If $0<\varphi_{n_k}\xrightarrow{k\to\infty}\varphi\ne0$ for some $(n_k)$, are we able to conclude $\inf_{n\in\mathbb N}\varphi_n(x)>0$ for all $x$?

Question

Let $d\in\mathbb N$ and $(\varphi_n)_{n\in\mathbb N}\subseteq C(\mathbb R^d)$ with $\varphi_n(\mathbb R^d)\subseteq(0,\infty)$ for all $n\in\mathbb N$. Assume $$\varphi_{n_k}\xrightarrow{k\to\infty}\varphi\tag1$$ for some increasing $(n_k)_{k\in\mathbb N}$ and $\varphi\in C(\mathbb R^d)$ with $0\not\in\varphi(\mathbb R^d)$.

Is this enough to conclude that $$\inf_{n\in\mathbb N}\varphi_n(x)>0\tag2$$ for all $x\in\mathbb R^d$?

Answer 1

Unfortunately not. You haven't quantified the convergence, but it's not true even for the strongest version. Let $\varphi_{2k}(x) = 1/k$, $\varphi_{2k+1}(x) = 1$ and $\varphi(x) = 1$. Then $\varphi_{2k+1} = \varphi$ for all $n$, so certainly $\varphi_{2k+1} \to \varphi$, regardless of your choice of convergence. However, $\inf_{k \in \mathbb N} \varphi_{2k}(x) = \inf_{k \in \mathbb N} 1/k = 0$ for all $x$.