# Is having critical points a necessary condition for a Morse function?

The definition of a Morse function requires that its critical points are all degenerate and no two of them share the same function value. Now, I'm wondering whether or not the criticality condition is a must or not. By definition, $$x$$ is a critical point of $$f$$ if

$$\dfrac{\partial f}{\partial x} = 0.$$

Instead, given a (very) small $$\epsilon >0$$, assume that

$$\dfrac{\partial f}{\partial x} = \epsilon.$$

Suppose also

$$\dfrac{\partial^2 f}{\partial x^2} = \epsilon$$

where $$\delta >0$$ is another very small real number close to zero. Obviously, $$x$$ is not a critical point, yet it is very close to that. So, are we authorized to call this function as a Morse one?

The intention to ask this question is function $$\psi: \mathbb{R}^2 \rightarrow \mathbb{R}^2$$ defined as below

$$\psi(q_i) = \displaystyle\sum_{j} \dfrac{q_i - q_j}{||q_i - q_j||^{3}}$$

in which $$q_i$$ is a vector. In particular, this kind of function is a candidate repulsive term of an artificial potential field to navigate robots. The first derivative of this function (as well as the higher order ones) can be very small but not exactly zero. Since being a Morse function is a requirement for a righteous candidate in this application, may one consider this function as a Morse one?

• Definitions in math mean exactly what they say. Whether an arbitrary Morse function really is suitable for your application is a separate question. – Eric Wofsey Oct 22 '19 at 18:01
• @EricWofsey: Can you please explain what you mean by an "arbitrary" Morse function? – Pinton Oct 22 '19 at 18:06