An inequality of Sobolev type
Consider the space $C_c^1(RR)$ of all the compactly-supported functions of class $C^1$ defined in $RR$: it is a real vector space and the application $u\mapsto ||u||_oo=max_(RR) |u|$ is a norm in $C_c^1(RR)$ (but $(C_c^1(RR), ||\cdot ||_oo)$ is not a Banach space).
For all $u \in C_c^1(RR)$, then $u' \in C_c(RR) \subseteq L^1(RR)$ and it is possible to evaluate $||u'||_1=\int_(-oo)^(+oo)|u'|" d"x$; the application $u \mapsto ||u'||_1$ is a norm on $C_c^1(RR)$.
***
Problem:
Show that the following inequality:
(*) $AA u in C_c^1(RR), \quad c*||u||_oo<=||u'||_1$
holds for some constant $c>0$.
Find the best constant in (*), put (*) in optimal form and show that equality holds at least for non-negative functions which are symmetric-decreasing about some point in $RR$*.
Can (*) be extended beyond $C_c^1(RR)$, i.e. to a space strictly containing $C_c^1(RR)$?
__________
* We say that a function $u$ is symmetric-decreasing about a point $x_0\in RR$ iff it satisfies the following conditions:
i) $AA xi \in RR, u(x_0+xi)=u(x_0-xi)$ (i.e. $u$ is even about $x_0$);
ii) $u$ is increasing in $]-oo,x_0]$ and decreasing in $[x_0,+oo[$;
or equivalently iff:
j) exists a decreasing function $v:[0,+oo[ \to RR$ s.t. $AA x \in RR, u(x)=v(|x-x_0|)$.
For example, the function $u(x)=1-(x-3)^2$ is symmetric-decreasing about $x_0=3$.
For all $u \in C_c^1(RR)$, then $u' \in C_c(RR) \subseteq L^1(RR)$ and it is possible to evaluate $||u'||_1=\int_(-oo)^(+oo)|u'|" d"x$; the application $u \mapsto ||u'||_1$ is a norm on $C_c^1(RR)$.
***
Problem:
Show that the following inequality:
(*) $AA u in C_c^1(RR), \quad c*||u||_oo<=||u'||_1$
holds for some constant $c>0$.
Find the best constant in (*), put (*) in optimal form and show that equality holds at least for non-negative functions which are symmetric-decreasing about some point in $RR$*.
Can (*) be extended beyond $C_c^1(RR)$, i.e. to a space strictly containing $C_c^1(RR)$?
__________
* We say that a function $u$ is symmetric-decreasing about a point $x_0\in RR$ iff it satisfies the following conditions:
i) $AA xi \in RR, u(x_0+xi)=u(x_0-xi)$ (i.e. $u$ is even about $x_0$);
ii) $u$ is increasing in $]-oo,x_0]$ and decreasing in $[x_0,+oo[$;
or equivalently iff:
j) exists a decreasing function $v:[0,+oo[ \to RR$ s.t. $AA x \in RR, u(x)=v(|x-x_0|)$.
For example, the function $u(x)=1-(x-3)^2$ is symmetric-decreasing about $x_0=3$.
Risposte
Your try (the one with $W=C_c^1(RR)+"span" \{ f\}$ and $f \in C_0^oo(RR)$ fast-decaying with all its derivatives at infinity) was a nice one. 
Perhaps it wasn't what I expected, mostly because spaces like $W$ aren't commonly used in the applications of Sobolev inequalities... In fact inequalities of Sobolev type, i.e. inequalities like:
$||u||_q <=c*||u'||_p$ (or $||u||_q <=c*||\nabla u||_p$),
are used to embed Sobolev spaces $W^(1,p)$ into $L^q$ with $q>p$, so that one can "gain" summability on $u$ assuming that $u$ is weakly differentiable and that it's derivative is in $L^p$.
