To fix the ideas, let's work on the flat periodic torus $\mathbf{T}^d:=\mathbf{R}^d/\mathbf{Z}^d$.


My question is the following.



> 
> Does there exist an infinite dimensional Banach (sub-)algebra $A \subset \mathscr{C}^\infty(\mathbf{T}^d)$ stable by derivation ?
> 
> 
> 


I am almost certain that the answer is no, but I do not manage to prove it.


My original interrogation concerns nonlinear evolution equations of the form $\partial\_t u = \Delta\_x F(u)$, where $F$ is some (non degenerative) nonlinearity. The only method I know to prove that starting from $u\_0\in\mathscr{C}^\infty(\mathbf{T}^d)$ such equations have at least a local smooth solution is by bootstraping Sobolev-like estimates (differentiating the equation itself). I wonder how much one can use the (Banach-valued) Picard-Lindelöf Theorem, starting from $u\_0$ in some algebra as above to produce directly a non-trivial smooth (local) solution. It is the Fréchet structure of $\mathscr{C}^\infty(\mathbf{T}^d)$ which forbids to use this type of Picard-based theorem.


But I have a strong feeling that if such a proof is possible, it would have been written somewhere !


**EDIT** : several comments below made me realize that my question was not precise enough, sorry for that. I wanted to simplify as much as possible but I think it will be more clear if I explicit what I precisely need. Consider the following non linear system of PDE (named " cross-diffusion system " in the literature) on $\mathbf{R}\_{\geq 0}\times \mathbf{T}^d$


\begin{align\*}
\partial\_t u &= \Delta\_x((d\_1+a\_{12}v)u),\\
\partial\_t v &= \Delta\_x((d\_2+a\_{21}u)v).
\end{align\*}


The parameters $d\_i$ and $a\_{ij}$ are all positive numbers, the unkowns being $u$ and $v$.


My goal is to have the most simple proof of existence of one local (non constant) solution, starting from a smooth initial data (I want it to be smooth all the way down to $t=0$). Existence of such smooth solution can be achieved differentiating the equation and iterating [a big hammer](https://www.math.uzh.ch/amann/files/nlq.pdf) (Theorem 17.1) but this is precisely what I would like to avoid. This type of results give more information than I need : existence is obtained in Sobolev spaces, together with a criterion of explosion in these spaces. To be completely honest I also prefer to avoid the use of a long article that I obviously will never read in detail in my whole academic life.


Let me now precise my question in this very setting. I am wondering if there exists an infinite dimensional algebra $A\subset \mathscr{C}^\infty(\mathbf{T}^d)$ equipped with a norm $\|\cdot\|\_A$ such that $(A,\|\cdot\|\_A)$ is Banach, with $\Delta\_x(A)\subset A$ (and thus, in particular, $\Delta\_x : A\rightarrow A$ is continuous for $\|\cdot\|\_A$). If such a structure were to exist, then the previous system could be written as $U'(t) = F(U(t))$, where $U:=(u,v)\in A\times A$ and $F:A\times A\rightarrow A\times A$ is locally lipschitz : we would get in this way existence of local smooth solutions by the Picard-Lindelöf Theorem.


In line with suggestions below : note that such a strategy will produce a solution in a neighboorhood $(-\varepsilon,\varepsilon)$ of $t=0$. The simple case of the heat equation ($d\_2=a\_{ij}=0$) suggests such an algebra $A$ is necessarily contained in $\mathscr{C}^\omega(\mathbf{T}^d)$ (analytic regularity), for the heat equation cannot be solved backward otherwise.



Banach algebra of smooth functions
I am not 100% sure that I am correctly interpreting your hypotheses, so let me write something which is more cumbersome but which hopefully applies to the situation you have in mind.


FACT 1. Let $A$ be a commutative Banach algebra and let $D:A\to A$ be a continuous derivation, i.e. a bounded linear map satisfying $D(ab)=aD(b)+D(a)b$ for all $a,b\in A$. Then every element in $D(A)$ is quasinilpotent, i.e. has spectral radius zero; equivalently, if we adjoin an identity to $A$ then $1+D(a)$ is invertible for all $a\in A$.


Proof: This is the Singer–Wermer theorem, whose proof can be found in various sources such as Bonsall and Duncan's book, and probably Rickart's, and almost certainly in the Big Book of Dales. I'll sketch the standard proof (a more elementary proofs was later given by Sinclair, IIRC).


The key observation is that the power series $\exp(D) :A \to A$ is a homomorphism, so replacing $D$ with $\lambda D$ for a complex scalar $\lambda$ one obtains a one-complex-parameter group of automorphisms of $A$. Composing with any fixed homomorphism $\varphi:A\to\mathbb C$, one obtains a family of homomorphisms $\psi\_\lambda :=\varphi \circ \exp(\lambda D) : A\to \mathbb C$, and standard book-keeping shows that $\lambda\mapsto \psi\_\lambda$ is holomorphic as a function $\mathbb C \to A^\*$. But homomorphisms $A\to \mathbb C$ must all have norm $\leq 1$, so $\lambda\mapsto\psi\_\lambda$ is a bounded entire function and hence constant by Liouville. By differentiating, one can deduce that for all $a\in A$ we have $\varphi(D(a))=0$; since this holds for all characters $\varphi$, Gelfand theory now tells us that $D(a)$ has spectral radius zero. **QED.**


REMARK 2. In fact, the previous result remains true without the assumption that the derivation $D$ is continuous. This is (much)$^{10}$ harder in my opinion and was a result of M. P. Thomas in Acta Mathematica in the 1980s.


COROLLARY 2. Equip $C^\infty({\mathbb T}^d)$ with its usual Fr'echet algebra structure, let $A$ be a commutative Banach algebra, and let $\Psi: A\to C^\infty({\mathbb T}^d)$ be an injective continuous homomorphism. Suppose that $\partial\_j \Psi(A) \subseteq \Psi(A)$ for $j=1,\dots, d$ where $\partial\_j$ is the partial derivative in the variable $\theta\_j$. Then $\Psi(A)$ consists only of constant functions, and so $A$ is at most one-dimensional.


Proof. Let $D\_j:A\to A$ be the unique linear map satisfying $\Psi\circ D\_j = \partial\_j \circ\Psi$. Since $\partial\_j$ is continuous (for the Frechet algebra topology) and $\Psi$ is continuous, the closed graph theorem can be used to show that $D\_j:A\to A$ is continuous. Clearly $D\_j$ is also a derivation, since $\partial\_j$ is and since $\Psi$ is an injective homomorphism.


Let $a\in A$. By Fact 1 (the Singer–Wermer theorem), $D\_j(a)$ has spectral radius zero for each $j$, and hence $\partial\_j\Psi(a)$ has spectral radius zero for each $j$. But the only element of $C^\infty({\mathbb T}^d)$ with spectral radius $0$ is the zero function. We conclude that $\Psi(a)$ has all partial derivatives equal to zero and by smoothness $\Psi(a)$ must be constant.
**QED.**


