Functions in Sobolev space












0












$begingroup$


How to prove that $x_1=signt notin W{}_2^1[-1;1] $ ?



I know that signt=$frac{d}{dt}mid tmid$ and $ mid tmid '=frac{t}{mid tmid}$, derivative of module is not defined in t=0. Am I thinking right?



Thanks for any help






functional-analysis







share|cite|improve this question











$endgroup$












  • $begingroup$
    By $W_2^1$ do you mean $W^{1,2}$?
    $endgroup$
    – Zachary Selk
    Dec 24 '18 at 22:04










  • $begingroup$
    @ZacharySelk yes
    $endgroup$
    – T. Elmira
    Dec 24 '18 at 22:09










  • $begingroup$
    What do you mean by $x_1$? Is that the name of your function depending on $t$? However, you need to prove that either your function does not live in $L^2[-1,1]$ or its weak derivative does not. Hint: Its the second case.
    $endgroup$
    – YoungMath
    Dec 24 '18 at 22:28












  • $begingroup$
    @YoungMath $x_1$ is the name of my function depending on t. Can you please explain how to prove that signt does not live in $L^2[-1;1]?
    $endgroup$
    – T. Elmira
    Dec 24 '18 at 22:33






  • 3




    $begingroup$
    Well, $text{sign}$ is indeed in $L^2[-1,1]$. But its weak derivative is not. Hint two: The weak derivative of $text{sign}$ only exists in a distributional manner. It is (twice) the famous Dirac-distribution.
    $endgroup$
    – YoungMath
    Dec 24 '18 at 22:40


















0












$begingroup$


How to prove that $x_1=signt notin W{}_2^1[-1;1] $ ?



I know that signt=$frac{d}{dt}mid tmid$ and $ mid tmid '=frac{t}{mid tmid}$, derivative of module is not defined in t=0. Am I thinking right?



Thanks for any help






functional-analysis







share|cite|improve this question











$endgroup$












  • $begingroup$
    By $W_2^1$ do you mean $W^{1,2}$?
    $endgroup$
    – Zachary Selk
    Dec 24 '18 at 22:04










  • $begingroup$
    @ZacharySelk yes
    $endgroup$
    – T. Elmira
    Dec 24 '18 at 22:09










  • $begingroup$
    What do you mean by $x_1$? Is that the name of your function depending on $t$? However, you need to prove that either your function does not live in $L^2[-1,1]$ or its weak derivative does not. Hint: Its the second case.
    $endgroup$
    – YoungMath
    Dec 24 '18 at 22:28












  • $begingroup$
    @YoungMath $x_1$ is the name of my function depending on t. Can you please explain how to prove that signt does not live in $L^2[-1;1]?
    $endgroup$
    – T. Elmira
    Dec 24 '18 at 22:33






  • 3




    $begingroup$
    Well, $text{sign}$ is indeed in $L^2[-1,1]$. But its weak derivative is not. Hint two: The weak derivative of $text{sign}$ only exists in a distributional manner. It is (twice) the famous Dirac-distribution.
    $endgroup$
    – YoungMath
    Dec 24 '18 at 22:40
















0












0








0





$begingroup$


How to prove that $x_1=signt notin W{}_2^1[-1;1] $ ?



I know that signt=$frac{d}{dt}mid tmid$ and $ mid tmid '=frac{t}{mid tmid}$, derivative of module is not defined in t=0. Am I thinking right?



Thanks for any help






functional-analysis







share|cite|improve this question











$endgroup$




How to prove that $x_1=signt notin W{}_2^1[-1;1] $ ?



I know that signt=$frac{d}{dt}mid tmid$ and $ mid tmid '=frac{t}{mid tmid}$, derivative of module is not defined in t=0. Am I thinking right?



Thanks for any help






functional-analysis




functional-analysis






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 24 '18 at 21:45







T. Elmira

















asked Dec 24 '18 at 21:38









T. ElmiraT. Elmira

174




174












  • $begingroup$
    By $W_2^1$ do you mean $W^{1,2}$?
    $endgroup$
    – Zachary Selk
    Dec 24 '18 at 22:04










  • $begingroup$
    @ZacharySelk yes
    $endgroup$
    – T. Elmira
    Dec 24 '18 at 22:09










  • $begingroup$
    What do you mean by $x_1$? Is that the name of your function depending on $t$? However, you need to prove that either your function does not live in $L^2[-1,1]$ or its weak derivative does not. Hint: Its the second case.
    $endgroup$
    – YoungMath
    Dec 24 '18 at 22:28












