Closed subset of $W^{1,2}([a, b], mathbb{R})$












2












$begingroup$


In the context of weak solutions of boundary value problems, I want to show that the set $${u in W^{1, 2}([a, b], mathbb{R}) ; : ; u(a) = 0 = u(b) }$$ is closed in $W^{1,2}([a, b], mathbb{R})$.



Is there any elementary way of proving it?



I might know one way by using the, due Sobolev's embedding theorem, well-defined and continuous dirac delta $delta_a$ and $delta_b$ on $W^{1,2}([a, b], mathbb{R})$.
However, I want to refrain from distribution theory.



Hints are appreciated.






functional-analysis sobolev-spaces self-learning







share|cite|improve this question









$endgroup$












  • $begingroup$
    How do you define $u(a)$ and $u(b)$ to begin with? I don't think you can avoid using the Sobolev embedding theorem.
    $endgroup$
    – MaoWao
    Dec 6 '18 at 19:04


















2












$begingroup$


In the context of weak solutions of boundary value problems, I want to show that the set $${u in W^{1, 2}([a, b], mathbb{R}) ; : ; u(a) = 0 = u(b) }$$ is closed in $W^{1,2}([a, b], mathbb{R})$.



Is there any elementary way of proving it?



I might know one way by using the, due Sobolev's embedding theorem, well-defined and continuous dirac delta $delta_a$ and $delta_b$ on $W^{1,2}([a, b], mathbb{R})$.
However, I want to refrain from distribution theory.



Hints are appreciated.






functional-analysis sobolev-spaces self-learning







share|cite|improve this question









$endgroup$












  • $begingroup$
    How do you define $u(a)$ and $u(b)$ to begin with? I don't think you can avoid using the Sobolev embedding theorem.
    $endgroup$
    – MaoWao
    Dec 6 '18 at 19:04
















2












2








2





$begingroup$


In the context of weak solutions of boundary value problems, I want to show that the set $${u in W^{1, 2}([a, b], mathbb{R}) ; : ; u(a) = 0 = u(b) }$$ is closed in $W^{1,2}([a, b], mathbb{R})$.



Is there any elementary way of proving it?



I might know one way by using the, due Sobolev's embedding theorem, well-defined and continuous dirac delta $delta_a$ and $delta_b$ on $W^{1,2}([a, b], mathbb{R})$.
However, I want to refrain from distribution theory.



Hints are appreciated.






functional-analysis sobolev-spaces self-learning







share|cite|improve this question









$endgroup$




In the context of weak solutions of boundary value problems, I want to show that the set $${u in W^{1, 2}([a, b], mathbb{R}) ; : ; u(a) = 0 = u(b) }$$ is closed in $W^{1,2}([a, b], mathbb{R})$.



Is there any elementary way of proving it?



I might know one way by using the, due Sobolev's embedding theorem, well-defined and continuous dirac delta $delta_a$ and $delta_b$ on $W^{1,2}([a, b], mathbb{R})$.
However, I want to refrain from distribution theory.



Hints are appreciated.






functional-analysis sobolev-spaces self-learning




functional-analysis sobolev-spaces self-learning






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 6 '18 at 17:27









TaufiTaufi

570411




570411












  • $begingroup$
    How do you define $u(a)$ and $u(b)$ to begin with? I don't think you can avoid using the Sobolev embedding theorem.
    $endgroup$
    – MaoWao
    Dec 6 '18 at 19:04




















  • $begingroup$
    How do you define $u(a)$ and $u(b)$ to begin with? I don't think you can avoid using the Sobolev embedding theorem.
    $endgroup$
    – MaoWao
    Dec 6 '18 at 19:04


















$begingroup$
How do you define $u(a)$ and $u(b)$ to begin with? I don't think you can avoid using the Sobolev embedding theorem.
$endgroup$
– MaoWao
Dec 6 '18 at 19:04






