I am struggling with the following problem:
Let $f$ be a real function such that:
- $f\in\mathcal{C}^\infty(\mathbb{R},\mathbb{R})$,
- $f$ is strictly convex on $(-\infty,0)$, strictly concave on $(0,\infty)$, strictly increasing on $\mathbb{R}$,
- (exponential growth ($f(x) = o(e^{D|x|})$)? just to suppose the following integral well defined).
Let $a_1>a_2>0$ two real numbers.
Can one prove (or give a counter-example to) the following statment:
The function $g$ defined by
$$g(x) := \int_\mathbb{R}\left(f(x+a_1s)-f(x+a_2s)\right)e^{-s^2/2}ds$$
has a unique $0$ on $\mathbb{R}$.
I am (numerically) convinced that the statement is true. Any hint, counter-example or help will be highly appreciated ! Thank you very much.
Numerical example
Here a picture showing the function $f = \arctan$ and the function $g$. The difficulty is that $g$ is not increasing function...
Note that the function $f$ is not necessarly odd.
Possible hints
The function $g$ can be as a DoG (difference of gaussians) function: $$g = f*(G_{a_1}-G_{a_2}).$$
DoG is known to be a possible approximation of gaussian laplacian, which is a smoothed version of the Laplacian. If there is one inflection point, we could imagine that the DoG function will have one zero point.
One way to proceed is to show two points:
- If the Gaussian of Laplacian of $f$ (denoted $\text{LoG}(f)$) has two zeros, the Laplacian of $f$ $\Delta f$ has two zeros. (This point is already proved)
- If the $\text{DoG}$ function has two zeros, $\text{LoG}(f)$ must have two zeros. (This has to be proved). If one manage to prove the second point, the result follows by contradiction (single 0 of $\Delta f$).
We will adapt the method well described by @fedja in this post. The second important result is the property of the $\text{LoG}$ (can be found here). Let $\sigma_1, \sigma_2>0$ two real numbers, we denote $I_f(x;\sigma) $ the gaussian convolution of $f$. We have the following relation:
$$ \partial_\sigma I_f(x, \sigma) = \sigma \text{LoG}(x, \sigma). $$
From the mean value theorem, it exists $\sigma\in(\sigma_1,\sigma_2)$ such that:
$$\text{DoG} (x;\sigma_1,\sigma_2)=(\sigma_1-\sigma_2)\sigma\text{LoG}(x;\sigma)$$
The Laplacian of Gaussian is:
$$\text{LoG}_f(x;\sigma) := \frac{1}{\sigma\sqrt{2\pi}}\int_\mathbb{R} f''(s)e^{-\frac{(x-s)^2}{2\sigma^2}}ds.$$
We define $\mathscr{C}^+$ and $\mathscr{C}^-$ as:
\begin{cases} \displaystyle \mathscr{C}^+(x) := \frac{1}{\sqrt{2\pi}} \int_0^\infty f''(s)e^{-\frac{(x-s)^2}{2\sigma^2}}ds,\\ \displaystyle \mathscr{C}^-(x) := -\frac{1}{\sqrt{2\pi}} \int_{-\infty}^0f''(s)e^{-\frac{(x-s)^2}{2\sigma^2}}ds.\\ \end{cases}
We can rewrite the RHS as:
$$\text{DoG}_f(x;\sigma_1, \sigma_2) = \mathscr{C}^+(x) - \mathscr{C}^-(x)$$
with $\mathscr{C}^+(x), \mathscr{C}^-(x) >0$. By taking the derivatives:
$$\partial_x\mathscr{C}^+(x) = \frac{1}{\sqrt{2\pi}} \int_0^\infty\frac{s-x}{\sigma^2}f''(s)e^{-\frac{(x-s)^2}{2\sigma^2}}ds >-x\frac{1}{\sigma^2\sqrt{2\pi}} \int_0^\infty f''(s)e^{-\frac{(x-s)^2}{2\sigma^2}}ds$$
and by definition: $$\partial_x\mathscr{C}^+(x)>-\frac{x}{\sigma^2}\mathscr{C}^+(x).$$
The other derivative:
$$\partial_x\mathscr{C}^-(x) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^0 \frac{(s-x)}{\sigma^2} (-f''(s))e^{-\frac{(x-s)^2}{2\sigma^2}}ds<-x\frac{1}{\sigma^2\sqrt{2\pi}} \int_{-\infty}^0(-f''(s))e^{-\frac{(x-s)^2}{2\sigma^2}}ds$$
and by definition: $$\partial_x\mathscr{C}^-(x)<-\frac{x}{\sigma^2}\mathscr{C}^-(x).$$
It follows from the two inequalities:
$$\frac{\partial_x\mathscr{C}^+(x)}{\mathscr{C}^+(x)}>\frac{\partial_x\mathscr{C}^-(x)}{\mathscr{C}^-(x)}$$
and therefore:
$$\frac{\partial_x\mathscr{C}^+(x)}{\mathscr{C}^+(x)}-\frac{\partial_x\mathscr{C}^-(x)}{\mathscr{C}^-(x)}>0.$$
The very same computation can be done for $x<0$, thus we finaly recognize the $\log$-derivative and then:
$$\forall x \in \mathbb{R},~ \partial_x\left[\log\left(\frac{\mathscr{C}^+(x)}{\mathscr{C}^-(x)}\right)\right]>0.$$
Hence the ratio can only cross 1 one time. We conclude that $\text{DoG}$ has only one root.