Let $f(\mathbf{x})$ be a twice differentiable function, where $\mathbf{x} \in \mathbb{R}^n$. Let $\mathbf{x}^*$ be a local minimizer of $f(\mathbf{x})$. Consider a differentiable and invertible function $\mathbf{h}:\mathbb{R}^n \to \mathbb{R}^n$. Let $\mathbf{h}(\mathbf{y}^*)=\mathbf{x}^*$.
Is $\mathbf{y}^*$ a local minimizer of $f(\mathbf{h}(\mathbf{y}))$?
Is $\mathbf{y}^*$ a strict local minimzer?
I know that:
- $\mathbf{a}$ is a local minima if $f(\mathbf{x}) \leq f(\mathbf{a})$ in the neighbourhood of $\mathbf{a}$.
- $\mathbf{a}$ is a strict local minima if $f(\mathbf{x}) < f(\mathbf{a})$ in the neighbourhood of $\mathbf{a}$.
$\mathbf{h}(\mathbf{y}^*)=\mathbf{x}^*$ which means is the local minim point of f(x).
According of what they tell me, $f(\mathbf{h}(\mathbf{y}))=f(\mathbf{x})$ so that should be a local minimum. But I don't know how to determine if $\mathbf{y}^*$ is a local minimizer of $f(\mathbf{h}(\mathbf{y}))$. Could you give me some clue on how to start? Thank you!
$x^*$ is a local minimum of $f(x)$ means that there exists some $r>0$ such that $$f(x^*)\le f(x),\quad \forall x\in B(x^*,r).$$ Now, using that $h$ is continuous and $h(y^*)=x^*,$ there exists $s>0$ such that $h(B(y^*,s))\subset B(x^*,r).$ Thus,we have that
$$(f\circ h)(y^*)=f(h(y^*))=f(x^*)\le f(h(y))=(f\circ h)(y), \forall y\in B(y^*,s).$$ So, $y^*$ is a local minimum of $f\circ h,$ only assuming continuity of $h.$
Can you show that if $h$ is invertible and differentiable and $x^*$ is a strict local minimum of $f$ then $y^*$ is a strict local minimum of $f\circ h?$