Htykuut

Let $Esubseteqmathbb{R}^n$ an open connected set and let $f:Erightarrow mathbb{R}$ 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 $k$.

For any $piin{1,...,n}^n$ 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 $c_iinmathbb{R}, forall 1leq i leq n$. I feel that this proof doesn't hold but I don't know where is the mistake and what is the right way. Thanks

Abridged solution. Proceed by induction on $k.$ The case $k = 0$ is very standard; 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 $k = 0$ is proved.

For the general case, assume $f$ satisfies $f^{(k+1)}=0$ on $mathrm{E}.$ The case $k = 0$ 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 the result. 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.

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 $x_nin E$ 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 $E$.

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
$$
\ f(x_1,...,x_n)=sum_{i_1=1}^infty ...sum_{i_n=1}^infty frac{(x_1-a_1)^{i_1}cdot...cdot(x_n-a_n)^{i_n}}{i_1!cdot...cdot i_n!}cdotfrac{partial ^{i_1+...+i_n}f}{partial x_1^{i_1}...partial x_n^{i_n}}(a_1,...,a_n)
$$
Let $0leq i_1,...,i_nleq k$. 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 $k$.