To formulate my question I need the construction of the algebra $J^n_M(K)$ of jets of degree $n$ on a compact set $K$ of a smooth manifold $M$. I'll describe it for the simplest case of $M={\mathbb R}$, $K=[0,1]$ and $n=1$. This is done in two steps.

  1. First, we need the construction of the algebra of polynomials of degree 1 with the coefficients in $C[0,1]$. Let us consider the Banach algebra $C[0,1]$ of continuous functions on $[0,1]$ (with the usual sup-norm), and let us endow its cartesian square $C[0,1]^2$ with the multiplication $$ (x_0,x_1)\cdot(y_0,y_1)=(x_0\cdot y_0,x_0\cdot y_1+x_1\cdot y_0),\qquad x_i,y_i\in C[0,1]. $$ and the norm $$ ||(x_0,x_1)||=||x_0||+||x_1||,\qquad x_i\in C[0,1]. $$ After that $C[0,1]^2$ becomes a Banach algebra. If we introduce an "external variable" $\tau$ with the properties $$ a\cdot \tau=\tau\cdot a, \quad \tau^2=0,\qquad a\in C[0,1], $$ then each element $(x_0,x_1)$ of $C[0,1]^2$ can be represented as a sum $$ (x_0,x_1)=x_0+x_1\cdot\tau, $$ and the multiplication of these sums is exactly the multiplication in $C[0,1]^2$, that is why we call $C[0,1]^2$ the algebra of polynomials of degree 1 with the coefficients in $C[0,1]$.

  2. Let us consider now the algebra $C^\infty(\mathbb R)$ of smooth functions on $\mathbb R$, and the map $$ \varphi:C^\infty(\mathbb R)\to C[0,1]^2\quad\Big|\quad \varphi(x)=(x\big|_{[0,1]},x'\big|_{[0,1]}),\qquad x\in C^\infty[0,1] $$
    ($x'$ means the usual derivative, and $\cdot\big|_{[0,1]}$ are the restrictions to $[0,1]$). This is a homomorphism of algebras, hence its image $\varphi(C^\infty[0,1])$ is a subalgebra in $C[0,1]^2$. Let us call its closure the algebra of jets of degree 1 of smooth functions on $\mathbb R$ on the subset $[0,1]$ and denote it by $$ J^1_{\mathbb R}[0,1]=\overline{\varphi(C^\infty(\mathbb R))} $$

Being a closed subalgebra in the Banach algebra $C[0,1]^2$, $J^1_{\mathbb R}[0,1]$ itself is a Banach algebra. It is easy to see that $$ J^1_{\mathbb R}[0,1]\ne C[0,1]^2. $$ For example, we can take the pair of functions $$ x_0(t)=t,\quad x_1(t)=0, \qquad t\in [0,1], $$ and we'll see that there is no $x\in C^\infty(\mathbb R)$ such that $$ ||\varphi(x)-(x_0,x_1)||<\frac{1}{4} $$ (if this $x$ exists, then $||x'\big|_{[0,1]}||=||x'\big|_{[0,1]}-x_1||<\frac{1}{4}$, hence $|x(1)-x(0)|=|\int_0^1x'(t)d t|\le||x'\big|_{[0,1]}||<\frac{1}{4}$, but on the other hand, $||x\big|_{[0,1]}-x_0||<\frac{1}{4}$, so $|x(0)|<\frac{1}{4}$ and $|x(1)-1|<\frac{1}{4}$, and as a corollary, $|x(1)-x(0)|>\frac{1}{2}$).

In what I am doing now the following question becomes unexpectedly important:

does $J^1_{\mathbb R}[0,1]$ have the (classical) approximation property?

(In fact, I need the answer for the general case of $J^n_M(K)$, but I think this special question can clarify everything.)

  • $\begingroup$ This isn't a full answer to your question, but your space reminds me of the algebra $C^k[0,1]$ with a standard norm that makes it a Banach algebra. With that norm $C^k[0,1]$ is isomorphic (i.e. linearly homeomorphic) to to $C[0,1]$. Possibly a similar idea works for your construction. $\endgroup$ – Yemon Choi Nov 12 '16 at 15:16
  • $\begingroup$ Yemon, Sigurd's answer inspires an impression that everything is very simple here... I think there must be a simple trick for bigger $n$ and $\text{dim} M$. If this is so, I think we could delete this question (or move it to MSE). $\endgroup$ – Sergei Akbarov Nov 12 '16 at 15:29
  • $\begingroup$ Yemon, I have a feeling that for $k\ne m$ the space $C^k([0,1]^m)$ is not isomorphic to $C[0,1]$... (I think our doubts must be funny for specialists, so I am sorry. :) $\endgroup$ – Sergei Akbarov Nov 12 '16 at 15:40
  • $\begingroup$ Well $C[0,1]^m$ is isomorphic as a Banach space (but not isometrically so) to $C[0,1]$... $\endgroup$ – Yemon Choi Nov 12 '16 at 15:41
  • $\begingroup$ Yes, actually the question is if the space $C^k([0,1]^m)$ has the approximation property. I am sure, @BillJohnson knows the answer... $\endgroup$ – Sergei Akbarov Nov 12 '16 at 20:41

Is $x_0+\tau x_0' \mapsto (x_0'(\cdot), x_0(0))$ a linear homeomorphism from $J^1$ to $C([0,1])\times \mathbb R$? Wouldn't that imply that $J^1$ inherits the approximation property from $C^0([0,1])$?

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  • $\begingroup$ Hm... It seems, I asked a stupid question... $\endgroup$ – Sergei Akbarov Nov 12 '16 at 14:09
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    $\begingroup$ sigurd, if we increase the dimension (i.e. consider $[0,1]^m$ instead of $[0,1]$) and the degree $n$ of the jets (i.e. take more derivatives), is there a trick like this in this case? $\endgroup$ – Sergei Akbarov Nov 12 '16 at 14:17

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