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](https://arxiv.org/abs/1303.2424) the following question becomes unexpectedly important:



> 
> does $J^1\_{\mathbb R}[0,1]$ have the (classical) [approximation property](https://en.wikipedia.org/wiki/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.)



Does the Banach algebra of jets have the approximation property?
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 $C^0([0,1])$?