  • $begingroup$
    @YoungMath $x_1$ is the name of my function depending on t. Can you please explain how to prove that signt does not live in $L^2[-1;1]?
    $endgroup$
    – T. Elmira
    Dec 24 '18 at 22:33






  • 3




    $begingroup$
    Well, $text{sign}$ is indeed in $L^2[-1,1]$. But its weak derivative is not. Hint two: The weak derivative of $text{sign}$ only exists in a distributional manner. It is (twice) the famous Dirac-distribution.
    $endgroup$
    – YoungMath
    Dec 24 '18 at 22:40




















  • $begingroup$
    By $W_2^1$ do you mean $W^{1,2}$?
    $endgroup$
    – Zachary Selk
    Dec 24 '18 at 22:04










  • $begingroup$
    @ZacharySelk yes
    $endgroup$
    – T. Elmira
    Dec 24 '18 at 22:09










  • $begingroup$
    What do you mean by $x_1$? Is that the name of your function depending on $t$? However, you need to prove that either your function does not live in $L^2[-1,1]$ or its weak derivative does not. Hint: Its the second case.
    $endgroup$
    – YoungMath
    Dec 24 '18 at 22:28












  • $begingroup$
    @YoungMath $x_1$ is the name of my function depending on t. Can you please explain how to prove that signt does not live in $L^2[-1;1]?
    $endgroup$
    – T. Elmira
    Dec 24 '18 at 22:33






  • 3




    $begingroup$
    Well, $text{sign}$ is indeed in $L^2[-1,1]$. But its weak derivative is not. Hint two: The weak derivative of $text{sign}$ only exists in a distributional manner. It is (twice) the famous Dirac-distribution.
    $endgroup$
    – YoungMath
    Dec 24 '18 at 22:40


















$begingroup$
By $W_2^1$ do you mean $W^{1,2}$?
$endgroup$
– Zachary Selk
Dec 24 '18 at 22:04




$begingroup$
By $W_2^1$ do you mean $W^{1,2}$?
$endgroup$
– Zachary Selk
Dec 24 '18 at 22:04












$begingroup$
@ZacharySelk yes
$endgroup$
– T. Elmira
Dec 24 '18 at 22:09




$begingroup$
@ZacharySelk yes
$endgroup$
– T. Elmira
Dec 24 '18 at 22:09












$begingroup$
What do you mean by $x_1$? Is that the name of your function depending on $t$? However, you need to prove that either your function does not live in $L^2[-1,1]$ or its weak derivative does not. Hint: Its the second case.
$endgroup$
– YoungMath
Dec 24 '18 at 22:28






$begingroup$
What do you mean by $x_1$? Is that the name of your function depending on $t$? However, you need to prove that either your function does not live in $L^2[-1,1]$ or its weak derivative does not. Hint: Its the second case.
$endgroup$
– YoungMath
Dec 24 '18 at 22:28














$begingroup$
@YoungMath $x_1$ is the name of my function depending on t. Can you please explain how to prove that signt does not live in $L^2[-1;1]?
$endgroup$
– T. Elmira
Dec 24 '18 at 22:33




$begingroup$
@YoungMath $x_1$ is the name of my function depending on t. Can you please explain how to prove that signt does not live in $L^2[-1;1]?
$endgroup$
– T. Elmira
Dec 24 '18 at 22:33




3




3




$begingroup$
Well, $text{sign}$ is indeed in $L^2[-1,1]$. But its weak derivative is not. Hint two: The weak derivative of $text{sign}$ only exists in a distributional manner. It is (twice) the famous Dirac-distribution.
$endgroup$
– YoungMath
Dec 24 '18 at 22:40






$begingroup$
Well, $text{sign}$ is indeed in $L^2[-1,1]$. But its weak derivative is not. Hint two: The weak derivative of $text{sign}$ only exists in a distributional manner. It is (twice) the famous Dirac-distribution.
$endgroup$
– YoungMath
Dec 24 '18 at 22:40












1 Answer
1






active

oldest

votes


















1












$begingroup$

Problem. A weak derivative $v(t)$ of $sign(t)$ will satisfy, for all smooth functions $phi$ such that $phi(-1) = phi(1) = 0$,
$$int_{-1}^1sign(t)phi'(t),dt = -int_{-1}^1v(t)phi(t),dt$$ so we want to see that it can't exist.




