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Consider a function $u$ in $L^2([0,1]; H^k(S^2))$ where $k$ is a positive integer.

Where would $u(0)$ live (or $u(r)$ for some fixed $r \in [0,1]$)? Is there a version of the trace theorem saying that $u(0) \in H^{k-1/2}(S^2)$? Or does it hold that $u(0) \in H^{k}(S^2)$?

What if we in addition know that $u' \in L^2[0,1]; H^{k'}(S^2))$ for some $k'\leq k$?

If you know any references that study the trace theorem for such spaces, please send them over.

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  • $\begingroup$ For the question in your title: the constant function $1$ on $S^2$ is in any $H^k$. So any $L^2$ function $u$ can be realized as an $L^2([0,1]; H^k(S^2))$ function. But there is no "trace theorem" allow you to take pointwise values of an arbitrary $L^2$ function. // For the motivation: when $F \equiv 0$, your ODE is equivalent to $(e^r u(r))' = 0$. It is absolutely false that for arbitrary $u_0, u_1$ you can find a solution. Any solution must satisfy $u_0 = e \cdot u_1$. // Considering that you are studying a first order ODE, it doesn't make sense to give two boundary values. $\endgroup$
    – Willie Wong
    Commented Apr 25 at 22:48
  • $\begingroup$ Oh I didn't mean to add the "$u_1$". Thanks for pointing that out. I edited the post. I see, so we cannot make sense of $u(r)$ for a given $r$. If we in addition know that $\partial_r u \in L^2([0,1];H^{k-1})$, then is there a trace theorem saying something about $u(0)$? $\endgroup$
    – Laithy
    Commented Apr 26 at 0:43
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    $\begingroup$ Yes there is, by interpolation theory, see e.g. here. This will give you $u(\tau) \in H^\ell(S^2)$ with $\ell = \frac12 k + \frac12 k'$ for every $\tau \in [0,1]$ with a uniform-in-$\tau$ norm dependence against $\|u\|_{L^2(H^k)} + \|u'\|_{L^2(H^{k'})}$. $\endgroup$
    – Hannes
    Commented Apr 26 at 7:59
  • $\begingroup$ @Hannes Thank you! That was what I was looking for. I guess "trace theorem" is not the right phrase to use. $\endgroup$
    – Laithy
    Commented Apr 26 at 15:59
  • $\begingroup$ @Laithy Actually, I would say "trace theorem" is just fine and in fact regularly used. :-) $\endgroup$
    – Hannes
    Commented Apr 26 at 19:49

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