$begingroup$
How do you define $u(a)$ and $u(b)$ to begin with? I don't think you can avoid using the Sobolev embedding theorem.
$endgroup$
– MaoWao
Dec 6 '18 at 19:04












1 Answer
1






active

oldest

votes


















1












$begingroup$

Usually Sobolev spaces are defined in open sets, so it should be $W^{1,2}((a,b),mathbb{R})$. Anyway, if $u$ is a smooth function in $W^{1,2}((a,b),mathbb{R})$, by the fundamental theorem of calculus,
$$u(x)-u(y)=int_x^y u'(s),ds.$$
Using Holder's inequality you have that
$$|u(x)-u(y)|le |x-y|^{1/2}left( int_x^y |u'(s)|^2,dsright)^{1/2}$$
for all $a<x<y<b$. If now $u$ is a function in $W^{1,2}((a,b),mathbb{R})$, using density of smooth functions you can find a sequence of smooth functions $u_n$ that converge to $u$ in $W^{1,2}((a,b),mathbb{R})$. By taking a subsequence you can find a subsequence which converges to $u$ a.e. in $(a,b)$. Since $$|u_n(x)-u_n(y)|le |x-y|^{1/2}left( int_x^y |u'_n(s)|^2,dsright)^{1/2},$$ letting $ntoinfty$ you get
that
$$|u(x)-u(y)|le |x-y|^{1/2}left( int_x^y |u'(s)|^2,dsright)^{1/2}$$ for a.e. $x,yin (a,b)$. This is telling you that $u$ has a representative $bar u$ which is Holder continuous. In particular, $bar u$ can be extended in $[a,b]$. So it makes now sense to talk about $bar u(a)$ and $bar u(b)$ (to answer MaoWao's question).
Now you can use the Ascoli-Arzela theorem to prove that your set is closed. I am skipping the details.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    You mean $y$, not $x$, in the LHS?
    $endgroup$
    – Taufi
    Dec 7 '18 at 12:28










  • $begingroup$
    Moreover, trying fill in the details you skipped, I need to show that the elements of the subset are equicontinuous and uniformly bounded. I would then conclude with Arzela-Ascoli that the set is compact and hence closed?
    $endgroup$
    – Taufi
    Dec 7 '18 at 12:34












  • $begingroup$
    Yes, I corrected the misprints. Exactly, you need to show equi-continuity and uniform boundedness.
    $endgroup$
    – Gio67
    Dec 7 '18 at 12:50










  • $begingroup$
    @Gio67 Of course you just proved the Sobolev embedding theorem in the one-dimensional case, so it's not essentially different from the approach suggested in the op (as I said, I have a hard time imagining one).
    $endgroup$
    – MaoWao
    Dec 7 '18 at 13:05










  • $begingroup$
    Yeah, I don't think there is any other alternative
    $endgroup$
    – Gio67
    Dec 7 '18 at 17:14











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1 Answer
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1 Answer
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1












$begingroup$

Usually Sobolev spaces are defined in open sets, so it should be $W^{1,2}((a,b),mathbb{R})$. Anyway, if $u$ is a smooth function in $W^{1,2}((a,b),mathbb{R})$, by the fundamental theorem of calculus,
$$u(x)-u(y)=int_x^y u'(s),ds.$$
Using Holder's inequality you have that
$$|u(x)-u(y)|le |x-y|^{1/2}left( int_x^y |u'(s)|^2,dsright)^{1/2}$$
for all $a<x<y<b$. If now $u$ is a function in $W^{1,2}((a,b),mathbb{R})$, using density of smooth functions you can find a sequence of smooth functions $u_n$ that converge to $u$ in $W^{1,2}((a,b),mathbb{R})$. By taking a subsequence you can find a subsequence which converges to $u$ a.e. in $(a,b)$. Since $$|u_n(x)-u_n(y)|le |x-y|^{1/2}left( int_x^y |u'_n(s)|^2,dsright)^{1/2},$$ letting $ntoinfty$ you get
that
$$|u(x)-u(y)|le |x-y|^{1/2}left( int_x^y |u'(s)|^2,dsright)^{1/2}$$ for a.e. $x,yin (a,b)$. This is telling you that $u$ has a representative $bar u$ which is Holder continuous. In particular, $bar u$ can be extended in $[a,b]$. So it makes now sense to talk about $bar u(a)$ and $bar u(b)$ (to answer MaoWao's question).
Now you can use the Ascoli-Arzela theorem to prove that your set is closed. I am skipping the details.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    You mean $y$, not $x$, in the LHS?
    $endgroup$
    – Taufi
    Dec 7 '18 at 12:28










