Cauchy-Schwarz Inequality

By Cauchy-Schwarz inequatlity, for any integrable function f, g,

|∫fg| <= √(∫|f|^2) √(∫|g|^2)
<= √(∫|| f ||∞ ^2) √(∫|| g ||∞ ^2)
= || f ||∞ || g ||∞ (b-a)

Hence,
| a(u,v) | <= |∫u'v' | + |c∫uv |
<= (|| u' ||∞ || v' ||∞ + |c| || u ||∞ || v ||∞ )(b-a)

Now take M = (b-a)max{1, |c|}, then
| a(u,v) | <= M(|| u' ||∞ || v' ||∞ + || u ||∞ || v ||∞)
<= M(|| u' ||∞ + || u ||∞)(|| v' ||∞ + || v ||∞)
= M ||u||1 ||v||1

A linear functional on a Banach space is bounded iff it is continuous. The same is also true for bilinear forms, by uniform boundedness principle.

< 仍然是亂碼收場

>>Is there any bilinear form continuous but not bounded ?
唔係 Banach space 好易 construct 的:
Take V = countably infinite dimensional vector space (e.g. l^2 sequences), {e_n} = basis
D a linear subspace of V, D = {Finite linear combination of e_n}, with usual inner product.

2008-03-23 12:28:56 補充：
Then D is normed but not complete.

Take bilinear form B, defined on basis as B(e_n, e_m) = {0 if n=/m, n if n=m}

Then this is an unbounded bilinear form on D, but it is continuous (because every Cauchy Sequence of D is trivial: either the index n will stay fix after sometime, or limit does not exist.)

2008-03-23 12:32:37 補充：
maximal_ideal_space:
我唔係好睇到點由 Uniform Boundedness Principle => continuous = bounded.
可否解釋清楚D?