Perhaps it wasn't what I expected, mostly because spaces like $W$ aren't commonly used in the applications of Sobolev inequalities... In fact inequalities of Sobolev type, i.e. inequalities like:
$||u||_q <=c*||u'||_p$ (or $||u||_q <=c*||\nabla u||_p$),
are used to embed Sobolev spaces $W^(1,p)$ into $L^q$ with $q>p$, so that one can "gain" summability on $u$ assuming that $u$ is weakly differentiable and that it's derivative is in $L^p$.
"Gugo82":
since $C_c^1 \subset W^(1,1)$, (*) can actually be extended to a space strictly containing $C_c^1$.
This completes the answer to question #2.
hope not to be too polemic (I'm tired because it's all day that I study for exams...) ... but if this was the question I had answered with my $W$, which u didn't consider....
(didn't show that the inequality works, but it should be an easy (straightforword) job, more or less involving the properties of approximation u used but in a more concrete context)...
boh ok... gugo82 is nice and instructive to read so I'll pass over if he doesn't accept my solutions!
Ok...
First, yes it was a typo; on the second line a $C_0$ (and not a $C_c$) is needed.
Second, the fact that each $u \in W^(1,1)$ is continuous and vanishes at infinity implies that (*) has a meaning also if $u \in W^(1,1)$; since $C_c^1 \subset W^(1,1)$, (*) can actually be extended to a space strictly containing $C_c^1$.
This completes the answer to question #2.
Third, the embedding $W^(1,1)\subseteq C_0$ is continuous because of (*): infact if $u,v in W^(1,1)$ then $u, v in C_0$ and:
$||u-v||_oo<= 1/2 ||u'-v'||_1<=1/2 ||u-v||_(1,1)$
so that the embedding $u \in W^(1,1) \mapsto u \in C_0$ is Lipschitz with constant $c=1/2$.
First, yes it was a typo; on the second line a $C_0$ (and not a $C_c$) is needed.
Second, the fact that each $u \in W^(1,1)$ is continuous and vanishes at infinity implies that (*) has a meaning also if $u \in W^(1,1)$; since $C_c^1 \subset W^(1,1)$, (*) can actually be extended to a space strictly containing $C_c^1$.
This completes the answer to question #2.
Third, the embedding $W^(1,1)\subseteq C_0$ is continuous because of (*): infact if $u,v in W^(1,1)$ then $u, v in C_0$ and:
$||u-v||_oo<= 1/2 ||u'-v'||_1<=1/2 ||u-v||_(1,1)$
so that the embedding $u \in W^(1,1) \mapsto u \in C_0$ is Lipschitz with constant $c=1/2$.
"Gugo82":
$\{(u_n\to u, " in " L^1(RR)),(u_n \to v, " in C_c(RR)} \quad$,
I read it quickly... just here $C_0(RR)$ seems to follow your argument right?...
- so the question was: use what u've done above to prove that $W^(1,1)(RR)$ embeds continuously in $C_0(R)$.. from the first post "Can (*) be extended beyond $C_c(RR)$, i.e. to a space strictly containing $C_c(RR)$?" I was supposed to imagine that?
anyway didn't know which space was the completion of which other so I couldn't solve this problem
... by the way, a nice one! especially the last result makes the exercise interesting for me (I learnt something):!: )
Ok, I'm going to explain better the question with the answer... Maybe it's a little bit technical, I don't konw.
The function $u\mapsto ||u||_(1,1):=||u||_1+||u'||_1$ is a norm (of Sobolev type) on $C_c^1(RR)$ and, if we equip $C_c^1(RR)$ with this norm, the normed space is not complete.
But $C_c^1(RR)$ has a completion w.r.t. $||\cdot ||_(1,1)$: this completion turns out to be the Sobolev space $W^(1,1)(RR)$, i.e. the space of weakly differentiable functions with weak derivative in $L^1(RR)$.
