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Show that H_0^1(R) is an algebra.

  1. Apr 24, 2013 #1
    1. The problem statement, all variables and given/known data
    Let [itex]u,v\in H_0^1(\mathbb{R})[/itex], the closure of smooth [itex]\mathbb{R}[/itex]-valued functions with compact support with respect to the norm defined by [tex]||v||_{1}^2=||v||^2+||v'||^2,[/tex] where [itex]||\cdot||[/itex] is the standard L2 norm. Show that [itex]uv\in H_0^1(\mathbb{R})[/itex].

    3. The attempt at a solution
    Pretty much stuck on this one. Density of smooth functions with compact support likely won't be super helpful since they're also dense in L2, and as far as I know, that space isn't closed under multiplication (if it is, then this exercise is trivial, since then we can just apply Cauchy-Schwartz and use the multiplication in L2 to bound [itex] ||uv|| [/itex]).

    I would appreciate a starting point... :smile:
    Thanks!
     
  2. jcsd
  3. Apr 24, 2013 #2

    Fredrik

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    I expect it to be very easy to show that the set of smooth functions with compact support is closed under multiplication. The problem here is that ##H_0^1(\mathbb R)##, as you have defined it, also contains all the functions that are limits of sequences of such functions. So if u and v are such limits, can you show that uv is too?
     
  4. Apr 24, 2013 #3
    I think I'm close... By density, let [itex]u_n\rightarrow u, v_n\rightarrow v[/itex] in [itex]H^1(\mathbb{R})[/itex]. We can write [itex] u_nv_n-uv=v_n(u_n-u)+u(v_n-v)[/itex]. Then, to show that [itex]uv\in H_0^1(\mathbb{R})[/itex], it suffices to show
    [tex]||v_n(u_n-u)+u(v_n-v)||_1\leq ||v_n(u_n-u)||_1+||u(v_n-v)||_1 \rightarrow 0. [/tex] The first expression to the right of the inequality reads, after expanding,
    [tex] \left [\int (v_n^2+v_n'^2)(u_n-u)^2 + 2v_n'v_n(u_n-u)(u_n'-u')+v_n^2(u_n'-u')^2dx \right]^{1/2}[/tex] By the Sobolev imbedding theorem, we have that [itex]v_n[/itex] is uniformly bounded, so we can bound the above integral by
    [tex]\int (A+v_n'^2)(u_n-u)^2dx +\int 2Bv_n' |(u_n-u)(u_n'-u')| dx + \int A(u_n'-u')^2dx \\
    = A||u_n-u||+\int v_n'^2(u_n-u)^2dx+2B\int |v_n' (u_n-u)(u_n'-u')| dx +A||u_n'-u'||[/tex]for some constants [itex]A,B[/itex]. The problem is now the two terms in the middle. The left and right-most terms converge to zero because [itex]u_n\rightarrow u[/itex] in H1. If we knew that [itex]v_n'[/itex] was uniformly bounded, then we would be done. Unfortunately, this does not hold (as far as I know); all we know is that [itex]v_n'\rightarrow v'[/itex] in L2.

    What next?
     
    Last edited: Apr 24, 2013
  5. Apr 24, 2013 #4
    OK, think I got it.

    The left and right-most quantities converge to zero since [itex]u_n\rightarrow u[/itex] in H1. By Cauchy-Schwartz and the uniform boundedness of [itex](u_n-u)[/itex] (by Sovolev imbedding), [tex]2B\int |v_n' (u_n-u)(u_n'-u')| dx \leq 2B'||v_n'||\cdot||u_n'-u'||\rightarrow 0[/tex] since [itex]v_n'\rightarrow v'[/itex] in L2 and [itex]||u_n'-u'||\rightarrow 0[/itex]. We also have that since [itex]u_n-u[/itex] is continuous and uniformly bounded (Sobolev imbedding theorem), there exists [itex]C(n)\rightarrow 0[/itex] with [itex]||u_n-u||_{∞}\leq C(n)[/itex]. Hence, [tex]\int v_n'^2(u_n-u)^2dx\leq \int v_n'^2\cdot C(n)^2dx = C(n)^2||v_n'||\rightarrow 0\cdot ||v'||=0.[/tex] Therefore, [itex]||v_n(u_n-u)||_1\rightarrow 0[/itex]. Similar computations show that [itex]||u(v_n-v)||_1 \rightarrow 0.[/itex] The result follows.

    Is this correct?
     
    Last edited: Apr 24, 2013
  6. Apr 24, 2013 #5

    Fredrik

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    The strategy I had in mind starts exactly the way you did it. I was thinking that if the L_2 norm satisfies ##\|fg\|\leq\|f\|\|g\|## for all f,g, then we might be able to use that to show that the other norm satisfies this condition too. This is either impossible, or a significant simplification. (I haven't thought about it enough to know which).

    I don't have time to look at the details of your proof right now. If you think you have solved it, you probably have.
     
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