exponential boundedness of components given exponential boundedness of the norm

Question

Let $v:[0,\infty)\rightarrow \mathbb{R}^n$ be a function such that $\forall t\ge 0$, $v_i(t)\ge 0$ and $$ ||v(t)||\le \beta ||v(0)||e^{-at}, t\ge 0$$ with $\beta,a>0$ can I conclude that for all $i=1,...,n$ there exist $\xi_i,\eta_i>0$ s.t. $$v_i(t) \le \xi_iv_i(0)e^{-\eta_it}$$?

Answer 1

No. Counterexample: $$ v(t)=\Bigl(e^{-t},\frac{t}{1+t}\,e^{-t}\Bigr). $$ Since $v_2(0)=0$ and $v_2(t)>0$ if $t>0$, the inequality is not possible.

Answer 2

I would say yes. Just pick $||.||_\infty$ as the norm. Then the individual components are bounded by above by $\beta ||v(0)||_\infty e^{-at}$, so you can chose $\eta_i = a$, and $\xi_i = \beta \frac{||v(0)||_\infty}{v(0)}$. Because all norms in $\mathbb{R}^n$ are equivalent, you can scale $\xi_i$ to get a proof for other norms too.

I'm assuming that $\mathbb{R}^n_+$ refers those $\mathbb{R}^n$ whose components are all greater than 0.