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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 (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.

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    $\begingroup$ The answer is: yes. Take the algebra of constant functions. $\endgroup$ – Zero Nov 10 '20 at 9:22
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    $\begingroup$ Hum, you're right. I will specify that I want it to be infinite dimensionnal. The goal was to be able to produce non trivial solutions ! Thanks for the correction ! $\endgroup$ – Ayman Moussa Nov 10 '20 at 9:28
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    $\begingroup$ I assume that the answer will be no i you assume that the algebra is rich enough (at least that might be a criterion to kick out something). I would think that something like the classical arguemnt why $C^\infty (\mathbb{R},\mathbb{R})$ is not a Banach space making differentiation continuous would work: n this case one finds that differentiation is a continuous linear operator with a countable unbounded family of eigenvalues (the exponentials). Perhaps this could be adapted... $\endgroup$ – Alexander Schmeding Nov 10 '20 at 9:30
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    $\begingroup$ My suspicion is that 'being stable under differentiation' isn't compatible with how fast the Fourier coefficients have to decay for $A\subset C^\infty$ (thinking about the $d=1$ case). Stability under multiplication isn't inconsistent with $A\subset C^\infty$ (some Beurling algebras and some 'Dales-Davie' algebras consist entirely of $C^\infty$ functions). Shilov might be a good name to search if you're looking for results like this (although you might know that already). $\endgroup$ – DCM Nov 10 '20 at 20:09
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    $\begingroup$ With your assumptions, you still allow for the subalgebra of functions that do not depend, say on $x_1$. $\endgroup$ – Romain Gicquaud Nov 10 '20 at 21:40
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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.

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  • $\begingroup$ Maxime Breden (who's not able to post for some reason) suggests to apply directly your FACT 1 to the hypothetic algebra A of my post. If instead of $\Delta_x(A)\subset A$ we have $\partial_j(A)\subset A$ then it works (which seems fairly reasonnable), so I accept your answer : thanks ! $\endgroup$ – Ayman Moussa Nov 12 '20 at 10:45
  • $\begingroup$ @AymanMoussa Yes, as long as you want the derivation to be continuous on A then the Singer-Wermer theorem (Fact 1) forces $A$ to be one-dimensional or the zero algebra. The only reason I stated Corollary 2 in the form I did was because I wasn't sure if you needed the derivation to be continuous, and the easiest way to force this to hold was to require the inclusion $A\hookrightarrow C^\infty$ to be continuous. $\endgroup$ – Yemon Choi Nov 12 '20 at 18:41
  • $\begingroup$ @AymanMoussa In fact, Thomas's result tells us that we don't even need to assume continuity of the derivation, but I prefer to avoid appealing to that result since it is very hard and since continuity of the derivation is usually "clear from inspection" in most cases of interest. $\endgroup$ – Yemon Choi Nov 12 '20 at 18:42
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For those who, like me, are not acquainted too much with algebras and Gelfand theory, here is a quick translation of the proof given by Yemon up there. When I tried reading the proof from the textbooks given above, all of them seemed to use AC in a way or another (wether it is for Krull's theorem or Hahn-Banach's extension). Since I am quite reluctant about AC, I followed thoroughly the lines of the proof to understand the argument in the very setting I was interested in (smooth functions on the flat torus). Unsurprisingly, AC is not needed here as the setting is extremely specific compared with the scope covered by the Singer–Wermer theorem.

Nothing's new in the following lines, I just present the argument using crucially only two results :

  • Normal convergence of a series implies convergence in a Banach algebra
  • Liouville's theorem for entire functions from $\mathbf{C}$ to itself

PROPOSITION : In a Banach algebra $A$ any algebra homomorphism $\chi : A\rightarrow\mathbf{C}$ is continuous, with operator norm not exceeding $1$. Such maps are called the characters of $A$.

Proof : Fix $\chi:A\rightarrow\mathbf{C}$ an algebra homomorphism. For $f \in A$ and $z\in \mathbf{C}$ whose modulus is larger than $\|f\|$, the sum $\sum_{n\in\mathbf{N}} \frac{f^{n+1}}{z^n}$ converges normally towards an element of $g$ solving the implicit equation $z g=zf+fg$. Applying $\chi$ to the previous equality, we get $\chi(g)(z-\chi(f))=z\chi(f)$ which forbids $\chi(f)= z$ : we have proven $|\chi(f)|\leq \|f\|$. QED

THEOREM : Consider a subalgebra $A\subset \mathscr{C}^\infty(\mathbf{T}^d)$ stable by differentiation. If there exist a norm $\|\cdot\|$ making $A$ complete and for which differentiation is continuous, then $A$ solely contains constant functions.

Proof : We prove the result only for $d=1$ ; the arguments below can then be used for each partial derivative operator in dimension $d>1$.

Since the linear map $f\mapsto f'$ is assumed continuous for the hypothetic norm $\|\cdot\|$, we have $M>0$ such that for all $k\in\mathbf{N}$, $\|f^{(k)}\|\leq M^k \|f\|$. In particular for all $z\in\mathbf{C}$ the taylor expansion expression \begin{align*} \text{TE}_z(f) := \sum_{k\in\mathbf{N}} \frac{z^kf^{(k)}}{k!}, \end{align*} defines a vector space endomorphism of the Banach algebra $A$, by normal convergence. By Leibniz formula and sommability we have \begin{multline*} \text{TE}_z(fg) = \sum_{k\in\mathbf{N}} \frac{z^k (fg)^{(k)} }{k!} =\sum_{k\in\mathbf{N}} \frac{z^k}{k!}\sum_{\ell=0}^k \begin{pmatrix}k\\\ell\end{pmatrix}f^{(\ell)}g^{(k-\ell)}\\ =\sum_{k\in\mathbf{N}} \sum_{\ell=0}^k \frac{z^\ell f^{(\ell)}}{\ell!}\frac{z^{k-\ell}g^{(k-\ell)}}{(k-\ell)!} = \text{TE}_z(f)\text{TE}_z(g). \end{multline*} The operators $\text{TE}_z$ are thus algebra endomorphisms.

For all $x\in\mathbf{T}$, the evaluator $\chi_x:f\mapsto f(x)$ is a character of $A$. For all $z\in\mathbf{C}$, $\chi_x \circ \text{TE}_z$ is therefore also a character of $A$. On the one hand the previous proposition implies continuity of $\chi_x$ and therefore \begin{align*} \chi_x \circ \text{TE}_z(f) = \sum_{n\geq 0} \frac{z^n f^{(n)}(x)}{n!}. \end{align*} On the other hand, the same proposition invoked for the character $\chi_x\circ \text{TE}_z$ ensures that the previous sum is bounded independently of $x$ and $z$ by $\|f\|$ : Liouville's theorem shows that this expression does not depend on the variable $z$ and therefore $f^{(n)}(x)=0$ for $n\geq 1$. The previous being valid for any $x\in \mathbf{T}$, we obtain $f'=0$ for all $f\in A$. QED

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