2

Let $E, F$ be Fréchet spaces and $f: E \to F$. We say that $f$ is differentiable at $x$ in the direction of $h \in E$ if the following limit exists. \begin{eqnarray} Df[x](h) := \lim_{t \to 0}\frac{f(x+th)-f(x)}{t} \tag{1}\label{1} \end{eqnarray} If $Df[x](h)$ exists for all $x, h \in E$ we say that $f$ is differentiable. Also, $f$ is called $C^{1}$ if $Df: E\times E \to F$ is continuous.

Take $E= \mathcal{S}(\mathbb{R}^{d})$ and $F = \mathbb{C}$ and suppose $f: \mathcal{S}(\mathbb{R}^{d}) \to \mathbb{C}$ is $C^{1}$. If $x \in E$ is fixed, the map $Df[x]: \mathcal{S}(\mathbb{R}^{d}) \to \mathbb{C}$ given by $h \mapsto Df[x](h)$ is continuous by hypothesis. Also, it can be proved that this map is linear, so that $Df[x] \in \mathcal{S}'(\mathbb{R}^{d})$.

It is possible that $Df[x]$ is the distribution induced by some function, say, $\delta f/\delta x$, so that: \begin{eqnarray} Df[x](h) = \int \frac{\delta f}{\delta x}(x)h(x)dx \tag{2}\label{2} \end{eqnarray} However, to me there is nothing that assures $Df[x]$ to be in fact induced by some kernel. However, it is very common (specially in the physics literature) to write $Df[x]$ as in (\ref{2}).

I'd like to know if I'm missing something here. Can $Df[x]$ always be written in terms of a kernel, like in (\ref{2}) or is it just a matter of notation?strong text

JustWannaKnow
  • 4,075
  • 3
    You may have hypotheses here that you don't need, but, also, not all Frechet spaces behave as Schwartz spaces do. The latter are "nuclear Frechet", guaranteeing that there is a "Schwartz kernel" for maps $S\to S'$. The proof is not trivial. Is this the sort of thing you're asking about? In particular, the kernel $K(x,y)=\delta'(x-y)$ (with the latter expression suitably interpreted) is a/the kernel for differentiation. Somewhat anti-climactic. – paul garrett Apr 08 '20 at 23:46
  • Yes! Perfect! Do you have any reference on this result? I'd like to, at least, know the statement which garantees the existence of "Schwartz kernel". – JustWannaKnow Apr 08 '20 at 23:48
  • 2
    Googling "Schwartz Kernel Thm" gives lots of hits, including one of my own functional analysis note fragments. There's no shortage of references... – paul garrett Apr 09 '20 at 00:13
  • 1
    Seconding Paul's comment above, trying to be more general and work in the setting of Frechet spaces is a red herring. One needs to work with more concrete spaces of distributions. The proof of the (strong version) of the Kernel Thm is not that hard and it is sketched in https://math.stackexchange.com/questions/3512357/understanding-the-proof-of-schwartz-kernel-theorem/3512932#3512932 – Abdelmalek Abdesselam Apr 09 '20 at 14:06
  • @AbdelmalekAbdesselam thanks! I'm only using Fréchet spaces to define derivatives but then I'm concentrating on $\mathcal{S}(\mathbb{R}^{d})$. If I defined the derivative directly on $\mathcal{S}(\mathbb{R}^{d})$ nothing would change, this is why I took a more general path. – JustWannaKnow Apr 09 '20 at 14:22
  • My point here whether there was an integral kernel. Schwartz KT states the existence of a kernel, but nothing is said about an integral kernel. As far as I understand, the integral representation is only a matter of notation. – JustWannaKnow Apr 09 '20 at 14:27
  • 1
    correct. It depends what one means by "kernel" a true function (analogue of a matrix or tensor seen as a function of the indices) or a generalized function as in the Kernel Thm. – Abdelmalek Abdesselam Apr 09 '20 at 14:33

0 Answers0