More precisely, I ask whether there is a "simple elementary" way to prove that there is a linear homeomorphism $E\to E\times E$ when for $I=[0,1]$ we have $E=C(I)$, the topological vector space of continuous functions $x:I\to\mathbb R$ topologized by the supremum norm. By "simple elementary" I refer to an argument which is approximatively "as simple" as the following allowing to prove $E\cong E\oplus F$ where $F=C_{\rm o}(I)$ is the subspace of $E$ formed by the functions $x$ with $x(0)=0$.
Argument. Define $i_0:E\to E\times F$ by $z\mapsto(x,y)$ where $x(t)=z(\frac 12 t)$ and $y(t)=z(\frac 12(1+t))-z(\frac 12)$. It is trivial to check that $i_0$ is linear, continuous and bijective, and then the open mapping theorem implies that it is a linear homeomorphism.
The above argument in fact shows that there is a split short exact sequence of Banach(able topological vector) spaces: $$ {\rm O}\longrightarrow E\overset i\longrightarrow E\overset p\longrightarrow F\longrightarrow{\rm O} $$ where $i$ is defined by $(iu)(t)=u(2\,t)$ for $0\le t\le\frac 12$ and $(iu)(t)=u(1)$ for $\frac 12 < t\le 1$, and $p$ is defined by $(pv)(t)=v(\frac 12(1+t))-v(\frac 12)$ for $t\in I$. Then ${\rm Im\,}i={\rm Ker\,}p$, and we have the direct sum decomposition $E\cong E\oplus F$ since a topological linear complement of ${\rm Im\,}i$ in $E$ is the set of all $v$ in $E$ with $v(t)=0$ for $0\le t\le\frac 12$. I would like to have a split short exact sequence $$ {\rm O}\longrightarrow E\longrightarrow E\longrightarrow E\longrightarrow{\rm O} $$ by a "simple elementary" argument as above.
My basic motivation behind all this is that I would like to have a representation $C^\infty(\mathbb R)\cong(C^\infty(I))^{\kern.7mm\mathbb N_0}$ without using any sequence space representation of $C^\infty(I)$ by the space $s$ of rapidly decreasing sequences. I am just seaching for any ideas whether this is possible by first considering an overly simplified case.