endgroup$
– BAYMAX
Dec 31 '18 at 21:28












$begingroup$
Usually, it is better to ask a separate question. For this time I addressed a sketch of prove in the edit.
$endgroup$
– Davide Giraudo
Dec 31 '18 at 21:56




$begingroup$
Usually, it is better to ask a separate question. For this time I addressed a sketch of prove in the edit.
$endgroup$
– Davide Giraudo
Dec 31 '18 at 21:56


















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Ida-Boy-Ed-Garten

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Die Lage des Ida-Boy-Ed-Gartens, rot markiert Der Ida-Boy-Ed-Garten Der Ida-Boy-Ed-Garten, Blick von der Wakenitzmauer durch das Tor in der Stadtmauer Der Ida-Boy-Ed-Garten ist eine Straße der Lübecker Altstadt. Inhaltsverzeichnis 1 Lage 2 Geschichte 3 Literatur 4 Weblinks Lage | Der etwa 90 Meter lange Ida-Boy-Ed-Garten befindet sich an der nördlichen Spitze der Altstadtinsel, vor dem Burgtor. Die in einem Bogen durch eine Grünanlage hangaufwärts führende Straße verbindet die Wakenitzmauer mit der Burgtorbrücke. Geschichte | Der heutige Ida-Boy-Ed-Garten wurde ursprünglich als Verlängerung der Wakenitzmauer angelegt. 1903 wurde in den erhaltenen Teil der mittelalterlichen Stadtmauer, an dem die Wakenitzmauer bei der Einmündung der Kaiserstraße begann, ein Tor mit einem großen Bogen für die Durchführung der Straße und zwei kleineren seitlichen Öffnungen für Fußgänger gebrochen, um für die neu errichteten Wohnhäuser in diesem Bereich
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