Let $A\subset\mathbb{R}$ and $f_n:A\rightarrow\mathbb{R}$ are continuous functions such that $\liminf_{n\to\infty}f_n$ is finite and continuous. Is it true that $$g_n=\inf_{k\ge n}f_k$$ is continuous for all $n\in\mathbb{N}$ ?
Are these functions continuous assuming $\liminf_{n\to\infty} f_n$ is continuous?
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real-analysis
sequences-and-series
analysis
1 Answers
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No. Let $A = [0,1]$, and let $f_n$ be the piecewise linear function whose "turning points" are given by $$ f_n(0) = 1\\ f_n(1-\tfrac1n) = 1\\ f_n(1-\tfrac1{2n}) = 0\\ f_n(1) = 1 $$ I.e., it is a function which has value 1 except for a steep dip to 0 near 1. Each $f_n$ is continuous, and for each $x \in A$ we have $\lim_{n \to \infty}f_n(x) = 1$, so the (pointwise) limit of this sequence of functions exists and is continuous. However, for each $n$, the sequence of points $$ 1-\frac1{2n}, 1-\frac1{2(n+1)}, 1-\frac1{2(n+2)},... $$ gives a sequence of points where $g_n$ takes the value 0, converging to 1, where $g_n$ takes the value 1. Hence $g_n$ is not continuous.