Take $u in W^(1,1)(RR)$ and $(u_n) \subseteq C_c^1(RR)$ s.t. $u_n to u$ in $W^(1,1)(RR)$ (i.e. $||u_n-u||_(1,1)\to 0$); since $W^(1,1)(RR)$ is $||\cdot||_(1,1)$-complete, $(u_n)$ is a Cauchy sequence and the trivial inequality:
$||u_n'-u_m'||_1<=||u_n-u_m||_(1,1)$
implies that $(u_n')$ is a Cauchy sequence in $L^1(RR)$ (which obviously converges to $u'$).
Now (*) comes into play: infact, since $(u_n')$ is $||\cdot ||_1$-Cauchy's, by (*) $(u_n)$ is $||||_oo$-Cauchy's, i.e. $(u_n)$ converges uniformly in $RR$ to a continuous function; let $v$ be the uniform limit of $(u_n)$: this function trivially belongs to the $||\cdot ||_oo$-completion of $C_c$, which is the space $C_0(RR)$ of the functions which vanish at infinity.
On the other hand, $(u_n)$ is a Cauchy sequence in $L^1(RR)$, for:
$||u_n-u_m||_1<=||u_n-u_m||_(1,1) \quad$,
and $u_n\to u$ in $L^1(RR)$.
Summarizing, we have:
$\{(u_n\to u, " in " L^1(RR)),(u_n \to v, " in "C_0(RR)) :} \quad$,
therefore $u=v$ a.e. in $RR$ for known results of Measure Theory.
We have proven that each $u in W^(1,1)$ is continuous and vanishes at infinity*, therefore we have the continuous embedding $W^(1,1)(RR) \subseteq C_0(RR)$ (continuity is a consequence of (*), since $u_n \to u$ in $W^(1,1)$ implies $u_n to u$ in $C_0$).
__________
* This is an abuse of notation: each $u in W^(1,1)(RR)$ is, infact, a sort of equivalence class of functions (w.r.t. equality a.e.); the statement has to be read as: "Each $u in W^(1,1)(RR)$ has a representative element $v \in C_0(RR)$".
The function $u\mapsto ||u||_(1,1):=||u||_1+||u'||_1$ is a norm (of Sobolev type) on $C_c^1(RR)$ and, if we equip $C_c^1(RR)$ with this norm, the normed space is not complete.
But $C_c^1(RR)$ has a completion w.r.t. $||\cdot ||_(1,1)$: this completion turns out to be the Sobolev space $W^(1,1)(RR)$, i.e. the space of weakly differentiable functions with weak derivative in $L^1(RR)$.
Take $u in W^(1,1)(RR)$ and $(u_n) \subseteq C_c^1(RR)$ s.t. $u_n to u$ in $W^(1,1)(RR)$ (i.e. $||u_n-u||_(1,1)\to 0$); since $W^(1,1)(RR)$ is $||\cdot||_(1,1)$-complete, $(u_n)$ is a Cauchy sequence and the trivial inequality:
$||u_n'-u_m'||_1<=||u_n-u_m||_(1,1)$
implies that $(u_n')$ is a Cauchy sequence in $L^1(RR)$ (which obviously converges to $u'$).
Now (*) comes into play: infact, since $(u_n')$ is $||\cdot ||_1$-Cauchy's, by (*) $(u_n)$ is $||||_oo$-Cauchy's, i.e. $(u_n)$ converges uniformly in $RR$ to a continuous function; let $v$ be the uniform limit of $(u_n)$: this function trivially belongs to the $||\cdot ||_oo$-completion of $C_c$, which is the space $C_0(RR)$ of the functions which vanish at infinity.
On the other hand, $(u_n)$ is a Cauchy sequence in $L^1(RR)$, for:
$||u_n-u_m||_1<=||u_n-u_m||_(1,1) \quad$,
and $u_n\to u$ in $L^1(RR)$.
Summarizing, we have:
$\{(u_n\to u, " in " L^1(RR)),(u_n \to v, " in "C_0(RR)) :} \quad$,
therefore $u=v$ a.e. in $RR$ for known results of Measure Theory.
