A non-trivial derivation on $C^{k}(\mathbb{R})$ for $k\geqslant 1$?

Question

Recall that a derivation on a commutative algebra $A$ is a linear operator $D:A\to A$ which satisfies the Leibniz rule for products $D(fg)=fDg+gDf$. The standard differentiation $f\mapsto f'$ is certainly a derivation on $C^{\infty }(\mathbb{R})$ and it is known that it is essentially the only one because any derivation is of the form $f\mapsto hf'$ with a $h\in C^{\infty }(\mathbb{R})$. But these are not derivations on $C^{k}(\mathbb{R})$ for $k\geqslant 1$ as $f'\notin C^{k}(\mathbb{R})$ for some $f\in C^{k}(\mathbb{R})$. I suspect there is only the trivial one. Is this true? And if so, how can I prove that? I know it is true for $C^{0}(\mathbb{R})$ (proof).

Answer 1

The following proof is based on Theorem 3.1 from the book of König and Milman: Operator Relations Characterizing Derivatives, Birkhäuser (2018):

Theorem: Let $k \in \mathbb{N}$ and let $D:C^k(\mathbb{R}) \to C(\mathbb{R})$ be a mapping with the property $$ \forall f,g \in C^k(\mathbb{R}): ~ D(f\cdot g)=Df \cdot g+f\cdot Dg. $$ Then there exist functions $c,d \in C(\mathbb{R})$ such that $$ \forall f \in C^k(\mathbb{R}): ~ Df = c\cdot f\cdot \log|f| + d \cdot f'. $$

Now, if $D$ is in addition linear, then inserting the constant function $g=e$ yields $$ (2e(\log(2)+1))\cdot c=(2e\log(2e))\cdot c = D(2g)=2Dg = (2e\log e)\cdot c = (2e)\cdot c, $$ hence $c=0$. If in adddition $D(C^k(\mathbb{R})) \subseteq C^k(\mathbb{R})$, then inserting $h(x)=x$ yields $Dh=d \in C^k(\mathbb{R})$. Now $$ \forall f \in C^k(\mathbb{R}): ~ Df=d\cdot f' \in C^k(\mathbb{R}). $$ If $d(x_0) \not= 0$ for some $x_0 \in \mathbb{R}$ choose $f \in C^k(\mathbb{R})$ such that $f^{(k)}$ is not differentiable in $x_0$. Then $df'$ is in $C^{k-1}(\mathbb{R})$ but not in $C^k(\mathbb{R})$, a contradiction. Summing up, $D=0$.

Answer 2

The fact that no such derivation exists is a special case of theorem 3.1 in Operator Relations Characterizing Derivatives by König and Milman.

Theorem 3.1. Let $k\geq0$ and $I\subset\mathbb{R}$ be open. The general solution to the operator equation $$T[fg]=T[f]g+fT[g]$$ for $T:C^k(I,\mathbb{R})\rightarrow C^0(I,\mathbb{R})$ is given by $$T[f]=\begin{cases}cf\ln|f|+df' & k > 0\\ cf\ln|f| & k=0\end{cases}$$ for any $c,d\in C^0(I,\mathbb{R})$.

A derivation $D$ on $C^k(\mathbb{R})$ is a linear operator satisfying the equation above with image in $C^k(\mathbb{R})$. Linearity forces $c=0$. On the other hand, having image in $C^k(\mathbb{R})$ also forces $d=0$ (as we will see). We can prove a version of this theorem specifically for derivations quite easily using similar ideas to those presented by König and Milman. In what follows, let $D:C^k(\mathbb{R})\rightarrow C^k(\mathbb{R})$ be a derivation with $k>0$.

Lemma 1. If $f_1,f_2\in C^k(\mathbb{R})$ agree on an open set $I\subseteq \mathbb{R}$, then $D[f_1],D[f_2]$ agree on $I$ as well.

For any open $I\subseteq\mathbb{R}$, $p\in I$, and $f_1,f_2\in C^k(\mathbb{R})$ that agree on $I$, we may choose a $g\in C^k(\mathbb{R})$ with $g(p)=1$ and $\text{supp}(g)\subset I$. Then, since $(f_1-f_2)g=0$, the Leibniz rule yields $$(D[f_1]-D[f_2])g+(f_1-f_2)D[g]=0$$ Evaluating at $p$, we get $D[f_1](p)-D[f_2](p)=0$. Since $p$ was arbitrary, $D[f_1]$ and $D[f_2]$ also agree on $I$. $\square$

Lemma 2. $D$ is an order-$k$ linear differential operator.

Let $f\in C^k(\mathbb{R})$, $p\in \mathbb{R}$, and $q\in \mathbb{R}[x]$ be the degree $k$ Taylor polynomial of $f$ at $p$. Define $$g(x)=\begin{cases}f(x)& x\leq p\\ q(x) & x\geq p\end{cases}$$ By lemma 1, $D[g](x)=D[f](x)$ for $x<p$ and $D[g](x)=D[q](x)$ for $x>p$. By continuity, $D[f](p)=D[g](p)=D[q](p)$. Thus, $D[f](p)$ depends only on $f(p),\ldots,f^{(k)}(p)$. Since $p$ was arbitrary and $D$ is linear, $D[f]$ is linear in $f,\ldots,f^{(k)}$. $\square$

Lemma 3. $D[f]$ is of the form $df'$ for some $d\in C^k(\mathbb{R})$.

In light of lemma 2, we may write $$D[f]=\sum_{n=0}^k\alpha_nf^{(n)}$$ for some functions $(\alpha_n)$. Let $e_p^m$ be the map $x\mapsto (x-p)^m$. Then $$D[e_0^m]=\sum_{n=0}^m\frac{m!}{(m-n)!}\alpha_ne_0^{m-n}\in C^k(\mathbb{R})$$ Induction on $m$ shows that $\alpha_n\in C^k(\mathbb{R})$ for each $n$.

We know $\alpha_0=0$, as $D[1]=D[1^2]=2D[1]$ gives $D[1]=0$. On the other hand, for $m>1$ we have $$m!\alpha_m(p)=D[e_p^m](p)=me_p^{m-1}(p)D[e_p](p)=0$$ by the power rule (easily derivable from the Leibniz rule). Thus, $\alpha_m=0$ for $m>1$. The only remaining term in the formula for $D[f]$ is $\alpha_1f'$. $\square$

Theorem. $D$ is trivial.

Write $D[f]=df'$ as in lemma 3. If $d$ is not identically $0$, then with the choice of a non-degenerate interval $[a,b]\subset \text{supp}(d)$, we may take $$f(x)=\int_a^x\frac{g(t)}{d(t)}dt \quad\quad x\in[a,b]$$ for some $g$ that is exactly $k-1$ times differentiable in $[a,b]$. We can extend $f$ to $\mathbb{R}$ by defining it outside of $[a,b]$ to be the Taylor polynomial of degree $k$ of the closest endpoint. Then, $df'$ and $g$ are identical on $[a,b]$, so $df'\not\in C^k(\mathbb{R})$. $\square$