# Second order condition for a unconstrained minium of a function

long time reader first time writer

Okay my question is on the following statement:

Let $f :R^n\rightarrow \mathbb{R}$ be a twice differentiable function , $\bar{x} \in Dom(f)$ with $\nabla{f(\bar{x})}=0$ and $\nabla^2{f(\bar{x})}\succeq 0$ $\Rightarrow$ $\bar{x}$ is a local minimizer

I know that this statement is false because we can easily found a counter example:

1) $f(x)={x_1}^3+{x_2}^2$

2) $\nabla{f(\bar{x})}=[3{\bar{x}_1}^2\enspace 2\bar{x}_2]^T=[0 \enspace 0]^T \Leftrightarrow \bar{x}=[0\enspace 0]^T$

3) $\nabla^2{f(\bar{x})}=\begin{bmatrix} 6\bar{x}_1 & 0 \\ 0 & 2 \\ \end{bmatrix}=\begin{bmatrix} 0 & 0 \\ 0 & 2 \\ \end{bmatrix} \succeq 0$

4) $\bar{x}$ is not a local minimizer since $f(x^*)<f(\bar{x})$ for $x^*=[-\epsilon\enspace 0]^T$ with $\epsilon>0$ approaching $0$

But the problem is that initially i proved that the statement was true however i cannot understand were the my proof fails. I will post the original proof here and any help would be greatly appreciated.

1) Pick $\alpha>0$ and $\forall y \in R^n$

2) $f(\bar{x}+\alpha y)=f(\bar{x})+\alpha\nabla{f(\bar{x})}^ty+0.5\alpha^2y^T\nabla^2{f(\bar{x})}y+o(\alpha^2||y||^2)$

3) If $||y||^2=0 \Leftrightarrow y=0\Rightarrow$ $f(\bar{x}+0)= f(\bar{x})$ which is true

4) If $||y||^2\neq 0\Leftrightarrow y\neq0$ and since $\nabla{f(\bar{x})}=0$ and $\nabla^2{f(\bar{x})}\succeq 0\Rightarrow$ $f(\bar{x}+\alpha y) \geq f(\bar{x})+o(\alpha^2||y||^2) \Leftrightarrow \frac{f(\bar{x}+\alpha y)}{\alpha^2||y||^2} \geq \frac{f(\bar{x})+o(\alpha^2||y||^2)}{\alpha^2||y||^2}$

5) $\displaystyle{\lim_{\alpha \to 0}\frac{o(\alpha^2||y||^2)}{\alpha^2||y||^2}=0} \Rightarrow$ $\forall z \in B(\bar{x},\alpha):\enspace f(\bar{x})\leq f(z)\Leftrightarrow \bar{x}$ is a local minimizer of $f$

• I made a mistake in my original explanation which I therefore removed, I'll try to come back with one fixed! – tibL Sep 16 '16 at 11:49