As part of a different problem, I came across the following simplified question, for which I cannot exhibit a proof nor a counterexample. Note that the assumptions of smoothness and strict positivity of $f$ away from the $y$-axis are important.

Let $f(x,y): \mathbb{R}^2 \rightarrow \mathbb{R}$ be a smooth (${\cal C}^\infty$) function such that $f(0,y)=0$ for all $y \in \mathbb{R}$ and $f(x,y)>0$ for $x \neq 0$. Does the following hold:

There exists $y_0 \in \mathbb{R}$, $\varepsilon >0$ such that $f(x,y_0)+\int_0^x \frac{\partial f}{\partial y}(s,y_0)ds >0$ for $|x|<\varepsilon, x \neq 0$.


Carlos pointed out that this works for analytic functions, leaving the general case open.

I believe, in the general case, this is false. I think something like $e^{-1/x^2} \left(\sin\left( y- \frac{1}{x} \right)+1\right)+g(x)$, for $g(x)$ a sufficiently small positive smooth function, should do. This is smooth, because any factors that occur as we take the derivative will be easily dwarfed by $e^{1/x^2}$. For each value of $y$, $\sin\left( y- \frac{1}{x} \right) =-1$ for infinitely many $x$ near $0$. So no matter what $y_0$ is or how small $\epsilon$ is, you will always be able to find an $x$ with $f(x,y_0)=g(x)$ and $x>0$.

Now we integrate $\int_{0}^{x} e^{-1/s^2} \cos\left(y_0-\frac{1}{s}\right) ds$, where $y_0-\frac{1}{s} \equiv -\pi/2$ modulo $2\pi$. We can break the integral up into intervals of the form $\left[\frac{1}{n\pi+\pi/2+y_0},\frac{1}{n\pi-\pi/2+y_0}\right]$. Let the integral over this interval be $A_n$.

As $n$ increases, $e^{-1/s^2}$ goes to $0$, as does the width of the interval, so $|A_n|$ is a decreasing function of $n$. The sign of $\cos\left( y_0 -\frac{1}{s}\right)$ does not change. It is always $(-1)^n$, so the sign of $A_n$ is $(-1)^n$.


$$\int_{0}^{x} e^{-1/s^2} \cos\left(y_0-\frac{1}{s}\right) ds = \sum_{n=k}^\infty A_n= (A_k+A_{k+1}) + (A_{k+2} + A_{k+3})+\dots$$

In each pair, the first one is dominant, and has sign $(-1)^k$, so $\int_{0}^{x} e^{-1/s^2} \cos\left(y_0-\frac{1}{s}\right) ds$ has sign $(-1)^k$. Since $y_0-\frac{1}{s} \equiv -\pi/2$ modulo $2\pi$, $k$ is odd, and this is negative.

Clearly if we did a slightly messy calculation, we could find some uniform lower bound on the size of this integral. Since $f(x,y_0)=g(x)$ here, we just need to choose a $g(x)$ smooth, positive, and smaller than this bound to find a counterexample.


Taylor expansion in $x$ of $f$ around $x=0$:

$f(x,y)=x^{n} g(y) +$ terms of order $x^{n+1}$, with $g(y)$ not identically zero.

The requirement that $f(0,y)=0$ and $f(x,y)>0$ for all $y$ and for all $x\neq 0$ implies that $g(y)\geq 0$ for all $y$ and that $n=2p\geq 2$ is a positive even integer. Choose a $y_0$ where $g(y_0)=c>0$.

Set $y=y_0$ and make a Taylor expansion in $x$ of $f_y=\partial f/\partial y$ around $x=0$:

$f_y(x,y_0)=x^{m} c' + {\rm order}(x^{m+1})$ with $m\geq n$.

(The power $m$ may be larger than $n$ if $f_y=0$ at $y=y_0$.) We thus conclude that

$\lim_{x\rightarrow 0}x^{-2p}\left[f(x,y_0)+\int_0^x f_y(s,y_0)ds\right]=c>0$,

which amounts to your statement.

  • 3
    $\begingroup$ $e^{-1/x^2}$ ? $\endgroup$ – fedja May 26 '13 at 16:44
  • $\begingroup$ you're right, an analytic (not just smooth) function is required, thank you for the correction. $\endgroup$ – Carlo Beenakker May 26 '13 at 18:31

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