Let $f(x) = (x_1-x_2^2)(x_1-\frac{1}{2}x_2^2)$. Verify that $\overline{x} = (0,0)^t$ is a local minimizer of $\phi(\lambda) = f(\overline{x}+\lambda d)$ for all $d\in\mathbb{R}^2$ but $\overline{x}$ is not a local minimizer of $f$

$$\frac{\partial f}{\partial x_1} = 1\left(x_1-\frac{1}{2}x_2^2\right)+(x_1-x_2^2)$$

$$\frac{\partial f}{\partial x_2} = -2x_2\left(x_1-\frac{1}{2}x_2^2\right)-x_2^2(x_1-x_2^2)$$

$$\frac{\partial^2 f}{\partial x_1^2} = 2$$

$$\frac{\partial^2 f}{\partial x_2^2} = 6x^2$$

$$\frac{\partial^2 f}{\partial x_2\partial x_1} = -3x_2$$

$$\nabla^2 f = \begin{bmatrix} 2 & -3x_2 \\ -3x_2 & 6x_2^2 \end{bmatrix}$$

$$\nabla^2 f(0,0) = \begin{bmatrix} 2 & 0 \\ 0 & 0 \end{bmatrix}$$

$$\begin{bmatrix} a & b \\ \end{bmatrix}\begin{bmatrix} 2 & 0 \\ 0 & 0 \end{bmatrix}\begin{bmatrix} a \\ b \end{bmatrix} = a^2$$

for $(0,0)$ to be a local minimizer we should have $\nabla f = 0$ (we have) and $\nabla^2\ge 0$. But $\nabla^2$ is not always semipositive definite so $(0,0)$ is not a local minimizer of $f$.

Now let's analyze $f((0,0)+\lambda d)$.

$$\nabla(\lambda d_1,\lambda d_2) = \begin{bmatrix} \lambda d_1-\frac{1}{2}\lambda^2d_2^2 + \lambda d_1 -\lambda^2d_2^2 \\ -2\lambda d_2(\lambda d_1-\frac{1}{2}\lambda^2 d_2^2)-\lambda d_ 2^2(\lambda d_1-\lambda^2d_2^2) \end{bmatrix}\neq 0$$

and the hessian gives an even worse thing. So I think instead I should analyze

$$\phi(\lambda) = f(\lambda d_1,\lambda d_2) = (\lambda d_1-\lambda^2 d_2^2)(\lambda d_1-\frac{1}{2}\lambda^2 d_2^2)$$

but what does it mean for $\overline{x}$ to be a local minimizer of something that depends on $\lambda$?


I think I have to minimize

$$\phi(\lambda) = f(\lambda d_1,\lambda d_2) = (\lambda d_1-\lambda^2 d_2^2)(\lambda d_1-\frac{1}{2}\lambda^2 d_2^2) = \\\lambda^2 d_1^2 -\frac{1}{2}\lambda^3d_1d_2^2-\lambda^3d_2^2 d_1 + \frac{1}{2}\lambda^4d_2^4\implies \\ \phi'(\lambda) = 2d_1^2\lambda -\frac{3}{2}\lambda^2d_1d_2^2 -3\lambda^2d_2^2d_1 + 2\lambda^3d_2^4 = 0\implies \\2d_2^4\lambda^2+\lambda\left(-\frac{9}{2}d_2^2d_1\right) + 2d_1^2 = 0\implies\\ \lambda = \frac{\frac{9}{2}d_2^2d_1\pm\sqrt{\ (\frac{9}{2}d_2^2d_1)^2 -4\cdot 2d_2^4\cdot2d_1^2}}{4d_2^4}, \lambda = 0$$

It's unpratical to test these values on the function to see which of them is smaller. And even with this I don't know how to proceed. Maybe I should test the second derivative too, I don't know what it means to prove $\overline{x} = (0,0)$ is a minimizer of $\phi(\lambda) = f(\overline{x}+\lambda d)$. The only possible thing I can image is $\lambda$ being $0$ giving us the minimum of $\phi(\lambda)$.

So since we know $\lambda=0$ gives us derivative $0$, let's see how the second derivative is at this point (if it's positive, then $\lambda=0$ is a minimizer):

$$\phi''(\lambda) = 6\lambda^2 d_2^4 -6\lambda d_2^2 d_1 -\frac{6}{2}\lambda d_1d_2^2 + 2d_1^2 \implies\\\phi''(0) = 2d_1^2$$

So for any direction with $d_1\neq 0$, we'll have $\lambda=0$ as minimizer. What about when the direction is $(0,d_2)$ for any $d_2$? In this case we have $$\phi(\lambda) = (0-(\lambda d_2)^2)(0-\frac{1}{2}(\lambda d_2)^2 ) = \frac{1}{2}\lambda^4d_2^4$$

for which $\lambda= 0$ is also a minimizer.

