Norm / inner product.
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From: Chip Eastham on 11 Jul 2010 00:44 On Jul 10, 11:29 am, Joubert <trappedinthecloset9...(a)yahoo.com> wrote: > > ??? What does it _mean_ to say that two inner products > > are "equivalent"? > > The proof states that a particular functional, continuous w.r.t. to the > usual norm of (H^1)_0, is continuous also w.r.t. the norm defined by > some new scalar product (so as to use Riesz representation theorem with > this new scalar product and show the existence of a weak solution to a > Dirichlet problem). The continuity w.r.t. the new scalar product relies > on the "equivalence" of the scalar products. My guess is two scalar > products are equivalent iff they define equivalent norms, but how does > one show it "by hand"? Furthermore: is my interpretation correct? What you need to show is often referred to as "coerciveness" of a bilinear form. Recall that for a bilinear form to be a (real) inner product, certain properties must be satisfied: a(x,y) = a(y,x) [symmetry] a(x,x) > 0 for nonzero x [positivity] Coerciveness strengthens the latter property in requiring that there exists real c > 0 s.t. a(x,x) >= c ||x||^2 [coercivity] Continuity of a bilinear form requires it be bounded with respect to the norm, i.e. that there exists real C > 0 s.t. a(x,y) =< C ||x|| ||y|| for all vectors x,y Such continuity implies a(x,x) =< C ||x||^2, so we see that coercivity together with continuity imply: c ||x||^2 =< a(x,x) =< C ||x||^2 for all x Such inequalities bound the norm given by the "standard" inner product by that given by the "new" inner product, and furthermore the converse. It follows that the metric topologies are the same. Without knowing the specifics it would be hard to say much more of concrete details. regards, chip
From: Joubert on 11 Jul 2010 07:48
On 07/11/2010 06:44 AM, Chip Eastham wrote: > What you need to show is often referred to > as "coerciveness" of a bilinear form. Recall > that for a bilinear form to be a (real) inner > product, certain properties must be satisfied: > > a(x,y) = a(y,x) [symmetry] > a(x,x)> 0 for nonzero x [positivity] > > Coerciveness strengthens the latter property > in requiring that there exists real c> 0 s.t. > > a(x,x)>= c ||x||^2 [coercivity] > > Continuity of a bilinear form requires it be > bounded with respect to the norm, i.e. that > there exists real C> 0 s.t. > > a(x,y) =< C ||x|| ||y|| for all vectors x,y > > Such continuity implies a(x,x) =< C ||x||^2, > so we see that coercivity together with > continuity imply: > > c ||x||^2 =< a(x,x) =< C ||x||^2 for all x > > Such inequalities bound the norm given by > the "standard" inner product by that given > by the "new" inner product, and furthermore > the converse. It follows that the metric > topologies are the same. > > Without knowing the specifics it would be > hard to say much more of concrete details. > > regards, chip The inner products are pretty messy but this should be enough for me to fill in the details of my case by myself, thanks. |