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Fix $x \in \mathbb{R}$ and let $I_{[x]}$ be its indicator function. Does anyone know of a sequence of (obviously) discontinuous approximations $g_n$ to $I_{[x]}$ such that

  • $g_n$ converge uniformly to $I_{[x]}$ on $\mathbb{R}$,
  • $|g_n(y)-I_{[x,x+n^{-1})}(y)|\in (\frac1{2n},\frac1{n}]$? Is this possible?
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Such a sequence $(g_n)$ does not exist -- if by $[x]$ you mean $\{x\}$ and if you want $|g_n(y)-I_{[x,x+n^{-1})}(y)|\in (\frac1{2n},\frac1{n}]$ to hold for all real $y$.

Indeed, then $|g_n(x+1/(2n))-1|\in (\frac1{2n},\frac1{n}]$, so that $g_n(x+1/(2n))\to1$ (as $n\to\infty$) and hence $$\liminf_n\sup_{t\in\mathbb R}|g_n(t)-I_{\{x\}}(t)| \ge\liminf_n|g_n(x+1/(2n))-I_{\{x\}}(x+1/(2n))|=1>0.$$ So, $g_n$ does not converge uniformly to $I_{\{x\}}$ on $\mathbb R$.

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  • $\begingroup$ It think it works if we replace the first condition "for all real y" with "for all $y \in [x,x+n^{-1})$"; no? $\endgroup$
    – Bernard_Karkanidis
    Feb 1, 2021 at 14:44
  • $\begingroup$ @Bernard_Karkanidis : Yes, if by "it works" you mean "this counterexample works". $\endgroup$
    – Iosif Pinelis
    Feb 1, 2021 at 14:54

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