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For Sobolev spaces $H^s(R^d)$, with $s> \frac{d}{2}$ every element of $H^s(R^d)$ is an equivalence class $[f]$ and in every such a class there exists a unique continuous function $f^{*}$. Can we define the trace operator $tr_{R^{d_0}}: H^s(R^d)\to H^{s-\frac{d-d_0}{2}}(R^{d_0})$, on some $R^{d_0}$ with $d_0<d$ as:

$\hspace{3cm}tr_{R^{d_0}}[f]=[f^{*}|_{R^{d_0}}]$ $\ \ \ $? $\ \ \ $ What is wrong with this definition ?

As far as I know the trace operator is defined as a continuous extension of the restriction operator, initially defined from $C^{\infty}_c(R^d)$ to $C^{\infty}_c(R^{d_0})$

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  • $\begingroup$ Well, usually, it is of interest into which function space such a trace operator actually maps, right? $\endgroup$
    – Hannes
    Commented Feb 7, 2019 at 9:40
  • $\begingroup$ I forgot to add that, I added now. I can not find what is wrong with this definition. $R^{d_0}$ is a null set in $R^d$ but since $f^{*}$ is continuous $tr_{R^{d_0}}$ seems to be well defined. Functions with jump discontinuities are not in $H^s(R^d)$ ( for $s$ big enough ) and I can not find counterexamples to argue that $tr_{R^{d_0}}$ is not well-defined. $\endgroup$
    – Emanuel
    Commented Feb 7, 2019 at 10:26
  • $\begingroup$ How do you know that your operator actually maps into the given space and that it is continuous? (I presume you want to use your definition as the actual definition of the trace operator and not just show that it coincides with the trace operator as defined elsewhere.) $\endgroup$
    – Hannes
    Commented Feb 7, 2019 at 12:18
  • $\begingroup$ Do you see obstructions for continuity ? $\endgroup$
    – Emanuel
    Commented Feb 7, 2019 at 12:49
  • $\begingroup$ What kind of question is that? Do you see obstructions for the Riemann Hypothesis? You claim something, you need to prove it. An attempt to prove it will answer your question: Either everything works out, then nothing is wrong with your definition, or you cannot prove stuff and then that's what's wrong. $\endgroup$
    – Hannes
    Commented Feb 7, 2019 at 14:07

1 Answer 1

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Your definition is highly restrictive. One can show (cf. McLean, Lemma 3.35) that $\gamma$ has a unique extension to a bounded operator $\gamma : H^s(\mathbb{R}^d) \to H^{s-1/2}(\mathbb{R}^{d-1})$ for $s > 1/2$.

So in your case, you need that $s > d/2$, whereas the extension is valid for $s > 1/2$.

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  • $\begingroup$ I forgot to add the spaces. I this situation with $d_0=d-1$ can one use the above definition for the trace operator ? $\endgroup$
    – Emanuel
    Commented Feb 7, 2019 at 10:23
  • $\begingroup$ The question is about how $tr_{R^{d_0}}$ is constructed. Does it lead to the same trace operator as that one from the book of McLean or E.Taylor ? $\endgroup$
    – Emanuel
    Commented Feb 7, 2019 at 10:33
  • $\begingroup$ I'd suppose that your definition also continuously extends the standard definition, so by uniqueness it is contained in the extension to $s > 1/2$. $\endgroup$
    – mcd
    Commented Feb 8, 2019 at 14:56

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