Given two linear functions $f(x)$ and $g(x)$ defined on real values, let's say that I want to show that $f(x) > g(x)$ for all real $x > t > 0$. According to the order-1 Taylor expansion at the origin, these two functions can be written as \begin{equation*} f(x) = f(0) + (x-0)f'(0)\quad \text{and}\quad g(x) = g(0) + (x-0)g'(0) \end{equation*} because the higher-order derivatives will be zero for linear polynomials.

If I can assume that $f(0) = g(0)$, then $f'(0) > g'(0)$ will guarantee that $f(x) > g(x)$ for all $x>t>0$. (Well, it may guarantee $f(x)>g(x)$ for all $x > 0$, but let's just say that the lower bound of interest is some positive number $t$.)

I know multivariable Taylor expansion, but don't have much background on it. So I'd like to ask whether the above approach still works for real-coefficients multilinear polynomials $g,f:\mathbb{R}^n\to\mathbb{R}$. As you know, multilinear means that no variable $x_i$ appears squared, cubed, etc. And my goal is to show $f(x_1,\cdots,x_n)> g(x_1,\cdots,x_n)$ for $x_1 > t, x_2, > t, \cdots, x_n> t$ where $t$ is some positive constant assuming $f(0,\cdots,0) = g(0,\cdots, 0)$.

Extending from the simple case above, I may say the order-$(1,\cdots,1)$ Taylor expansion at the origin of these two polynomials are \begin{align*} f(x_1,\cdots,x_n) &= \left(\prod_{i=1}^n\left(1 + x_i\frac{\partial}{\partial z_i}\right)\right) f(z_1,\cdots,z_n)\big\vert_{z_1 = \cdots = z_n = 0} \\ g(x_1,\cdots,x_n) &= \left(\prod_{i=1}^n\left(1 + x_i\frac{\partial}{\partial z_i}\right)\right) g(z_1,\cdots,z_n)\big\vert_{z_1 = \cdots = z_n = 0} \\ \end{align*} because the simple case corresponds to $f(x) = (1+x \frac{\partial}{\partial z})f(z)\big\vert_{z=0}$.

  1. The simple case of linear polynomials is all about comparing the slopes $f'(0)$ and $g'(0)$. What about the multilinear cases? Similar to the linear case with one variable, is it OK just to compare the all the partial derivatives in each variable at the origin\begin{equation*}\frac{\partial}{\partial z_i}f(z_1,\cdots,z_n)\vert_{z_1=\cdots=z_n=0} > \frac{\partial}{\partial z_i}g(z_1,\cdots,z_n)\vert_{z_1=\cdots=z_n=0}\end{equation*}to conclude that $f(x) > g(x)$ in the domain of interest?
  2. If the above doesn't make sense at all, then what is the general way to compare two multilinear polynomials over some domain $x_1 > t, x_2 > t, \cdots, x_n >t$ for some positive constant $t$? Is there any textbook explaining this problem?

What you say at the beginning is not true. If $f(0) = g(0)$ and $f'(0) > g'(0)$, it is not true that $f(x)>g(x)$ for all $x>0$. One can only prove that we have the inequality on a small neighborhood $[0, \epsilon]$.

The second order Taylor expansion gives a quadratic form, whose positivity can be analyzed by finding a basis of orthonormal eigenvectors.

The higher order terms do not have an easy way to approach their positivity, as far as I am aware.

Also, I think that you may be unnecessarily limiting your domains. While looking at a ray is perfectly reasonable in $\mathbb{R}$, in multiple dimensions you might be interested in more general positive cones than just the ones generated by the standard basis vectors. In other words, instead of limiting your investigations to the domains $\{x_i>0\}$, you might consider the domains $\{\sum_i c_i \vec{v}_i : c_i \geq 0\}$ where $\vec{v}_i$ are some basis.

  • $\begingroup$ For the first paragraph in your answer, you're right for the general cases but in the special case of 'linear' functions I think I can guarantee the inequality between the two functions for $x>0$ not just for the neighborhood of the origin. $\endgroup$ – user19906 Jun 29 '15 at 23:19
  • $\begingroup$ For the second paragraph in your answer, I'm not quite sure about the eigenvectors. What eigenvectors are you talking about? Eigenvectors from which matrix? $\endgroup$ – user19906 Jun 29 '15 at 23:47
  • $\begingroup$ @user19906 The hessian matrix, which is the matrix which represents the quadratic form. For instance, the matrix associated to $ax^2 + 2bxy + cy^2$ would be $\begin{bmatrix} a & b \\ b & c \end{bmatrix}$ because $ax^2 + 2bxy + cy^2 = \begin{bmatrix} x & y \end{bmatrix} \begin{bmatrix} a & b \\ b & c \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix}$ $\endgroup$ – Steven Gubkin Jun 30 '15 at 1:02
  • $\begingroup$ By the real spectral theorem, this matrix has an orthonormal basis of eigenvectors. If the eigenvalues are nonnegative, then the quadratic form is always nonnegative. This is the reasoning behind the second derivative test in multivariable calculus. $\endgroup$ – Steven Gubkin Jun 30 '15 at 1:05

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