In this paper, J. Eells defines this notion of $C^r$-smoothness for Banach spaces:
A Banach space $E$ is $C^r$-smooth, $r \geq 0$, if there exists a nontrivial (that is, nonzero) $C^r$ function $\phi : E \rightarrow \mathbb{R}$ with bounded (not necessarily compact) support.
He then claims that the Banach space $$c_0 = \{~(a_1, a_2, a_3, \ldots) : \lim a_n = 0~\},~~\|(a_n)\|_{c_0} = \sup_n |a_n|,$$ is $C^{\infty}$-smooth. My question is: how do we construct the smooth function having bounded support here?
Here's what I've been trying; take your favorite smooth function $\phi : \mathbb{R} \rightarrow \mathbb{R}$ having compact support and such that:
- $\phi \equiv 1$ on [-1/4, 1/4];
- $\phi \equiv 0$ outside [-1, 1].
Now, given $(a_n)_{n \in \mathbb{N}}$ in $c_0$, define:
$$\Phi\big((a_n)_{n \in \mathbb{N}}\big) = \large\prod_{n = 1}^{\infty} \phi(a_n).$$
Note that the product is well-defined, since the terms $a_n$ are eventually all within $[-1/4, 1/4]$, and hence almost all terms in the product are equal to $1$. Moreover, this function has bounded support, since if $\|(a_n)\|_{c_0} > 1$, then at least one $a_n$ has to be greater than $1$ in modulus, and then the whole product is zero.
My problem is that I cannot show that this is differentiable, let alone smooth. Thank you for your input!
Let's name $\psi_n = \phi \circ \pi_n$, with the coordinate projections $\pi_n(a) = a_n$. Then $\pi_n$ is linear and continuous, hence smooth. $\phi$ is smooth, the composition of smooth functions is smooth, hence $\psi_n$ is smooth.
Finite products of smooth functions are smooth, hence for all $N \in \mathbb{N}$ the function
$$\Phi_N(a) = \prod_{n=1}^N \psi_n(a)$$
is smooth.
Smoothness is a local property, so let's fix an arbitrary $a \in c_0$. Choose an $N \in \mathbb{N}$ such that for all $n > N$ we have $\lvert a_n \rvert < \frac18$. On $B_{1/16}(a) = \{b \in c_0 : \lVert b - a\rVert < \frac{1}{16}\}$, we have $\Phi \equiv \Phi_N$, hence $\Phi$ is smooth on a neighbourhood of $a$.