We have proven that each $u in W^(1,1)$ is continuous and vanishes at infinity*, therefore we have the continuous embedding $W^(1,1)(RR) \subseteq C_0(RR)$ (continuity is a consequence of (*), since $u_n \to u$ in $W^(1,1)$ implies $u_n to u$ in $C_0$).
__________
* This is an abuse of notation: each $u in W^(1,1)(RR)$ is, infact, a sort of equivalence class of functions (w.r.t. equality a.e.); the statement has to be read as: "Each $u in W^(1,1)(RR)$ has a representative element $v \in C_0(RR)$".
"Gugo82":
What I had in mind was: "What if we take the completion of $C_c^1(RR)$ with respect to the Sobolev norm $||u||_(1,1) :=||u||_1+||u'||_1$, i.e. the Sobolev space $W^(1,1)(RR)$?"
In other words, can inequality (*) be used to find an embedding of $W^(1,1)(RR)$ into another space?
I don't understand... you want a maximal subspace (if it exists... I don't think it's obvious because the relation is not automatically "linear") in $W^(1,1)$ where (*) is true?
and what is the meaning of what you say after "In other words"? don't see how it connects to the first part...
...maybe this language is too technical for me...
What I had in mind was: "What if we take the completion of $C_c^1(RR)$ with respect to the Sobolev norm $||u||_(1,1) :=||u||_1+||u'||_1$, i.e. the Sobolev space $W^(1,1)(RR)$?"
In other words, can inequality (*) be used to find an embedding of $W^(1,1)(RR)$ into another space?
In other words, can inequality (*) be used to find an embedding of $W^(1,1)(RR)$ into another space?
I guess yes... for example with $f=exp(-x^2)$
$W=C^1_c(RR)\o+{\alpha f}_{\alpha in \RR}$
if the answer is "yes", I'll try to show in a second time that $W$ works (the idea is that the fast decreasing function should't change the norms with respect to a function with support compact approximating it, but I didn't try to see if it works)....
$W=C^1_c(RR)\o+{\alpha f}_{\alpha in \RR}$
if the answer is "yes", I'll try to show in a second time that $W$ works (the idea is that the fast decreasing function should't change the norms with respect to a function with support compact approximating it, but I didn't try to see if it works)....
"Thomas":
[quote="Gugo82"]A classical integral inequality implies:
$\quad |u(x)|=\int_(-oo)^(x) |u'(t)|" d"t \quad$ and $\quad |u(x)|=\int_x^(+oo)|u'(t)|" d"t \quad$
I guess here you wanted to write two inequality signs
[/quote]Yes, of course!
"Gugo82":
Can (*) be extended beyond $C_c^1(RR)$, i.e. to a space strictly containing $C_c^1(RR)$?
Now, what about this question?
"Gugo82":
A classical integral inequality implies:
$\quad |u(x)|=\int_(-oo)^(x) |u'(t)|" d"t \quad$ and $\quad |u(x)|=\int_x^(+oo)|u'(t)|" d"t \quad$
I guess here you wanted to write two inequality signs
"Gugo82":
It remains to solve the second part of the problem. in order to do this, it suffices to show that $2$ is the best constant in (*) and that equality holds for symmetric-decreasing functions. This part is very simple.
eheh... in the case of symmetric decreasing if $x=x_0$ in your proof then the equality signs are correct
Ok, it seems correct.
My solution is the following (and it relies on elementary Calculus facts).
Take $x_1
(a) $\quad u(x_2)-u(x_1)=\int_(x_1)^(x_2) u'(t)" d"t \quad$;
but it's $lim_(x\to pm oo) u(x)=0$ and then, putting $x_1=-oo,x_2=x$ and $x_1=x,x_2=+oo$ in (a), we get:
$\quad u(x)=\int_(-oo)^(x) u'(t)" d"t \quad$ and $\quad -u(x)=\int_x^(+oo)u'(t)" d"t \quad$.
