Let $S(\mathbb R^d)$ be the space of Schwartz functions and $S'(\mathbb R^d)$ be the space of tempered distributions (the continuous linear functionals from $S(\mathbb R^d)$ to $\mathbb R$) endowed with the weak topology.

Assume that $I \subset \mathbb R$ is a compact interval and that $u : I \rightarrow S'(\mathbb R^d)$ is a family of tempered distributions such that $$\langle u(t), \phi \rangle \in L^1(I) \ \forall \phi \in S(\mathbb R^d)$$

Does it follow that $\langle u(t), \phi(t, \cdot) \rangle \in L^1(I) \ \forall \phi \in S(\mathbb R^{1+d})$ ?

I've already asked this question in mathoverflow, but I've got no answer so far. This question is also related to this other question of mine.

I'm asking this question because it is proved in Chapter 4 of "Eléments de distributions et d'équations aux dérivées partielles" by Claude Zuily that if

$$\langle u(t), \phi \rangle \in C^0(I) \ \forall \phi \in S(\mathbb R^d)$$

then $\langle u(t), \phi(t, \cdot) \rangle \in C^0(I) \ \forall \phi \in S(\mathbb R^{1+d})$ as well. It is a consequence of Banach-Steinhaus. For the sake of completeness, I'm rewriting the proof for the $C^0$ case below.

I was wondering if we can do something similar for $L^1$ space. Any reference is welcomed.

One last remark : Observe that $\langle u(t), \phi(t, \cdot) \rangle$ is always the pointwise limit of $L^1(I)$ functions, hence it is Lebesgue measurable. Indeed, if $\phi(t,x) = a(t)b(x)$ for $a \in S(\mathbb R)$, $b \in S(\mathbb R^d)$, then

$$\langle u(t), \phi(t, \cdot) \rangle = \langle u(t), b \rangle a(t) \in L^1(I)$$

For a general $\phi(t,x)$, it can be shown (see here for example) that it is the limit in $S(\mathbb R^{1+d}$) of a sequence $\phi_n(t,x)$, where each $\phi_n$ is a finite linear combination of elements of the form $a(t)b(x)$, $a \in S(\mathbb R)$, $b \in S(\mathbb R^d)$.

Proof of Claude Zuily for the $C^0$ case :

We prove continuity of $\langle u(t), \phi(t, \cdot) \rangle$ at $t_0 \in I$. Let $(t_n)_{n=1}^{\infty} \subset I$ converge to $t_0$. We define $u_j := u(t_j)$ and $\phi_j := \phi(t_j, \cdot)$. By assumption, $u_j$ converges to $u(t_0)$ in $S'(\mathbb R^d)$. In particular $$\sup_{j \in \mathbb N} | \langle u_j, \psi \rangle | < +\infty \ \forall \psi \in S(\mathbb R^d)$$ Banach-Steinhaus theorem implies that there exists a continuous semi-norm $\rho : S(\mathbb R^d) \rightarrow \mathbb R$ such that $$| \langle u_j, \psi \rangle | \leq \rho(\psi) \ \forall j \in \mathbb N \ \forall \psi \in S(\mathbb R^d)$$ Moreover, it is easy to check that $\phi_j$ converges to $\phi(t_0, \cdot)$ in $S(\mathbb R^d)$.

Hence, $$| \langle u_j, \phi_j \rangle - \langle u(t_0), \phi(t_0, \cdot) \rangle | \leq | \langle u_j, \phi_j - \phi(t_0, \cdot) \rangle | + | \langle u_j, \phi(t_0, \cdot) \rangle - \langle u(t_0), \phi(t_0, \cdot) \rangle| $$

and everything goes to zero as $j \to +\infty$.

  • 2
    $\begingroup$ The idea that I have is to try to establish that the distributions $u(t)$ all share the same order (or lower) for almost every $t$. If you take distributions of arbitrary order in $u(t)$, you can have bad things I think, see math.stackexchange.com/questions/352392/… Then, if this is true, maybe there is some hope $\endgroup$ Jul 4, 2021 at 8:57


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.