Smooth function vanishing on axis

Question

suppose you have a smooth function $f : \mathbb{R}^{2} \rightarrow \mathbb{R}$ such that $f(x,y) = 0$ if either of the co-ordinates $x$ and $y$ are $0$ (i.e. it vanishes on both of the co-ordinate axes). Does there exist constants $k>0$ and $\epsilon>0$ so that $|f(x,y)| \leq k|x||y|$, when $|(x,y)| \leq\epsilon$?

Answer 1

Since $f$ is smooth and vanishes on the $x-$axis, we have $$f(x,y) = \int_0^y f_y(x,s) ds.\tag{1}$$ Because $f$ vanishes on the $y-$axis, we know that $f_y=0$ on the $y-$axis; so if we restrict to the closed ball $\overline{B(0,1)}$ so that $|f_{xy}|$ is bounded by some constant $k$ then we have $$|f_y(x,y)|=\left|\int_0^xf_{xy}(s,y)ds\right|\le \int_0^xk\,ds =k|x|.$$ Thus, applying the exact same kind of estimate to $(1)$ we conclude $$|f(x,y)|\le k|x||y|$$ as desired.

Answer 2

Taylor expanding with respect to $x$ gives $$ f(x, y)=\partial_x f (0,y) x + O(x^2). $$ Now Taylor expanding $\partial_x f(0, y)$ with respect to $y$ gives $$ \partial_x f(0,y)=\partial_x f(0,0)y+O(y^2)=O(y^2).$$ Here we used that both partials of $f$ must vanish at $(0,0)$. So $f(x, y)=O(x^2+xy^2).$ Redoing all of this with the roles of $x$ and $y$ reversed gives $f(x, y)=O(y^2+yx^2)$. Multiplying these relations we get $$ f(x, y)^2=O(x^2y^2+x^4y+xy^4+x^3y^3)=O(xy), $$ which is the sought asymptotic relation. (not exactly)