Equivalence of norm minimization

Question

Suppose $f:\mathbb{R}^n \rightarrow \mathbb{R}^m$ is some function and let $\|\cdot\|_a$, $\|\cdot\|_b$ be two norms on $\mathbb{R}^m$. Then is the following true: $x^* \in \mathbb{R}^n$ is a solution to $$\text{arg min}_{x \in \mathbb{R}^n} \|f(x)\|_a$$ if and only if $x^*$ is a solution to $$\text{arg min}_{x \in \mathbb{R}^n} \|f(x)\|_b.$$ I feel it should be true because of finite dimensional norm equivalence namely there exists a constants $C_1, C_2$ such that $$\frac{1}{C_1}\|f(x)\|_b \leq \|f(x)\|_a \leq C_1 \|f(x)\|_b$$ $$\frac{1}{C_2}\|f(x)\|_a \leq \|f(x)\|_b \leq C_2 \|f(x)\|_a$$ $\forall x \in \mathbb{R}^n$. But I can't seem to quite prove it. Any thoughts? Thanks.

Answer 1

There's a counterexample even when $f$ is continuous. Take a function that maps $\mathbb R^2 \to \{(x,y)\in \mathbb R^2\colon y\geq 0, x^2+y^2 \geq 1\}$. For example, one such map would be the composition of the functions $g(x,y)=(x,|y|)$ and $h(x,y) = \begin{cases}(x,y)&\text{for }|x|\geq 1\\(x,y+\sqrt{1-x^2}) &\text{for } |x|<1\end{cases}$.

We note that $\frac{1}{\sqrt{2}}(1,1)$ is a solution to $\arg\min\|f(x)\|_2$ but not to $\arg \min \|f(x)\|_\infty$.