proof of Riesz representation theorem

Existence - If f=0 we can just take u=0 and thereby have f(x)=0=x,0 for all x.

Suppose now f0, i.e. Kerf.

Recall that, since f is continuousMathworldPlanetmath (, Kerf is a closed subspace of (continuity of f implies that f-1(0) is closed in ). It then follows from the orthogonal decomposition theorem that


and as Kerf we can find z(Kerf) such that z=1.

It follows easily from the linearity of f that for every x we have


and since z(Kerf)

     0 = f(x)z-f(z)x,z
= f(x)z,z-f(z)x,z
= f(x)z2-x,f(z)¯z
= f(x)-x,f(z)¯z

which implies


The theorem then follows by taking u=f(z)¯z.

Uniqueness - Suppose there were u1,u2 such that for every x


Then x,u1-u2=0 for every x. Taking x=u1-u2 we obtain u1-u22=0, which implies u1=u2.

Title proof of Riesz representation theorem
Canonical name ProofOfRieszRepresentationTheorem
Date of creation 2013-03-22 17:32:37
Last modified on 2013-03-22 17:32:37
Owner asteroid (17536)
Last modified by asteroid (17536)
Numerical id 5
Author asteroid (17536)
Entry type Proof
Classification msc 46C99