In my understanding, Morse function just means the determinant of Hessian matrix is nonsingular at critical points.

So my claims are:

  1. the function itself should be continuous

  2. the reference to Hessian matrix in the definition implies Morse functions are twice differentiable - T/F?

  3. But Morse functions are not $C^2$ since the definition does not require continuity after twice differential - T/F?

Thank you very much!

n.b.1. A morse function is defined to be "A function for which all critical points are nondegenerate and all critical levels are different."

n.b.2. $f$ is of class $\mathscr{C}^k$ on $U$ if all iterated partial derivatives of $f$, of order at most $k$, exist and are continuous on $U$.

n.b.3. Differentiable functions are continuos. But the derivative of a differentiable function may not be continuous: $$f(x) = \left\{\begin{matrix} x^2 \sin \frac{1}{x} & x \neq 0\\ 0 & x=0 \end{matrix}\right.$$


1 Answer 1


In most introductory texts, a Morse function is defined to be smooth ($C^\infty$), since most such texts do not wish to bother themselves with the minutiae of degrees of regularity. However, the assertion that a Morse function is $C^2$ suffices, at least for the basics of the subject, i.e. results like the Morse Lemma do not require smoothness.

Although you can define the Hessian without $f$ being $C^2$, I can not find any text that develops the theory assuming only twice differentiability. If you read some proofs of the Morse Lemma, they use the fact that the entries of the Hessian are continuous, so I suspect that one must proceed with care when some, but not all, double partials are continuous.

  • $\begingroup$ Thank you very much Isaac! But the definition does not require continuity after twice differentiate. So I wonder if it is $C^2$? $\endgroup$ Commented May 26, 2013 at 21:16
  • 1
    $\begingroup$ Thank you so much for providing me with the insights! $\endgroup$ Commented May 26, 2013 at 21:56

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