Consider the function $f(x,y) = x^3 + e^{3y}-3xe^y$. Show that $f$ has exactly one critical point and that this point is a local minimizer, but not a global minimizer.

I have attempted this, but it seems that I have made a mistake somewhere along the way. Can someone please help me find where I went wrong?

My attempt:

\begin{align}\nabla f = \begin{bmatrix}3x^2 -3e^y \\3e^{3y} -3xe^y\end{bmatrix} = \bar{0}\end{align} Thus giving the following two homogeneous equations \begin{align}3x^2 - 3e^y &= 0\\ 3e^{3y} - 3xe^y &=0\end{align}

From the second equation we find that $$x = e^{2y}$$ and upon substitution into the first equation we find that $$e^y(e^{3y} -1) =0$$

Notice however that $e^y \neq 0$ and we must that have that $e^{3y}=1 \implies y=0$ and thus $x =1$

Our critical point is thus $(x,y) = (1,0)$.

How consider the Hessian

\begin{align}H_f &= \begin{bmatrix}6x & -3e^y \\ -3e^y & 9e^{3y}-3xe^y\end{bmatrix}\end{align} Evaluated at our critical point we obtain $$H = \begin{bmatrix}6 & -3 \\ -3& 6\end{bmatrix}$$

We can then deduce that $H_f$ is positive definite and thus our critical point must be a strict global minimizer. This, however is the exact opposite of what the question stated?

Edit: This is the theorem in my notes that I tried to use:

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  • $\begingroup$ You have shown that you have a local minimum. (Now you want to show that it is not a global minimum, by showing the function has smaller values.) $\endgroup$ – user84413 Aug 30 '15 at 17:41
  • $\begingroup$ @Thomas - How come? Did I not use everything as in the theorem? (Critical point, Hessian at that point and the found the Hessian and the point is positive definite)? $\endgroup$ – user860374 Aug 30 '15 at 17:51
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    $\begingroup$ The theorem you cited assumes something about the Hessian in every $x$. You only have the definiteness in one point (or did I miss something) $\endgroup$ – Thomas Aug 30 '15 at 17:52
  • $\begingroup$ @Thomas - I see what you mean now! :). Thank you! :). The Hessian is not positive definite over the whole of $\mathbb R^2$ and hence the minimizer is only local and not global :) $\endgroup$ – user860374 Aug 30 '15 at 17:54
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    $\begingroup$ (The really remarkable fact about the function which is discussed here is that it has a unique strict local minumum and therefore satisfies most prerequisites of the mountain pass theorem, while it does not satisfy the conclusion that theorem. This means it cannot be proper). $\endgroup$ – Thomas Aug 30 '15 at 18:21

Your conclusion is wrong. If the Hessian is positive definite, the critical point is a local minimizer.

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    $\begingroup$ I think I see now.! :). The Hessian is not positive definite over the WHOLE of $\mathbb R^2$ (as would be required in the Theorem) and hence the minimizer is only local and not global :). Is this correct? $\endgroup$ – user860374 Aug 30 '15 at 17:55

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