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The space $L^2([a,b];L^2(S^2))$ is a Banach space with respect to the norm $$\left\Vert f \right\Vert_1^2 = \int_{a}^b \left\Vert f(r) \right\Vert_{L^2(S^2)}^2 dr$$

The space $L^2([a,b]\times S^2)$ is a Banach space with respect to the norm

$$\left\Vert f \right\Vert_2^2 = \int_{a}^b \int_{S^2} |f|^2r^2 drd\sigma_{S^2}$$ where $d\sigma_{S^2}$ is the volume form on the unit sphere.

For each function $f \in L^2([a,b];L^2(S^2))$, we can interpret it as a function on $[a,b]\times S^2$ by in the obvious way: $f(r,x) = f(r)(x)$ for $r\in [a,b]$ and $x\in S^2$. This defines a map $\Phi$ from $L^2([a,b];L^2(S^2))$ to $L^2([a,b]\times S^2)$

Are these two spaces the same? Or does one live properly in the other? In other words, is $\Phi$ an isomorphism, or an embedding?

Suppose we pick a function $f$ in $L^2([a,b];H^1(S^2))$. We know that $f(a)$, $f(b)$ are not well-defined elements in $H^1(S^2)$. Is there a version of the trace theorem stating that $f(a)$,$f(b)$ are well-defined elements in $H^{1/2}(S^2)$?

Please send me any references on this. I didn't find papers or books that study these spaces.

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  • $\begingroup$ How do you define measureability of a function $[a, b] \to L^2(S^2)$? Just that the "uncurried" function $[a, b] \times S^2 \to \mathbb C$ is measureable? Does that notion of measureablity come from an honest $\sigma$-algebra on $L^2(S^2)$? $\endgroup$
    – LSpice
    Commented Apr 8 at 15:01
  • $\begingroup$ I think it's the function $r \mapsto \lVert f(r) \rVert_{L^2(S^2)}$ that is measurable. So $f$ is any function on $[a,b]$ taking values in $L^2(S^2)$ with the property that $r \mapsto \lVert f(r) \rVert_{L^2(S^2)}$ is a measurable $L^2$ function from $[a,b]$ to $\mathbb{R}$. $\endgroup$
    – Laithy
    Commented Apr 8 at 16:02
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    $\begingroup$ Yes, the spaces agree with one another and also with $L^2(a,b)\otimes L^2(S^2)$. An easy way to see this is to consider the natural map $L^2((a,b)\times S^2)\to L^2((a,b); L^2(S^2))$ on (say) smooth functions. It is isometric with dense domain and range and thus extends to a unitary map, still acting in the obvious way. $\endgroup$
    – Christian Remling
    Commented Apr 8 at 16:17
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    $\begingroup$ @LSpice Usually when one works with these spaces, what is required is "strong measurability" a.k.a. "Bochner measurability" which is formally stronger; roughly speaking, one takes good approximation by vector-valued simple functions as a definition, rather than a consequence. $\endgroup$
    – Yemon Choi
    Commented Apr 8 at 20:20
  • $\begingroup$ mathoverflow.net/questions/67434/… $\endgroup$
    – Martin Väth
    Commented Apr 8 at 20:21

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