Reason. You can evaluate $$int_{-1}^1 sign(t)phi'(t)dt = int_0^1phi'(t)dt - int_{-1}^0phi'(t)dt = -2phi(0)$$
As mentioned in the comments above this is a $delta$ distribution, $v(t) = 2delta_0$ which is not a function.




Finishing the proof. If you haven't done so already, the most straightforward way to prove that $delta$ distributions are not representable is by applying the dominated convergence theorem to a sequence $varphi_n$ of bump functions $0 leq varphi_n(t) leq 1$ whose support shrinks to a point. Someone wrote this down over here: Why Dirac's Delta is not an ordinary function?






share|cite|improve this answer











$endgroup$













  • $begingroup$
    That is not entirely true. As far as I know, $phi$ has to be a bump function, i.e. $phi$ has compact support in the interior of $[-1,1]$. Thus, $phi(-1)=phi(1)=0$.
    $endgroup$
    – YoungMath
    Dec 25 '18 at 11:25










  • $begingroup$
    @youngmath Sorry, I don’t understand your comment.
    $endgroup$
    – Ben
    Dec 25 '18 at 12:21










  • $begingroup$
    Long story short: $phi$ has to vanish on the boundary. See for instance aforementioned link to Wikipedia about weak derivatives.
    $endgroup$
    – YoungMath
    Dec 25 '18 at 12:32












  • $begingroup$
    @YoungMath Oh I see what you are saying - thanks! I'll fix that.
    $endgroup$
    – Ben
    Dec 25 '18 at 12:35










  • $begingroup$
    @YoungMath, if you shift the test function $varphi$ by a constant to $varphi + C$ it doesn't affect the left hand side but the right hand side changes by $C cdot int v,dt$, so yes what I wrote the first time was wrong.
    $endgroup$
    – Ben
    Dec 25 '18 at 12:39











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1 Answer
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oldest

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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes









1












$begingroup$

Problem. A weak derivative $v(t)$ of $sign(t)$ will satisfy, for all smooth functions $phi$ such that $phi(-1) = phi(1) = 0$,
$$int_{-1}^1sign(t)phi'(t),dt = -int_{-1}^1v(t)phi(t),dt$$ so we want to see that it can't exist.




Reason. You can evaluate $$int_{-1}^1 sign(t)phi'(t)dt = int_0^1phi'(t)dt - int_{-1}^0phi'(t)dt = -2phi(0)$$
As mentioned in the comments above this is a $delta$ distribution, $v(t) = 2delta_0$ which is not a function.




Finishing the proof. If you haven't done so already, the most straightforward way to prove that $delta$ distributions are not representable is by applying the dominated convergence theorem to a sequence $varphi_n$ of bump functions $0 leq varphi_n(t) leq 1$ whose support shrinks to a point. Someone wrote this down over here: Why Dirac's Delta is not an ordinary function?






share|cite|improve this answer











$endgroup$













  • $begingroup$
    That is not entirely true. As far as I know, $phi$ has to be a bump function, i.e. $phi$ has compact support in the interior of $[-1,1]$. Thus, $phi(-1)=phi(1)=0$.
    $endgroup$
    – YoungMath
    Dec 25 '18 at 11:25










  • $begingroup$
    @youngmath Sorry, I don’t understand your comment.
    $endgroup$
    – Ben
    Dec 25 '18 at 12:21










  • $begingroup$
    Long story short: $phi$ has to vanish on the boundary. See for instance aforementioned link to Wikipedia about weak derivatives.
    $endgroup$
    – YoungMath
    Dec 25 '18 at 12:32












  • $begingroup$
    @YoungMath Oh I see what you are saying - thanks! I'll fix that.
    $endgroup$
    – Ben
    Dec 25 '18 at 12:35










  • $begingroup$
    @YoungMath, if you shift the test function $varphi$ by a constant to $varphi + C$ it doesn't affect the left hand side but the right hand side changes by $C cdot int v,dt$, so yes what I wrote the first time was wrong.
    $endgroup$
    – Ben
    Dec 25 '18 at 12:39
















1












$begingroup$

Problem. A weak derivative $v(t)$ of $sign(t)$ will satisfy, for all smooth functions $phi$ such that $phi(-1) = phi(1) = 0$,
$$int_{-1}^1sign(t)phi'(t),dt = -int_{-1}^1v(t)phi(t),dt$$ so we want to see that it can't exist.




