Show that no matter what the value of $a$ is chosen, the function $f(x_1,x_2)=x_1^3-3ax_1x_2+x_2^3$ has no global maximizers. Determine the nature of the critical points of this function for all values of $a$.

Here $\nabla f=(3x_1^2-3ax_2,-3ax_1+3x_2^2)$
Hessian$(f)=\begin{pmatrix}6x_1&-3a\\-3a&6x_2\\ \end{pmatrix}$

The critical points I found are $(0,0)$ and $(a,a)$.

  • at $(0,0)$: Hessian is indefinite so, $(0,0)$ is a point of inflection.

  • at $(a,a)$: if $a>0$, then Hessian is positive definite, hence a strict local minimizer exists, and if$ a<0 $then Hessian is negative definite hence a strict local maximizer exists.

Is this correct?
But I do not understand how to prove that a global maximizer does not exist.

  • 1
    $\begingroup$ I'm sorry if I'm missing something, but this seems equivalent to asking why $x^2+ax$ has no global maximizers. It's simply because the degree of the function is positive, and so it might have a critical point, or local max, but not a global maximum. $\endgroup$ Dec 25 '15 at 18:08
  • 1
    $\begingroup$ a global maximizer would certainly be a local maximizer and therefore a critical point, but you found all critical points and none are local maximizers $\endgroup$
    – math635
    Dec 25 '15 at 18:13
  • $\begingroup$ @math635 At the critical point (a,a) when a<0, doesn't it give local maximizers $\endgroup$
    – clarkson
    Dec 25 '15 at 18:21
  • $\begingroup$ @clarkson oh i overlooked that. But it can't be a global maximizer because if $x_1,x_2$ got to positive infinity then f goes to positive infinity. $\endgroup$
    – math635
    Dec 25 '15 at 18:23
  • 1
    $\begingroup$ Both off-diagonal entries in the Hessian should be $-3a$ $\endgroup$
    – Empy2
    Dec 25 '15 at 18:24

Set e.g. $x_2=0$ to get the restriction of the function on the first axis to be $$ f(x_1,0)=x_1^3. $$ It is not bounded from above ($x_1^3\to +\infty$ as $x_1\to +\infty$), then the function $f(x_1,x_2)$ is unbounded from above too, and the global maximum does not exist.


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