Uniformly tiny perturbations on larger and larger shells of a noncompact manifold can create new critical points far out

Since $\beta$ has compact support, $\mathrm{\parallel }{g}_n-f{\mathrm{\parallel }}_{\mathrm{\infty }}\le \mathrm{\parallel }\beta {\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{/}n\to 0$, so the perturbations are uniformly tiny in value. Their supports lie in $[n-1\mathrm{/}{n}^2,n+1\mathrm{/}{n}^2]$, hence for every fixed compact set $K\subseteq \mathbb{R}$ one has $g_n=f$ on $K$ once $n$ is large enough.

Differentiating gives $g_n^{\mathrm{\prime }}(x)=1+n{\beta }^{\mathrm{\prime }}(n^2(x-n))$. In particular $g_n^{\mathrm{\prime }}(n)=1+n{\beta }^{\mathrm{\prime }}(0)=1-n\le 0$, with equality only when $n=1$, while outside the support interval one has $g_n^{\mathrm{\prime }}(x)=1$. Thus $g_1^{\mathrm{\prime }}(1)=0$, and for every $n\ge 2$ continuity gives some $x_n\in [n-1\mathrm{/}{n}^2,n+1\mathrm{/}{n}^2]$ with $g_n^{\mathrm{\prime }}(x_n)=0$. Thus each $g_n$ has a new critical point near $x=n$.

Step 1.1 shows that the perturbations are tiny on every fixed compact set, while step 2.1 shows that they still create new far-out critical points. This is exactly the noncompact failure mode recorded in [L1]. Therefore the displayed claim is false.