1
$\begingroup$

Let $X$ be a compact metric space, $\{\delta_n\}_{n=1}^{\infty}$ be a strictly monotonically decreasing sequence in $[0,1]$ converging to $0$, and $\{h_n\}_{n=1}^{\infty}$ be a uniformly convergence sequence of continuous functions on $X$ converging to $h:X\rightarrow [0,1]$. Distinguish a non-empty compact subset $A\subseteq X$. If:

  • $h^{-1}[0]=A$,
  • $h_n(A)\subseteq [0,\delta_n)$ (for every $n$),

Can we guarantee that $$ I_{[0,\delta_n)}\circ h_n \mbox{ converges point-wise to } I_{\{0\}}\circ h? $$ If not, what additional conditions am I missing?


It seems that $I_{[0,\delta_n)}\circ h_n(A)=1$ for all $n$; which is good. So the convergence is uniform on $A$. However, I'm most worried about the pointwise convergence on $X-A$...this I can't manage to control...

$\endgroup$
1
$\begingroup$

$\newcommand\de\delta\newcommand\N{\mathbb N}\newcommand\R{\mathbb R}$The answer is yes. Indeed, fix any $x\in X$. We need to show that $$l_n:=I\{0\le h_n(x)<\de_n\}\to r:=I\{h(x)=0\}\tag1$$ as $n\to\infty$. Let $$N:=\{n\in\N\colon0\le h_n(x)<\de_n\}.$$

If $N\ni n\to\infty$, then $l_n=1$, $h_n(x)\to0$, and hence $h(x)=\lim_n h_n(x)=0$, so that $r=1$ and $l_n=1\to1=r$, i.e., (1) holds.

If $N\not\ni n\to\infty$, then $l_n=0$ and, by the condition $h_n(A)\subseteq [0,\de_n)$, we have $x\notin A$, so that, by the condition $h^{-1}(\{0\})=A$, we have $h(x)\ne0$ and hence $r=0$, so that $l_n=0\to0=r$, i.e., (1) again holds.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.