Prove that if $D^kf=0$ then $f$ is polynomial

Question

Let $E\subseteq\mathbb{R}^n$ an open connected set and let $f:E\rightarrow \mathbb{R}$ a $k+1$ times differentiable function such that $D^kf=0$. Show that $f$ is a polynomial in $x_1,...,x_n$ at a degree of at most $k$.

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For any $\pi\in\{1,...,n\}^n$ we're given that $\frac{\partial^{k+1}f}{\partial x_{\pi (1)}...\partial x_{\pi (k+1)}}=0$. Thus $\frac{\partial^{k}f}{\partial x_{\pi (1)}...\partial x_{\pi(k)}}=c_k$ and by induction $f=\partial^0 f=x_{\pi(1)}(x_{\pi(2)}(...(x_{\pi(k)}c_k+c_{k-1})...)+c_1)+c_0$ for some $c_i\in\mathbb{R}, \forall 1\leq i \leq n$. I feel that this proof doesn't hold but I don't know where is the mistake and what is the right way. Thanks

Answer 1

Another alternative proof (which I guess is worng as well because I don't use the fact that $E$ is open and connected):

From Taylor's formula we know that $$ \\ f(x_1,...,x_n)=\sum_{i_1=1}^\infty ...\sum_{i_n=1}^\infty \frac{(x_1-a_1)^{i_1}\cdot...\cdot(x_n-a_n)^{i_n}}{i_1!\cdot...\cdot i_n!}\cdot\frac{\partial ^{i_1+...+i_n}f}{\partial x_1^{i_1}...\partial x_n^{i_n}}(a_1,...,a_n) \ $$ Let $0\leq i_1,...,i_n\leq k$. If $\sum_{j=1}^n i_j>k$ then $\frac{\partial ^{i_1+...+i_n}f}{\partial x_1^{i_1}...\partial x_n^{i_n}}(a)=0$. Otherwise, $\frac{(x_1-a_1)^{i_1}\cdot...\cdot(x_n-a_n)^{i_n}}{i_1!\cdot...\cdot i_n!}$ is polynom of a degree at most $k$, and $\frac{\partial ^{i_1+...+i_n}f}{\partial x_1^{i_1}...\partial x_n^{i_n}}(a_1,...,a_n)$ is a constant linear tranformation. Thus $f(x_1,...,x_n)$ is a polynom of degree at most $k$.

Answer 2

Abridged solution. Proceed by induction on $k.$ The case $k = 1$ is very standard and often appears in books. The aim is to show $f$ is constant. It goes like this, consider a "base point" $a \in E$ and the set $\mathrm{X}$ of points in $x \in \mathrm{E}$ such that $f(x)=f(a).$ Then $\mathrm{X}$ is closed in $\mathrm{E}$ as $f$ is continuous. Now, $\mathrm{X}$ is also open in $\mathrm{E}$ for if $x \in \mathrm{X}$ then there is a small enough ball centred at $x$ contained in $\mathrm{E}$ and for such ball the mean value theorem (applied to the line segment between $x$ and another point in the ball) gives that $f$ is constant in such ball, so such ball is subset of $\mathrm{X}$ and then $\mathrm{X}$ is open as claimed. Being $\mathrm{E}$ connected, $\mathrm{X} = \mathrm{E}$ and the base case $k = 1$ is proved.

For the general case, assume $f$ satisfies $f^{(k+1)}=0$ on $\mathrm{E}.$ The case $k = 1$ shows that $f^{(k)}$ is a constant $k$-linear function on $\mathrm{E},$ call it $M.$ We show that $\varphi:x \mapsto f(x) - \dfrac{1}{k!}M(x, \ldots, x)$ satisfies $\varphi^{(k)}=0$ on $\mathrm{E}$ and hence, by induction, $\varphi$ is a polynomial function of degree $\leq k-1$ and then $f$ is a polynomial function of degree $\leq k.$ By standard results about derivatives of multilinear functions and the chain rule, we get at once that $\varphi^{(k)}=f^{(k)}-\dfrac{1}{k!} k!M = 0.$ Q.E.D.

Answer 3

Note that $f$ is smooth since if $D^k f = 0$ then $D^l f = 0$ for $ l \ge k$.

Fix some $x_0 \in E$, and suppose $C$ is an open convex set containing $x_0$ and contained in $E$. Then Taylor's theorem shows that if $y \in C$ then we have $f(y) = \sum_{|\alpha| \le k-1} {1 \over \alpha !} D^\alpha f(x_0) (y-x_0)^\alpha$.

Let $p(y) = \sum_{|\alpha| \le k-1} {1 \over \alpha !} D^\alpha f(x_0) (y-x_0)^\alpha$ be the polynomial, note that $p$ is defined everywhere and has maximum degree $k-1$. We wish to show that $f=p$ on $E$.

As an aside, note that if $q$ is a polynomial and $q(x) = 0$ for $x$ in some open set then $q=0$.

Pick some $x_n\in E$ and note that since $E$ is open and connected there are $x_k \in E$ such that the straight line curve $x_0 \to x_1 \to \cdots \to x_n$ is contained in $E$.

Since the segment $[x_0,x_1] \subset E$, there is some open convex set $C_1$ such that $[x_0,x_1] \subset C_1 \subset E$ and we have $f=p$ on $C_1$.

Now let $p'$ be the Taylor expansion based at $x_1$. As above, there is some open convex set with $[x_1,x_2] \subset C_2 \subset E$ and we have $f=p'$ on $C_2$.

Since $x_1 \in C_1 \cap C_2$, we see that $p=p'$ on an open set, and hence $p'=p$ everywhere.

It is clear that we can repeat this process to see that $f(x_n) = p(x_n)$.

Since $x_n \in E$ was arbitrary, we see that $f=p$ on $E$.