Is the following operator (related to the Dirac's delta) bounded?

Question

Consider $\eta \colon [-1,0] \to \mathbb{R}$ of bounded variation and define the (delay) operator $\Phi \colon \mathcal{C}[-1,0] \to \mathbb{R}$ by $$\Phi f = \int_{-1}^{0} f d \eta$$ for $f \in \mathcal{C}[-1,0]$ where the integral is the Riemann-Stieltjes integral.
$\Phi$ is a bounded linear operator acting on $\mathcal{C}[-1,0]$ and so by continuous embeddings it is a bounded linear operator acting on the Sobolev space $W^{1,2}[-1,0]$.

Now [Bátkai, András, and Susanna Piazzera. Semigroups for delay equations. CRC Press, 2005.] p.69 example 3.28 claims that $$\Phi f= f(-1), f \in W^{1,2}[-1,0]$$ is representable in the integral form before and it is a bounded linear operator.

But $\eta$ in this case should be the Dirac's delta in $-1$, i.e. $\eta = \delta_{-1}$.

It feels weird that it is a bounded linear operator on $W^{1,2}$ when usually distributions are defined as linear continuous functionals on $C_{0}^{\infty}$ which is a much smaller space? I mean ok this should be another way to define functionals from distributions right?

From the point of view of calcultions it seems like that: $$|\Phi f|= |f(-1)| \leq |f|_{\mathcal{C}^0} \leq C |f|_{W^{1,2}}$$ where $|\cdot|_{\mathcal{C}^0}$ is the sup norm.

Anyway since $\eta= \delta_{-1}$ is not a function can you define the Riemann-Stieltjes integral correctly?

Answer 1

Actually, in terms of Stieltjes integral, the evaluation at a point is represented by the Heaviside function and not the Dirac delta. More precisely, if you have $H_{-1}(x) = \mathbf{1}_{x>-1}$ then $$ f(-1) = \int_{-\infty}^\infty f(x)\,\mathrm d H_{-1}(x) = \int_{-\infty}^\infty f(x)\,\delta_{-1}(\mathrm d x) $$ The Dirac delta $\delta_{-1}$ is only the derivative (in the sense of distributions) of your function of bounded variation $H_{-1}$ ... As I explained more precisely here, the ambiguity is due to the abusive notation $∫ f\,\mathrm d \mu$ when $\mu$ is a measure, which should rather be $∫ f\,\mu = ∫ f\,\mathrm d \eta$ or $∫ f(x)\,\mu(\mathrm d x) = ∫ f(x)\,\mathrm d \eta(x)$ to be coherent with the Stieltjes integral.

Distributions are usually defined on $C^\infty_c$ when one wants to be able to take a lot of derivatives. The distributions of order $0$ are isomorphic to measures, and so to every distribution of order $0$ one associates a measure such that the formula $$ \langle\mu,\varphi\rangle = ∫ \varphi(x)\,\mu(\mathrm d x) $$ holds for every test function in $C^\infty_c$. But the right hand-side can be defined through the usual measure and integration theory, so one can take any $\varphi\in L^1(\mu)$. And so in particular, any function $\varphi\in C^0$.

Still, it is indeed a bit weird here to use the fact that one can write $f(-1)$ as an integral to deduce the result, and your computation $|f(-1)| ≤ \|f\|_{C^0} ≤ C\|f\|_{W^{1,2}}$ makes much more sense.