Main Question

I have a class of high-dimensional functions which obey certain constraints (of value or partial derivative) on specific submanifolds of their domain. I am quite confident that there is a more specific class of functions that matches these functions on these submanifolds and approximates them in the vicinity. I want to prove this or find out what additional assumptions I need to make. To this end, I am looking for a mathematical subfield or formalism that deals with such questions.


A simple statement along the lines of what I wish to prove is:

If $f(0,y) = f(x,0) = 0$ and $f$ is sufficiently benign, there exist functions $g$, $h$, and $r$ such that:

  • $f(x,y) = g(x) h(y) + r(x,y)$,
  • $g(0) = h(0) = 0$,
  • $r(x,0) = r(0,y) = 0$,
  • $∂_1 r(0,y) = ∂_2 r(x,0) = 0$.

In this case, the answer is to choose $g(x) = \frac{∂_2 f(x,0)}{∂_1 ∂_2 f(0,0)}$, $h(x) = ∂_1 f(0,y)$, and $r(x,y) = f(x,y) - g(x) h(y)$.

What I considered so far

Taylor series allow me to perform pointwise function approximations. If done sufficiently general, this may allow me to find the shape of the desired approximation, but the process is rather tedious and suggests that there is a more general variant of it.

Approximation theory seems to focus on numerically finding best approximations on the entire domain without paying special attention to certain regions – which is not what I want.

  • 1
    $\begingroup$ After two weeks and one bounty without result, I posted this on MathOverflow. $\endgroup$ – Wrzlprmft Jun 25 '18 at 9:21
  • $\begingroup$ I don't know if this will help you, but I suggest you look closer into approximation theory or numerical analysis books. An example could be Brenner and Scott (2010?) where they have a chapter on averaged Taylor polynomials. These are useful to work with Sobolev or Bezov functions that don't have well defined point wise values, but good local integrability properties. $\endgroup$ – Oskar Limka Jul 16 '18 at 9:00
  • $\begingroup$ @OskarLimka: I already mentioned approximation theory in my last paragraph. Can you address my concerns mentioned there? I also took a brief look at the topics you mentioned and fail to see how to translate them to my problem (but that may be just me). Finally, an answer is probably better suited for this. $\endgroup$ – Wrzlprmft Jul 16 '18 at 13:49

You can prove such statements employing the Mather division theorem.

For example, you can use the following statement derived from the theorem:

Let $f$, $f_1$ be smooth functions satisfying the conditions of the example and let additionally $∂_x ∂_y f_1(0,0)\ne0$. Then there exist a neighbourhood of the origin $U$ and a smooth function $q$ s.t. $f(x,y)=q(x,y)f_1(x,y)$ in $U$.

To see that, we'll make an orthogonal change of coordinates $u=(x+y)/2$,$v=(x−y)/2$, so $xy=u^2−v^2$, and apply the Mather division theorem. Overloading notation we have: $f_1(0,0)=\partial_uf_1(0,0)$. To get $k=2$ in the theorem one needs $\partial_u^2f_1(0,0)\ne0$. But this derivative is a linear combination of $\partial_x^2f_1(0,0)$, $\partial_y^2f_1(0,0)$ and $\partial_x\partial_yf_1(0,0)$. Since the first two are equal to zero, the condition $\partial_u^2f_1(0,0)\ne0$ is equivalent to $∂_x ∂_y f_1(0,0)\ne0$.

By the Mather division theorem: $$f(u,v)=q(u,v)f_1(u,v)+r_0(v)+r_1(v)u.$$ From $$\begin{alignat*}{1} f(v,v)=r_0(v)+r_1(v)v=0,\\ f(-v,v)=r_0(v)-r_1(v)v=0,\end{alignat*}$$ it follows that $r_0≡r_1≡0$ and $f(x,y)=q(x,y)f_1(x,y)$.

Taking $f_1(x,y)=xy$, one has $f(x,y)=q(x,y)xy$ and the solution for your example can be rewritten as $g(x)h(y)=\frac{xyq(x,0)q(0,y)}{q(0,0)}$.

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  • $\begingroup$ Thank you. Two further questions: 1) What’s the point of the sentence starting with “Overloading”? I don’t think it’s needed. 2) Can you specify under what conditions, $U=ℝ^2$? $\endgroup$ – Wrzlprmft Jul 17 '18 at 11:39
  • $\begingroup$ @Wrzlprmft 1) It's just checking the theorem conditions. 2) For $f_1(x,y)=xy$ if $f\in C^\infty(\mathbb R)$ defining $q(x,y)=f(x,y)/xy$ one gets $q\in C^\infty(\mathbb R)$. It is enough to check it at points $(x,0)$, $x\ne0$ and $(0,y)$, $y\ne0$. Say $f(x,y)/x$ will be smooth at $x=0$, $y\ne0$ because using Taylor formula of any order for $f$ at $(0,y)$ and dividing on $x$ one will get a Tailor formula for $f(x,y)/x$. For general $f_1$ the same holds if to require $f_1(x,y)\ne0$ when $xy\ne0$ plus $\partial_y f_1(x,0)\ne0$, $x\ne0$ and $\partial_x f_1(0,y)\ne0$, $y\ne0$. $\endgroup$ – Andrew Jul 17 '18 at 13:14
  • $\begingroup$ @Wrzlprmft 2) or perhaps again refer the Mather theorem. For $y_0\ne0$ in some neighbourhood of $(0,y_0)$ we have $f(x,y)=q_1(x,y)xy +r_0(y)$ and $f(0,y)=r_0(y)=0$ so $q_1=q$. $\endgroup$ – Andrew Jul 19 '18 at 9:16

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