If a distribution $f$ 's all $k$ order weak derivatives are zero on $\mathbb R^n$,Then $f$ must be the polynomal with at most $k-1$ order?

Question

When $k=1$ it must be right,see This But one can't extend this method to other situation. May someone give me some hint or solution? Thanks in advance.

Answer 1

One option is to proceed by induction on $k$. Let's illustrate the induction step for $k=2,3$, the general case is similar with more notation.

So suppose all second derivatives of $f$ vanish and we have only two variables. By the hypothesis and the case $k=1$, we know that $f_x \equiv a$ and $f_y \equiv b$ are both constants. Thus $g=f-ax-by$ satisfies $g_x=g_y=0$, so $g$ is a constant, whence $f$ is linear.

Now for $k=3$, with two variables, Suppose all third partials of $f$ vanish. This hypothesis implies that $f_{xx} \equiv c$ and $f_{xy} \equiv f_{yx} \equiv m$ and and $f_{yy} \equiv \ell$ are all constant. Then $h:=f-cx^2/2-mxy-\ell y^2/2$ has all its second partials vanish, so $h$ is linear, whence $f$ is a polynomial of degree at most 2.