Consider a bivariate polynomial $$f(x,y)=\sum_{i=0}^n \sum_{j=0}^m b_{ij}x^iy^j$$ defined on $[0,1]^2$. Suppose that on $[0,1]^2$ this polynomial weakly increases in direction $d=(d_1,d_2) \neq 0$: $$d_1 \frac{\partial f(x,y)}{\partial x} + d_2 \frac{\partial f(x,y)}{\partial y} \geq 0.$$

My question is whether it is possible to have a situation when such $d$ is the only non-trivial direction of monotonicity of $f$? That is, if there is another $\tilde{d} \neq 0$ such that $f$ is increasing in direction $\tilde{d}$ on $[0,1]^2$ then it has to be that $$\tilde{d}=\alpha d $$ for some $\alpha>0$?

  • $\begingroup$ Polynomials have continuous derivatives (of all orders), so if you perturb the direction $d=(d_1,d_2)$ the directional derivative will be positive not only instantaneously at $(x,y)$ but in a "directional neighborhood" (local cone) at that point. Perhaps I've misunderstood the problem, but it seems you assume locally increasing in direction $d$ at every point $(x,y) \in [0,1]^2$. $\endgroup$ – hardmath Dec 17 '16 at 18:38
  • $\begingroup$ Thanks @hardmath. Yes, I assume locally increasing in direction $d$. But the directional derivative could take the value of zero (potentially for many points $(x,y)$ so I am not sure whether the perturbation argument would apply in this case. $\endgroup$ – MerylStreep Dec 17 '16 at 18:44

I think I figured out an asnwer to my question. Yes, it can happen. E.g., it seem that the polynomial $$f(x,y)=x y(1-y), \quad (x,y) \in [0,1]^2$$ is increasing in the direction $d=(1,0)$ only.


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