  • $begingroup$
    Moreover, trying fill in the details you skipped, I need to show that the elements of the subset are equicontinuous and uniformly bounded. I would then conclude with Arzela-Ascoli that the set is compact and hence closed?
    $endgroup$
    – Taufi
    Dec 7 '18 at 12:34












  • $begingroup$
    Yes, I corrected the misprints. Exactly, you need to show equi-continuity and uniform boundedness.
    $endgroup$
    – Gio67
    Dec 7 '18 at 12:50










  • $begingroup$
    @Gio67 Of course you just proved the Sobolev embedding theorem in the one-dimensional case, so it's not essentially different from the approach suggested in the op (as I said, I have a hard time imagining one).
    $endgroup$
    – MaoWao
    Dec 7 '18 at 13:05










  • $begingroup$
    Yeah, I don't think there is any other alternative
    $endgroup$
    – Gio67
    Dec 7 '18 at 17:14
















1












$begingroup$

Usually Sobolev spaces are defined in open sets, so it should be $W^{1,2}((a,b),mathbb{R})$. Anyway, if $u$ is a smooth function in $W^{1,2}((a,b),mathbb{R})$, by the fundamental theorem of calculus,
$$u(x)-u(y)=int_x^y u'(s),ds.$$
Using Holder's inequality you have that
$$|u(x)-u(y)|le |x-y|^{1/2}left( int_x^y |u'(s)|^2,dsright)^{1/2}$$
for all $a<x<y<b$. If now $u$ is a function in $W^{1,2}((a,b),mathbb{R})$, using density of smooth functions you can find a sequence of smooth functions $u_n$ that converge to $u$ in $W^{1,2}((a,b),mathbb{R})$. By taking a subsequence you can find a subsequence which converges to $u$ a.e. in $(a,b)$. Since $$|u_n(x)-u_n(y)|le |x-y|^{1/2}left( int_x^y |u'_n(s)|^2,dsright)^{1/2},$$ letting $ntoinfty$ you get
that
$$|u(x)-u(y)|le |x-y|^{1/2}left( int_x^y |u'(s)|^2,dsright)^{1/2}$$ for a.e. $x,yin (a,b)$. This is telling you that $u$ has a representative $bar u$ which is Holder continuous. In particular, $bar u$ can be extended in $[a,b]$. So it makes now sense to talk about $bar u(a)$ and $bar u(b)$ (to answer MaoWao's question).
Now you can use the Ascoli-Arzela theorem to prove that your set is closed. I am skipping the details.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    You mean $y$, not $x$, in the LHS?
    $endgroup$
    – Taufi
    Dec 7 '18 at 12:28










  • $begingroup$
    Moreover, trying fill in the details you skipped, I need to show that the elements of the subset are equicontinuous and uniformly bounded. I would then conclude with Arzela-Ascoli that the set is compact and hence closed?
    $endgroup$
    – Taufi
    Dec 7 '18 at 12:34












  • $begingroup$
    Yes, I corrected the misprints. Exactly, you need to show equi-continuity and uniform boundedness.
    $endgroup$
    – Gio67
    Dec 7 '18 at 12:50