A classical integral inequality implies:
$\quad |u(x)|<=\int_(-oo)^(x) |u'(t)|" d"t \quad$ and $\quad |u(x)|<=\int_x^(+oo)|u'(t)|" d"t \quad$
so that an addiction term by term yelds:
(b) $\quad 2 |u(x)|<= \int_(-oo)^(+oo) |u'(t)|" d"t =||u'||_1$
with the LHS constant with respect to $x$; then taking the least upper bounds in (b) we have (*) with constant $c=2$.
Easy but nice.
It remains to solve the second part of the problem. in order to do this, it suffices to show that $2$ is the best constant in (*) and that equality holds for symmetric-decreasing functions. This part is very simple.
My solution is the following (and it relies on elementary Calculus facts).
Take $x_1
(a) $\quad u(x_2)-u(x_1)=\int_(x_1)^(x_2) u'(t)" d"t \quad$;
but it's $lim_(x\to pm oo) u(x)=0$ and then, putting $x_1=-oo,x_2=x$ and $x_1=x,x_2=+oo$ in (a), we get:
$\quad u(x)=\int_(-oo)^(x) u'(t)" d"t \quad$ and $\quad -u(x)=\int_x^(+oo)u'(t)" d"t \quad$.
A classical integral inequality implies:
$\quad |u(x)|<=\int_(-oo)^(x) |u'(t)|" d"t \quad$ and $\quad |u(x)|<=\int_x^(+oo)|u'(t)|" d"t \quad$
so that an addiction term by term yelds:
(b) $\quad 2 |u(x)|<= \int_(-oo)^(+oo) |u'(t)|" d"t =||u'||_1$
with the LHS constant with respect to $x$; then taking the least upper bounds in (b) we have (*) with constant $c=2$.
Easy but nice.
It remains to solve the second part of the problem. in order to do this, it suffices to show that $2$ is the best constant in (*) and that equality holds for symmetric-decreasing functions. This part is very simple.
This is my trial...
definitions:
$K=[x_(min),x_(max)]$ be a compact interval containg the support of the function
$C={x in K |u'(x)>0}$
$B={x in K |u'(x)<0}$
$f$ the point where the funcion has its maximum... wlog (we can change the sign of the function without changing the norms) we can imagine $f$ in the interior of $K$ and $u(f)=M>0$... otherwise we can say that $u$ is null...
$I=(x_min,f)$
$F=(f,x_max)$
and neglecting sets of null measure it holds $K=Iuuu F$... all the sets above are open (becaus $u'$ is $C^0$) and so Lesbege measurable)
now it is intuitive
$int_I u'=M => int_(I\cap C)|u'|>=M$ [1]
$int_F u'=-M => int_(F\cap B)|u'|>=M$ [2]
and using the results above
$||u'||=int_I|u'|+int_F|u'|>= int_(I\capC)|u'|+ int_(F\capB)|u'|>=2M$
giving the answer of $c=2$ to the problem... is it correct?
definitions:
$K=[x_(min),x_(max)]$ be a compact interval containg the support of the function
$C={x in K |u'(x)>0}$
$B={x in K |u'(x)<0}$
$f$ the point where the funcion has its maximum... wlog (we can change the sign of the function without changing the norms) we can imagine $f$ in the interior of $K$ and $u(f)=M>0$... otherwise we can say that $u$ is null...
$I=(x_min,f)$
$F=(f,x_max)$
and neglecting sets of null measure it holds $K=Iuuu F$... all the sets above are open (becaus $u'$ is $C^0$) and so Lesbege measurable)
now it is intuitive
$int_I u'=M => int_(I\cap C)|u'|>=M$ [1]
$int_F u'=-M => int_(F\cap B)|u'|>=M$ [2]
and using the results above
$||u'||=int_I|u'|+int_F|u'|>= int_(I\capC)|u'|+ int_(F\capB)|u'|>=2M$
giving the answer of $c=2$ to the problem... is it correct?
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