Reason. You can evaluate $$int_{-1}^1 sign(t)phi'(t)dt = int_0^1phi'(t)dt - int_{-1}^0phi'(t)dt = -2phi(0)$$
As mentioned in the comments above this is a $delta$ distribution, $v(t) = 2delta_0$ which is not a function.




Finishing the proof. If you haven't done so already, the most straightforward way to prove that $delta$ distributions are not representable is by applying the dominated convergence theorem to a sequence $varphi_n$ of bump functions $0 leq varphi_n(t) leq 1$ whose support shrinks to a point. Someone wrote this down over here: Why Dirac's Delta is not an ordinary function?






share|cite|improve this answer











$endgroup$













  • $begingroup$
    That is not entirely true. As far as I know, $phi$ has to be a bump function, i.e. $phi$ has compact support in the interior of $[-1,1]$. Thus, $phi(-1)=phi(1)=0$.
    $endgroup$
    – YoungMath
    Dec 25 '18 at 11:25










  • $begingroup$
    @youngmath Sorry, I don’t understand your comment.
    $endgroup$
    – Ben
    Dec 25 '18 at 12:21










  • $begingroup$
    Long story short: $phi$ has to vanish on the boundary. See for instance aforementioned link to Wikipedia about weak derivatives.
    $endgroup$
    – YoungMath
    Dec 25 '18 at 12:32












  • $begingroup$
    @YoungMath Oh I see what you are saying - thanks! I'll fix that.
    $endgroup$
    – Ben
    Dec 25 '18 at 12:35










  • $begingroup$
    @YoungMath, if you shift the test function $varphi$ by a constant to $varphi + C$ it doesn't affect the left hand side but the right hand side changes by $C cdot int v,dt$, so yes what I wrote the first time was wrong.
    $endgroup$
    – Ben
    Dec 25 '18 at 12:39














1












1








1





$begingroup$

Problem. A weak derivative $v(t)$ of $sign(t)$ will satisfy, for all smooth functions $phi$ such that $phi(-1) = phi(1) = 0$,
$$int_{-1}^1sign(t)phi'(t),dt = -int_{-1}^1v(t)phi(t),dt$$ so we want to see that it can't exist.




Reason. You can evaluate $$int_{-1}^1 sign(t)phi'(t)dt = int_0^1phi'(t)dt - int_{-1}^0phi'(t)dt = -2phi(0)$$
As mentioned in the comments above this is a $delta$ distribution, $v(t) = 2delta_0$ which is not a function.




Finishing the proof. If you haven't done so already, the most straightforward way to prove that $delta$ distributions are not representable is by applying the dominated convergence theorem to a sequence $varphi_n$ of bump functions $0 leq varphi_n(t) leq 1$ whose support shrinks to a point. Someone wrote this down over here: Why Dirac's Delta is not an ordinary function?






share|cite|improve this answer











$endgroup$



Problem. A weak derivative $v(t)$ of $sign(t)$ will satisfy, for all smooth functions $phi$ such that $phi(-1) = phi(1) = 0$,
$$int_{-1}^1sign(t)phi'(t),dt = -int_{-1}^1v(t)phi(t),dt$$ so we want to see that it can't exist.




Reason. You can evaluate $$int_{-1}^1 sign(t)phi'(t)dt = int_0^1phi'(t)dt - int_{-1}^0phi'(t)dt = -2phi(0)$$
As mentioned in the comments above this is a $delta$ distribution, $v(t) = 2delta_0$ which is not a function.




Finishing the proof. If you haven't done so already, the most straightforward way to prove that $delta$ distributions are not representable is by applying the dominated convergence theorem to a sequence $varphi_n$ of bump functions $0 leq varphi_n(t) leq 1$ whose support shrinks to a point. Someone wrote this down over here: Why Dirac's Delta is not an ordinary function?







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 25 '18 at 12:37

























answered Dec 25 '18 at 2:31









BenBen

4,048617




4,048617












  • $begingroup$
    That is not entirely true. As far as I know, $phi$ has to be a bump function, i.e. $phi$ has compact support in the interior of $[-1,1]$. Thus, $phi(-1)=phi(1)=0$.
    $endgroup$
    – YoungMath
    Dec 25 '18 at 11:25










  • $begingroup$
    @youngmath Sorry, I don’t understand your comment.
    $endgroup$
    – Ben
    Dec 25 '18 at 12:21










  • $begingroup$
    Long story short: $phi$ has to vanish on the boundary. See for instance aforementioned link to Wikipedia about weak derivatives.
    $endgroup$
    – YoungMath
    Dec 25 '18 at 12:32