  • $begingroup$
    @Gio67 Of course you just proved the Sobolev embedding theorem in the one-dimensional case, so it's not essentially different from the approach suggested in the op (as I said, I have a hard time imagining one).
    $endgroup$
    – MaoWao
    Dec 7 '18 at 13:05










  • $begingroup$
    Yeah, I don't think there is any other alternative
    $endgroup$
    – Gio67
    Dec 7 '18 at 17:14














1












1








1





$begingroup$

Usually Sobolev spaces are defined in open sets, so it should be $W^{1,2}((a,b),mathbb{R})$. Anyway, if $u$ is a smooth function in $W^{1,2}((a,b),mathbb{R})$, by the fundamental theorem of calculus,
$$u(x)-u(y)=int_x^y u'(s),ds.$$
Using Holder's inequality you have that
$$|u(x)-u(y)|le |x-y|^{1/2}left( int_x^y |u'(s)|^2,dsright)^{1/2}$$
for all $a<x<y<b$. If now $u$ is a function in $W^{1,2}((a,b),mathbb{R})$, using density of smooth functions you can find a sequence of smooth functions $u_n$ that converge to $u$ in $W^{1,2}((a,b),mathbb{R})$. By taking a subsequence you can find a subsequence which converges to $u$ a.e. in $(a,b)$. Since $$|u_n(x)-u_n(y)|le |x-y|^{1/2}left( int_x^y |u'_n(s)|^2,dsright)^{1/2},$$ letting $ntoinfty$ you get
that
$$|u(x)-u(y)|le |x-y|^{1/2}left( int_x^y |u'(s)|^2,dsright)^{1/2}$$ for a.e. $x,yin (a,b)$. This is telling you that $u$ has a representative $bar u$ which is Holder continuous. In particular, $bar u$ can be extended in $[a,b]$. So it makes now sense to talk about $bar u(a)$ and $bar u(b)$ (to answer MaoWao's question).
Now you can use the Ascoli-Arzela theorem to prove that your set is closed. I am skipping the details.






share|cite|improve this answer











$endgroup$



Usually Sobolev spaces are defined in open sets, so it should be $W^{1,2}((a,b),mathbb{R})$. Anyway, if $u$ is a smooth function in $W^{1,2}((a,b),mathbb{R})$, by the fundamental theorem of calculus,
$$u(x)-u(y)=int_x^y u'(s),ds.$$
Using Holder's inequality you have that
$$|u(x)-u(y)|le |x-y|^{1/2}left( int_x^y |u'(s)|^2,dsright)^{1/2}$$
for all $a<x<y<b$. If now $u$ is a function in $W^{1,2}((a,b),mathbb{R})$, using density of smooth functions you can find a sequence of smooth functions $u_n$ that converge to $u$ in $W^{1,2}((a,b),mathbb{R})$. By taking a subsequence you can find a subsequence which converges to $u$ a.e. in $(a,b)$. Since $$|u_n(x)-u_n(y)|le |x-y|^{1/2}left( int_x^y |u'_n(s)|^2,dsright)^{1/2},$$ letting $ntoinfty$ you get
that
$$|u(x)-u(y)|le |x-y|^{1/2}left( int_x^y |u'(s)|^2,dsright)^{1/2}$$ for a.e. $x,yin (a,b)$. This is telling you that $u$ has a representative $bar u$ which is Holder continuous. In particular, $bar u$ can be extended in $[a,b]$. So it makes now sense to talk about $bar u(a)$ and $bar u(b)$ (to answer MaoWao's question).
Now you can use the Ascoli-Arzela theorem to prove that your set is closed. I am skipping the details.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 7 '18 at 12:49

























answered Dec 7 '18 at 3:13









Gio67Gio67

12.5k1626




12.5k1626












  • $begingroup$
    You mean $y$, not $x$, in the LHS?
    $endgroup$
    – Taufi
    Dec 7 '18 at 12:28










