Sobolev spaces involving time

Question

In his notes on the Cauchy problem in general relativity, Ringstrom has the following lemma:

Lemma A.5. Let $I$ be an open interval. If $u \in L^\infty(I; H^k(\mathbb{R}^n))$ for some $k \geq 0$, then there is a function $U \in L^2_{\text{loc}}(I \times \mathbb{R}^n)$ such that for any $\phi \in C_0^\infty(I \times \mathbb{R}^n)$, $$ \langle \phi, u \rangle = \int_I (\phi(t), u(t))_{L^2(\mathbb{R}^n)}\, dt = \int_{I \times \mathbb{R}^n} \phi U\, d(x, t). $$ Furthemore, for every multi-index $\alpha$, there is a function $U_\alpha \in L^2_{\text{loc}}(I \times \mathbb{R}^n)$, such that for any $\phi \in C_0^\infty (I \times \mathbb{R}^n)$, $$ \int_{I \times \mathbb{R}^n} \partial^\alpha \phi U\, d(x, t) = (-1)^{|\alpha|}\int_{I \times \mathbb{R}^n} \phi U_\alpha \, d(x, t), $$ i.e. $U$ is weakly $k$-times differentiable in $x$. What follows is then a somewhat technical (though not super long) proof.

My question: (In brief: Why is this needed?) Intuitively, $U$ should just be $u$ itself. My question is, why isn't it? Should I view this as a technical lemma that basically lets us work with $U$ instead of $u$, since $u$ is a priori a "stranger" object, in that it actually takes a.e. $t \in I$ and spits out a function $u(t) \in H^k(\mathbb{R}^n)$? Is the issue here a kind of type-mismatch, in this sense? In that we are just trying to define a more easy-to-work-with object $U$ which is more or less $u$? Or is there something more going on that I'm not seeing?

Answer 1

Should I view this as a technical lemma that basically lets us work with $U$ instead of $u$, since $u$ is a priori a "stranger" object, in that it actually takes a.e. $t \in I$ and spits out a function $u(t) \in H^k(\mathbb{R}^n)$?

Probably this. There is danger afoot when you consider $L^\infty$ Bochner spaces: tread carefully. Consider the following example (exercise 1.11 in Robinson, Rodrigo & Sadowski's The Three-Dimensional Navier-Stokes Equations)

Show that the function defined on $(0,1) \times(0,1)$ by $$ f(x, t)=\left\{\begin{array}{ll} 0 & x<t \\ 1 & x \geq t \end{array}\right. $$ which is clearly an element of $L^{\infty}((0,1) \times(0,1))$, is not an element of $L_{t}^{\infty}\left((0,1) ; L_{x}^{\infty}(0,1)\right) .$ [Hint: define $F:(0,1) \rightarrow L_{x}^{\infty}(0,1)$ by setting $F(t)(x)=f(x, t)$, and show that $F^{-1}\left(B_{1 / 2}(F(t))\right)=\{t\}$ for any $t \in(0,1)$ Hence find an open subset $U$ of $L_{x}^{\infty}(0,1)$ such that $F^{-1}(U)$ is not measurable.] (This example is due to Juan Arias de Reyna.)

Not directly related, but 'switching the argument around' reminded me also of the following (Paraphrasing of example 1.3.1 in Hamilton's The Inverse Function Theorem of Nash and Moser.) Let $C^k_p$ denote the Banach space of $C^k$ $2\pi$-periodic functions, and consider $$\big( T (t)f\big)(x) = f(x+t)$$ Then

1. $T:\mathbb R\times C^0_p\to C^0_p$ is jointly continuous;
2. $T : \mathbb R \to L(C^0_p, C^0_p)$ is not continuous (wrt operator norm topology);
3. $T:\mathbb R \to L(C^1_p, C^0_p)$ is continuous;
4. $T:\mathbb R \to L(C^1_p, C^1_p)$ is not.

The lack of continuity in 2 stems from the lack of equicontinuity of the space all functions in $C^0_p$. So obvious sounding things still need to be checked.

Another (even less related) weird thing: if $X$ is infinite dimensional, then simple functions are not dense in $L^\infty(0,T; X)$...if you want to understand Bochner spaces carefully, the first few videos of this youtube lecture series by Amenta (which I thoroughly enjoyed) are relevant.