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Consider $C^\infty$ functions. For functions of one variable, we know that in general, at $x=c$ if the lowest order nonzero derivative is an even order say $2k$, then $f(c)$ is an extremum, depending on the sign of $f^{(2k)}(c)$. For example, if $f'(c)=f''(c)=f'''(c)=0$ and $f''''(c)>0$, then it is a minimum; if $f'(c)=f''(c)=0$, $f'''(c)\neq 0$, then it is not a local extremum.

Now for multivariable functions. We know the theory of Hessian matrix and eigenvalues. What if some of the second derivatives are zero? For example, consider $f(x,y)$ at $(a,b)$: $f_x=f_y=f_{xx}=f_{xy}=0$. Apparently if $f_{xxx}\neq 0$ the function cannot have a local extremum there, as the trend along the $x$-axis is monotone. Suppose $f_{xxxx},f_{yy}$ do not vanish. Should we examine the eigenvalues of \begin{equation} H=\begin{bmatrix} f_{xxxx} & f_{xxy}\\ f_{xxy} & f_{yy}\end{bmatrix}? \end{equation} What is the general theory about it? Any reference on this? thanks!

Haoran Chen
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  • The following reference is about this question. Maybe it is of some help: https://www.researchgate.net/publication/265638239_Extrema_in_Case_of_Several_Variables – Gerd Aug 10 '23 at 15:26
  • @Gerd Thanks. The necessary and sufficient conditions seem obvious. An example is $f(x,y)=x^2\pm y^4$. Suppose $f(x,y)$ has a minimum at $(0,0)$, and the $2k$-th (lowest)$x$ partial derivative is nonzero there. Does there exist $\epsilon>0$ such that $f(x,y)>\epsilon x^{2k}$ near $(0,0)$? – Haoran Chen Aug 11 '23 at 09:45
  • I think in this form it does not hold. Isn't $f(x,y)=(x-y)^4$ a counterexample? – Gerd Aug 11 '23 at 13:35
  • No, $(x-y)^4$ does not have a minimum at $(0,0)$. I assume $f(x,y)>f(0,0)$ near it. – Haoran Chen Aug 12 '23 at 03:14

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