  • $begingroup$
    @YoungMath Oh I see what you are saying - thanks! I'll fix that.
    $endgroup$
    – Ben
    Dec 25 '18 at 12:35










  • $begingroup$
    @YoungMath, if you shift the test function $varphi$ by a constant to $varphi + C$ it doesn't affect the left hand side but the right hand side changes by $C cdot int v,dt$, so yes what I wrote the first time was wrong.
    $endgroup$
    – Ben
    Dec 25 '18 at 12:39


















  • $begingroup$
    That is not entirely true. As far as I know, $phi$ has to be a bump function, i.e. $phi$ has compact support in the interior of $[-1,1]$. Thus, $phi(-1)=phi(1)=0$.
    $endgroup$
    – YoungMath
    Dec 25 '18 at 11:25










  • $begingroup$
    @youngmath Sorry, I don’t understand your comment.
    $endgroup$
    – Ben
    Dec 25 '18 at 12:21










  • $begingroup$
    Long story short: $phi$ has to vanish on the boundary. See for instance aforementioned link to Wikipedia about weak derivatives.
    $endgroup$
    – YoungMath
    Dec 25 '18 at 12:32












  • $begingroup$
    @YoungMath Oh I see what you are saying - thanks! I'll fix that.
    $endgroup$
    – Ben
    Dec 25 '18 at 12:35










  • $begingroup$
    @YoungMath, if you shift the test function $varphi$ by a constant to $varphi + C$ it doesn't affect the left hand side but the right hand side changes by $C cdot int v,dt$, so yes what I wrote the first time was wrong.
    $endgroup$
    – Ben
    Dec 25 '18 at 12:39
















$begingroup$
That is not entirely true. As far as I know, $phi$ has to be a bump function, i.e. $phi$ has compact support in the interior of $[-1,1]$. Thus, $phi(-1)=phi(1)=0$.
$endgroup$
– YoungMath
Dec 25 '18 at 11:25




$begingroup$
That is not entirely true. As far as I know, $phi$ has to be a bump function, i.e. $phi$ has compact support in the interior of $[-1,1]$. Thus, $phi(-1)=phi(1)=0$.
$endgroup$
– YoungMath
Dec 25 '18 at 11:25












$begingroup$
@youngmath Sorry, I don’t understand your comment.
$endgroup$
– Ben
Dec 25 '18 at 12:21




$begingroup$
@youngmath Sorry, I don’t understand your comment.
$endgroup$
– Ben
Dec 25 '18 at 12:21












$begingroup$
Long story short: $phi$ has to vanish on the boundary. See for instance aforementioned link to Wikipedia about weak derivatives.
$endgroup$
– YoungMath
Dec 25 '18 at 12:32






$begingroup$
Long story short: $phi$ has to vanish on the boundary. See for instance aforementioned link to Wikipedia about weak derivatives.
$endgroup$
– YoungMath
Dec 25 '18 at 12:32














$begingroup$
@YoungMath Oh I see what you are saying - thanks! I'll fix that.
$endgroup$
– Ben
Dec 25 '18 at 12:35




$begingroup$
@YoungMath Oh I see what you are saying - thanks! I'll fix that.
$endgroup$
– Ben
Dec 25 '18 at 12:35












$begingroup$
@YoungMath, if you shift the test function $varphi$ by a constant to $varphi + C$ it doesn't affect the left hand side but the right hand side changes by $C cdot int v,dt$, so yes what I wrote the first time was wrong.
$endgroup$
– Ben
Dec 25 '18 at 12:39




$begingroup$
@YoungMath, if you shift the test function $varphi$ by a constant to $varphi + C$ it doesn't affect the left hand side but the right hand side changes by $C cdot int v,dt$, so yes what I wrote the first time was wrong.
$endgroup$
– Ben
Dec 25 '18 at 12:39


















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Quarter-circle Tiles

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up vote 0 down vote favorite I have $99$ identical square tiles, each with a quarter-circle drawn on it like this: [asy] size(1.5cm); draw(Arc((2,0),1,90,180),red+1); draw((0,0)--(2,0)--(2,2)--(0,2)--(0,0)); [/asy] When I arrange the tiles in a $9times 11$ rectangular grid, each with a random orientation, what is the expected value of the number of full circles I form? I think this problem has to do with finding the chance any given 2x2 square has a circle, but I can't find it. expected-value share | cite | improve this question asked Nov 20 at 15:03 6minecraftninja 1 2 ...
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Mont Emei

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Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog...
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