/sci/ - Science & Math

>>10356408
Riesz representation theorem.
More extensively, assume [math] u \in ker f[/math] but [math] u \notin ker g[/math]. Then f(u)g(v)=0, but f(v)g(u)=/=0.
For the other way around, we use this:
Lemma: Any two functionals with the same kernel are scalar multiples of each other.
Proof of lemma: assume the cokernel of a linear functional f has more than one dimension, and thus contains a two-dimensional subspace spanned by a and b. Then [math]0= \lambda f(a) - f(b)= f( \lambda a -b)=0.
From this it follows trivially.