User talk:Behshour

1. The bilinear form does not define an inner product in $$H_0^1$$ since it does not satisfy the "positive definiteness of the inner product: For any inner product $$<.,.>$$ we must have: $$=0$$ if and only if $$u=0$$, but this is not the case here since $$=0$$ only implies that $$u'=0$$. For example consider the function $$u=constant$$. Then $$B(u,u)=0$$, but obviously $$u$$ is not zero. Furthermore, although this bilinear form does not generate an induced norm, but it generates a semi-norm, and it may be considered a "Minkowski functional" instead of an inner product. It's however noteworthy that another bilinear form on the space $$H_0^1$$, namely, $$\int_0^1 u'v'+uv$$ does define an inner product and its induced norm is the standar Sobolev space norm in this space. In this particular argument, whether the original bilinear form defines an inner product or not is irrelevant, therefore I suggest that that remark be omitted.

2. the Sobolev space $$H_0^1$$ is different from the space of functions of "bounded variations". I suggest that $$H_0^1(0,1)$$ be defined as: $$H_0^1(0,1)={u:u,u'\in L^2(0,1); u(0)=u(1)=0}$$.

Please note that it's important to verify the above as soon as possible, and to make corrections as needed since otherwise, it would be misinforming.

Best,Behshour 17:41, 26 June 2007 (UTC)


 * The bilinear form $\int_0^1 u' v'$ is positive definite on $H_0^1(0,1)$ by Poincaré's inquality. Indeed, $u'=0$ is sufficient, in that space, to imply that $u=0$. This is because of the boundary conditions $u(0)=u(1)=0$. The fact that $u'=0$ means that $u$ is constant, and the fact that $u(0)=0$ means that $u(x)=0$ for all $x \in (0,1)$.


 * For details, consult a book on Sobolev spaces. An excellent reference is Partial Differential Equations by Lawrence C. Evans. American Mathematical Society, Providence, RI, 1998. Graduate Studies in Mathematics 19.


 * Finally, I understand your curiosity, but like Corpx suggested, use the talk page of the article instead of the article itself for discussions like this one. Loisel 23:02, 26 June 2007 (UTC)


 * Oh and as to your second question, it turns out that exceptionally in 1d, H^1 has something to do with functions of bounded variation. This is not true in 2d. Loisel 23:05, 26 June 2007 (UTC)


 * I realized after I wrote this that bounded variation should have been absolutely continuous. Loisel 23:33, 26 June 2007 (UTC)

Thank you for the book reference. I am familiar with that book, and I can suggest a few more myself. However, your writings are incorrect!! Your state of mind is in the space of continuous functions but your writings are about $$L^2$$ which is a totally different world. Please walk along me step by step: Consider the following $$L^2(0,1)$$ function: $$ u=3$$ on THE OPEN INTERVAL $$(0,1) $$ and $$ u(0)=u(1)=0$$. Obviously this is an $$L^2(0,1)$$ function whose $$L^2(0,1)$$ norm equals 3. Its derivative $$u'$$ is identically zero with a zero $$L^2(0,1)$$ norm. Therefore $$u$$ is in the space $$H_0^1(0,1)$$. At the same time, $$B(u,u)=0$$, but $$u$$ is not identically zero. Again, I suggest that please make corrections.

In reference to the space $$H_0^1(0,1)$$, its possible that in DIMENSION 1, and due to embedding theorems of Sobolev, along with the reflexivity of the Hilbert space, this space (or its closure) may be associated with the space of functions of bounded variation. But as a mathematician, when one is defining a space, the definition must be as fundamental as possible, and associations come next. They should not replace definitions. The definition of this space is : $$H_0^1(0,1)={u:u,u'\in L^2(0,1); u(0)=u(1)=0}$$.Behshour 04:50, 27 June 2007 (UTC)


 * The function you propose is not absolutely continuous, so it is not in H^1, and certainly not in H^1_0.


 * A 2d analogue of your example would likewise not be weakly differentiable.


 * The fact that the bilinear form $\int \nabla u \cdot \nabla v$ is equivalent to the $\int \nabla u \cdot \nabla v + uv$ form is essential in showing that the Dirichlet problem for the Laplacian is well posed; see Elliptic boundary value problem.


 * Let's continue this discussion in the FEM article talk page. Loisel 18:09, 27 June 2007 (UTC)