# Uniqueness of the root for a DoG function (Difference of Gaussian)

I am struggling with the following problem:

Let $$f$$ be a real function such that:

1. $$f\in\mathcal{C}^\infty(\mathbb{R},\mathbb{R})$$,
2. $$f$$ is strictly convex on $$(-\infty,0)$$, strictly concave on $$(0,\infty)$$, strictly increasing on $$\mathbb{R}$$,
3. (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:

1. 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)
2. 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 $$\Delta f$$).
• try combining the facts that $$\begin{split} &(\lambda (x + a_1s) + (1 − \lambda)(x + a_2s)) < \lambda f (x + a_1s) + (1 − \lambda)f (x + a_2s) \hspace{.1in}\forall \lambda \in (0,1), x < -a_2s \\ & (\lambda (x + a_1s) + (1 − \lambda)(x + a_2s)) > \lambda f (x + a_1s) + (1 − \lambda)f (x + a_2s) \hspace{.1in}\forall \lambda \in (0,1), x > a_1s \\ \end{split}$$ with $a_1 > a_2 >0$, the properties of $e^{-s^2/2}$, and the fundamental theorem of calculus Jul 28, 2023 at 15:18
• Thanks @GeauxMath. I tried to use your indications but I didn't manage to obtain anything. I have to work with a difference in the integral and the convexity (or concavivity) gives me a relation with an addition. Could explain me a bit more your idea please ? Jul 28, 2023 at 16:23
• you want to use the above relations and rearrange them, for example if you multiply both sides of these inequalities by $e^{-s^2/2}$, integrate both sides, and rearrange you can get things like $$\lambda g(x) + \int_\mathbb{R} f(x+a_2s)ds$$ on the RHS of inequality 1 Jul 28, 2023 at 16:34
• I am trying to something with convexity and concavity, for example: $$\forall s>-x/a_2, ~\lambda f(x+a_2s) + (1-\lambda)f(x+a_1s) \leq f(\lambda (x+a_2s) + (1-\lambda)(x+a_1s)) = f(x+s(\lambda a_1 + (1-\lambda a_2))).$$ And I don't know how to handle the RHS. I can't go further then. Moreover, there is the region $[-x/a_2,x/a_1]$ where I can't say nothing. Jul 28, 2023 at 16:59
• factor out lambda and integrate both sides, you will see that $g$ pops out. Regarding the latter, you don't need to. You only NTS that $g$ changes sign only once. Jul 28, 2023 at 17:05

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.