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

[christoff]

Homework Statement

Let $u,v\in H_0^1(\mathbb{R})$, the closure of smooth $\mathbb{R}$-valued functions with compact support with respect to the norm defined by $$||v||_{1}^2=||v||^2+||v'||^2,$$ where $||\cdot||$ is the standard L2 norm. Show that $uv\in H_0^1(\mathbb{R})$.

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 $||uv||$).

I would appreciate a starting point...
Thanks!

 

[Fredrik]

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?

 

[christoff]

I think I'm close... By density, let $u_n\rightarrow u, v_n\rightarrow v$ in $H^1(\mathbb{R})$. We can write $u_nv_n-uv=v_n(u_n-u)+u(v_n-v)$. Then, to show that $uv\in H_0^1(\mathbb{R})$, it suffices to show
$$||v_n(u_n-u)+u(v_n-v)||_1\leq ||v_n(u_n-u)||_1+||u(v_n-v)||_1 \rightarrow 0.$$ The first expression to the right of the inequality reads, after expanding,
$$\left [\int (v_n^2+v_n'^2)(u_n-u)^2 + 2v_n'v_n(u_n-u)(u_n'-u')+v_n^2(u_n'-u')^2dx \right]^{1/2}$$ By the Sobolev imbedding theorem, we have that $v_n$ is uniformly bounded, so we can bound the above integral by
$$\int (A+v_n'^2)(u_n-u)^2dx +\int 2Bv_n' |(u_n-u)(u_n'-u')| dx + \int A(u_n'-u')^2dx \\ <br /> = A||u_n-u||+\int v_n'^2(u_n-u)^2dx+2B\int |v_n' (u_n-u)(u_n'-u')| dx +A||u_n'-u'||$$for some constants $A,B$. The problem is now the two terms in the middle. The left and right-most terms converge to zero because $u_n\rightarrow u$ in H1. If we knew that $v_n'$ was uniformly bounded, then we would be done. Unfortunately, this does not hold (as far as I know); all we know is that $v_n'\rightarrow v'$ in L2.

What next?

 

[christoff]

By density, let $u_n\rightarrow u, v_n\rightarrow v$ in $H^1(\mathbb{R})$. We can write $u_nv_n-uv=v_n(u_n-u)+u(v_n-v)$. Then, to show that $uv\in H_0^1(\mathbb{R})$, it suffices to show
$$||v_n(u_n-u)+u(v_n-v)||_1\leq ||v_n(u_n-u)||_1+||u(v_n-v)||_1 \rightarrow 0.$$ The first expression to the right of the inequality reads, after expanding,
$$\left [\int (v_n^2+v_n'^2)(u_n-u)^2 + 2v_n'v_n(u_n-u)(u_n'-u')+v_n^2(u_n'-u')^2dx \right]^{1/2}$$ By the Sobolev imbedding theorem, we have that $v_n$ is uniformly bounded, so we can bound the above integral by
$$\int (A+v_n'^2)(u_n-u)^2dx +\int 2Bv_n' |(u_n-u)(u_n'-u')| dx + \int A(u_n'-u')^2dx \\ <br /> = A||u_n-u||+\int v_n'^2(u_n-u)^2dx+2B\int |v_n' (u_n-u)(u_n'-u')| dx +A||u_n'-u'||$$for some constants $A,B$.

OK, think I got it.

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

Is this correct?

 

[Fredrik]

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.