I have the following problem : Let $f: \mathbb{R}^{2} \rightarrow \mathbb{R} ; f\left(x_{1}, x_{2}\right)=5 x_{2}$, let $\Omega=\left\{x \in \mathbb{R}^{2}: x_{1}^{2}+x_{2} \geq 1\right\}$, and consider the problem $\min _{x \in \Omega} f(x)$.
(c) Is the point $\left[\begin{array}{ll}0 & 1\end{array}\right]^{T}$ a local minimizer?
Sol: c. Consider points of the form $\boldsymbol{x}=\left[x_{1},-x_{1}^{2}+1\right]^{\top}, x_{1} \in \mathbb{R}$. Such points are in $\Omega$, and are arbitrarily close to $\boldsymbol{x}^{*}$. However, for such points $\boldsymbol{x} \neq \boldsymbol{x}^{*}$, $$ f(\boldsymbol{x})=5\left(-x_{1}^{2}+1\right)=5-5 x_{1}^{2}<5=f\left(\boldsymbol{x}^{*}\right) $$ Hence, $\boldsymbol{x}^{*}$ is not a local minimizer.
I checked the first order and second order necessary condition and they passed both. But I am not understanding the solution. Why does the solution work? For example, if I choose $x_1 = 100$, then $x = [10,-99]^{T} $. Then how is this point arbitrarily close to $[0,1]^{T}$? It seems very far from $x^*$. I know for a local minimizer the points need to be in an epsilon neighbourhood of $x^*$. Can someone explain?
Consider the point $z(x_1) = (x_1, 1 - x_1^2)$. Clearly it is in $\Omega$. Additionally $z(x_1)$ can be made arbitrarily close to $x^\star =(0, 1)$: $$ \mathrm{dist}(z(x_1), x^\star) = \sqrt{x_1^2 + (1 - (1-x_1^2))^2} = \sqrt{x_1^2(1+x_1^2)}\to 0, \quad \mbox{as}~x_1 \to 0. $$ Note that $$ f(z(x_1)) = 5 (1 - x_1^2) = 5 - 5x_1^2. $$ For $x^\star$ to be a local minimum, it should satisfy the following condition: $$ \text{there exists}~\epsilon > 0~\text{such that}\quad \min_{z : \mathrm{dist}(z, x^\star) \leq \epsilon} f(z) = f(x^\star). $$ However, let $\epsilon > 0$, and note that if $x_1^\star(\epsilon) = (\sqrt{4 \epsilon ^2 + 1} - 1)/2$, then $z(x_1^\star(\epsilon))$ is distance $\epsilon$ from $x^\star$, and therefore: $$ \min_{z : \mathrm{dist}(z, x^\star) \leq \epsilon} f(z) \leq f(z(x_1^\star(\epsilon)) = 5 - \frac{5}{4}\Big(\sqrt{4 \epsilon ^2 + 1} - 1\Big)^2 < 5 = f(x^\star). $$ Since this holds for arbitrary $\epsilon > 0$, it shows that $x^\star$ is not locally minimal.