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I got the intuition that $H^s := W^{s,2}$ is the Sobolev space of functions which are $k:=\lfloor s \rfloor$ times differentiable and $(s-k)$-hölder continuous. In this answer, the dual space of this sobolev space is aparently given by $H^{-s}$, which makes intuitive sense. It contains generalized functions which have to be roughly $s$ times integrated before they are actually functions. In particular it contains distriutions such as the dirac delta.

But now I am wondering how the Riez representation theorem should work here. I mean what continuous function $h\in H^s$ should possibly act the same as the dirac delta?

I.e. $\langle h, f\rangle = \int h(x) f(x) dx = f(0)$ for all $f\in H^s$?

This seems really wrong. So my understanding must be wrong in some place.

J. W. Tanner
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Felix Benning
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    It is not very clear what you are asking. First, Sobolev spaces a priori have nothing to do with Hölder regularity. Then it's not clear why you are looking for a continuous $h\in H^s$. What is true is the following: For $s>1/2$ there is an embedding $H^s\hookrightarrow C^0$ into continuous functions and hence $\delta_0\colon H^s\to \mathbb C$ makes sense. You can represent this as a member in $H^{-s}$ and indeed $\delta_0$ can be viewed as an element of $H^{t}$ for all $t<-1/2$. This is ignoring boundary conditions (or working on $\mathbb T^n$, say). Have a look at Taylor's PDE book (Ch. 4). – Jan Bohr Mar 20 '25 at 17:21
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    Here is an explicit construction of such an element when $k=1$ https://math.stackexchange.com/questions/538295/representing-the-dirac-distribution-in-h1-mathbb-r-through-the-scalar-produ – whpowell96 Mar 20 '25 at 17:58
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    One way to think about this is that the Dirac delta is the only pointwise evaluation map that works on arbitrary continuous functions. In Sobolev spaces of sufficient regularity, pointwise values are restricted by the integrals of derivatives, which are controlled by the norm. Since there are more restrictions on Sobolev functions, there are potentially more options for integration kernels that yield pointwise evaluation. This is similar to reproducing kernel Hilbert spaces where each space basically has its own different pointwise evaluation kernel. – whpowell96 Mar 20 '25 at 18:01
  • @whpowell96 The linked question considers the duality $H^1(\mathbb R)^{\ast} \cong H^1(\mathbb R)$ via the associated inner product, which I think is more confusing than helpful: one should take the pairing induced by the $L^2$ inner product, so one can really view $\delta_0 \in H^1(\mathbb R)^{\ast}$ as the evaluation map $f \to f(0)$ by taking the continuous representative. Also as a meta comment, I don't see why this is marked as a duplicate of said question. – ktoi Mar 21 '25 at 15:27
  • Hmm yes you are right. I should have been more careful when looking at the inner products. I believe that the comment in OP’s linked post may prove the most helpful. Because OP is specifically looking for an element of $H^s$ that works like the delta function, they may be simultaneously using the Riesz map identifying this space with its dual and the inclusion map $H^s \to H^{-s}$ which may not be compatible – whpowell96 Mar 21 '25 at 18:19

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