  • $begingroup$
    Moreover, trying fill in the details you skipped, I need to show that the elements of the subset are equicontinuous and uniformly bounded. I would then conclude with Arzela-Ascoli that the set is compact and hence closed?
    $endgroup$
    – Taufi
    Dec 7 '18 at 12:34












  • $begingroup$
    Yes, I corrected the misprints. Exactly, you need to show equi-continuity and uniform boundedness.
    $endgroup$
    – Gio67
    Dec 7 '18 at 12:50










  • $begingroup$
    @Gio67 Of course you just proved the Sobolev embedding theorem in the one-dimensional case, so it's not essentially different from the approach suggested in the op (as I said, I have a hard time imagining one).
    $endgroup$
    – MaoWao
    Dec 7 '18 at 13:05










  • $begingroup$
    Yeah, I don't think there is any other alternative
    $endgroup$
    – Gio67
    Dec 7 '18 at 17:14


















  • $begingroup$
    You mean $y$, not $x$, in the LHS?
    $endgroup$
    – Taufi
    Dec 7 '18 at 12:28










  • $begingroup$
    Moreover, trying fill in the details you skipped, I need to show that the elements of the subset are equicontinuous and uniformly bounded. I would then conclude with Arzela-Ascoli that the set is compact and hence closed?
    $endgroup$
    – Taufi
    Dec 7 '18 at 12:34












  • $begingroup$
    Yes, I corrected the misprints. Exactly, you need to show equi-continuity and uniform boundedness.
    $endgroup$
    – Gio67
    Dec 7 '18 at 12:50










  • $begingroup$
    @Gio67 Of course you just proved the Sobolev embedding theorem in the one-dimensional case, so it's not essentially different from the approach suggested in the op (as I said, I have a hard time imagining one).
    $endgroup$
    – MaoWao
    Dec 7 '18 at 13:05










  • $begingroup$
    Yeah, I don't think there is any other alternative
    $endgroup$
    – Gio67
    Dec 7 '18 at 17:14
















$begingroup$
You mean $y$, not $x$, in the LHS?
$endgroup$
– Taufi
Dec 7 '18 at 12:28




$begingroup$
You mean $y$, not $x$, in the LHS?
$endgroup$
– Taufi
Dec 7 '18 at 12:28












$begingroup$
Moreover, trying fill in the details you skipped, I need to show that the elements of the subset are equicontinuous and uniformly bounded. I would then conclude with Arzela-Ascoli that the set is compact and hence closed?
$endgroup$
– Taufi
Dec 7 '18 at 12:34






$begingroup$
Moreover, trying fill in the details you skipped, I need to show that the elements of the subset are equicontinuous and uniformly bounded. I would then conclude with Arzela-Ascoli that the set is compact and hence closed?
$endgroup$
– Taufi
Dec 7 '18 at 12:34














$begingroup$
Yes, I corrected the misprints. Exactly, you need to show equi-continuity and uniform boundedness.
$endgroup$
– Gio67
Dec 7 '18 at 12:50




$begingroup$
Yes, I corrected the misprints. Exactly, you need to show equi-continuity and uniform boundedness.
$endgroup$
– Gio67
Dec 7 '18 at 12:50












$begingroup$
@Gio67 Of course you just proved the Sobolev embedding theorem in the one-dimensional case, so it's not essentially different from the approach suggested in the op (as I said, I have a hard time imagining one).
$endgroup$
– MaoWao
Dec 7 '18 at 13:05




$begingroup$
@Gio67 Of course you just proved the Sobolev embedding theorem in the one-dimensional case, so it's not essentially different from the approach suggested in the op (as I said, I have a hard time imagining one).
$endgroup$
– MaoWao
Dec 7 '18 at 13:05












$begingroup$
Yeah, I don't think there is any other alternative
$endgroup$
– Gio67
Dec 7 '18 at 17:14




$begingroup$
Yeah, I don't think there is any other alternative
$endgroup$
– Gio67
Dec 7 '18 at 17:14


















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