So in all cases $\lambda=0$ is a minimizer of $f((0,0) + \lambda (d_1,d_2))$ which means $(0,0)$ is a minimizer of $\phi(\lambda)$ I guess?

  • $\begingroup$ It seems that you are requested to prove that $\phi(\lambda)$ takes local minimum at $\lambda = 0$. $\endgroup$
    – xbh
    Commented Aug 27, 2018 at 3:05
  • $\begingroup$ @xbh so I should substitute $(x_1+\lambda d_1, x_2+\lambda d_2)$ into $f$ and minimize the parabola over the variable $\lambda$? $\endgroup$ Commented Aug 27, 2018 at 3:22
  • $\begingroup$ If the question above is typed right, then for fixed $(d_1,d_2)$, $\phi(\lambda)$ is just the last equation you have written in the OP. Then minimize that over $\lambda$ as you thought. $\endgroup$
    – xbh
    Commented Aug 27, 2018 at 3:24
  • $\begingroup$ Your analysis based on the Hessian does not exclude the possibility of the origin being a minimum. It only matters what the Hessian is at the origin. The fact that it may be indefinite elsewhere is irrelevant. $\endgroup$
    – Asdf
    Commented Aug 27, 2018 at 13:47
  • $\begingroup$ @Asdf you're right, so a hessian analysis isn't going to give me anything. I should analyze the neighborhood of (0,0)? $\endgroup$ Commented Aug 29, 2018 at 19:28

1 Answer 1


This problem has a nice graphical solution.

Note that f(0)=0. If both of the factors $x_1-x_2^2$ and $x_1-\frac12x_2^2$ have the same sign then $f(x)>0$. If the signs differ then $f(x)<0$. Divide the $x_2x_1$-plane into three regions:

(i) if $x_1>x_2^2$ then $x_1>\frac12x_2^2$ and $f(x)>0$. This is the area in the plane above the quadratic $x_1=x_2^2$ (you let $x_1$ be the vertical axis and $x_2$ the horizontal).

(ii) if $x_1<\frac12 x_2^2$ then $x_1<x_2^2$ and $f(x)>0$. This is the area below $x_1=\tfrac12x_2^2$.

(iii) if $x_2^2>x_1>\frac12 x_2^2$ then $f(x)<0$. This is the area above the curve $x_1=\frac12x_2^2$ but below $x_1=x_2^2$.

The curve $x_1=\frac23x_2^2$ lies in (iii). On it, the objective value is strictly negative, $$(x_1-x_2^2)(x_1-\frac12x_2^2)=-\frac12x_2^2\frac16x_2^2.$$ Hence the origin with objective value 0 cannot be a minimum. Note however that you can only approach the origin from inside (iii) along a curved trajectory. If you consider a series of perturbations of the origin that all lie on a line, as in the case of $\phi(\lambda)$, then all points in some small ball around the origin will lie in region (i) or (ii). Graph the three regions and you will see this clearly.

  • $\begingroup$ what is an objective value? $\endgroup$ Commented Sep 2, 2018 at 4:34
  • $\begingroup$ doesn't it suffice to show that $x_1 = \frac{2}{3}x_2^2$ is always negative so there's a path on which small changes makes $f$ smaller than $0$? Why did you do all this analysis on the sign of $f$? $\endgroup$ Commented Sep 2, 2018 at 4:49
  • $\begingroup$ By objective value I refered to the value of the objective function $f$ at a point $(x_1,x_2)$. This is standard terminology in optimization $\endgroup$
    – Asdf
    Commented Sep 2, 2018 at 15:03
  • $\begingroup$ Yes, the last equation suffices. The analysis provides an intuitive understanding of the problem which in turn leads to the answer. You can substitute any relation of the form $x_1=cx_2^2$, where $c\in(\frac12,1)$, into the objective function $f$. Here, I picked $c=\frac23$. Please accept the answer if you like it. $\endgroup$
    – Asdf
    Commented Sep 2, 2018 at 15:04
  • $\begingroup$ Thank you for your answer. I'm gonna accept it. Could you just take a look at my argument about why $(0,0)$ is a minimizer for $\phi(\lambda)$? $\endgroup$ Commented Sep 2, 2018 at 19:01

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