# Uniform approximation of indicator function of a point

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?

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 $$\mathbb R$$.
• It think it works if we replace the first condition "for all real y" with "for all $y \in [x,x+n^{-1})$"; no? Feb 1, 2021 at 14:44