# Riesz (Hilbert-space) representation theorem and dirac delta on $\mathcal{C}_{0}$

I am thinking about this for a while now, but don't get near an understanding, so I must have gotten something important wrong.

I look at $\mathcal{C}_{0}$, the space of countinuous (bounded) functions from $\mathbb{R}$ to $\mathbb{R}$, that vanish at infinity. I can define an inner product on this space by setting $<f,g>:=\int fg \ d\lambda$, where $\lambda$ is the Lebesgue measure, so this makes $\mathcal{C}_{0}$ into a Hilbert space (or take some other measure, that has a nonzero density everywhere on $\mathbb{R}$, like one with a normal distribution).

Now the The Hilbert space representation theorem gives me that for every linear functional $\psi$ on $\mathcal{C}_{0}$ there is an $f_{\psi}\in\mathcal{C}_{0}$ so that $\psi=<\cdot,f_{\psi}>$.

On the other hand the Riesz–Markov–Kakutani representation theorem states that the space of all the linear, continuous functionals on $\mathcal{C}_{0}$ is isomorphically isometric to the space of countably additive Borel-measures on $\mathbb{R}$. The dirac-delta measure $\delta$ (that assigns the value 1 to the zero and 0 for everything else) is such a measure and also is a linear functional by $\delta(h)=\int h\ d\delta=h(0)$, but there is no element $f_{\delta}\in \mathcal{C}_{0}$ such that $\delta=<\cdot,f_{\delta}>$ (it seems to me there are only such representations for measures that have a Radon-Nikodym derivative with respect to the Lebesgue measure).

That is a contradiction, so where am I wrong?

• In that norm, you can approximate functions that are not continuous, so what you have defined is a normed space, not a Hilbert space. In order to get a Hilbert space, you have to take a closure, and then the dirac delta distribution loses its meaning in your new space. – Jonas Dahlbæk Aug 9 '14 at 14:30

The space $\mathcal{C}_{0}$ is not a Hilbert space. Consider, for example, a sequence of functions which is zero on $(-\infty,-1/n)$, goes linearly from 0 to 1 on $[-1/n,1/n]$, is one on $(1/n,1)$, and goes linearly from 1 to 0 on $[1,2]$, and is 0 on $[2,\infty]$. This sequence is Cauchy, but it cannot converge to a continuous function. In the completion of $\mathcal{C}_{0}$, the Dirac measure is no longer a well-defined linear functional.