Let $f:(0, +\infty)\to(0, +\infty) $ be a monotone decreasing, right-continuous function. Can we find a sequence $\{f_{n}\}_{n\in \mathbb{N}}$ of strictly monotone decreasing, continuous functions, such that $f_{n}$ converges pointwise to $f$, that is, $f(x)=\lim_{n\to \infty}f_{n}(x)$ for all $x\in (0, +\infty)?$ To achieve strict monotonicity is trivial, but I have problems with the continuity. I came across this question while reading this about convergence of generalized inverse functions.

[Edit]: Additional question: Is it possible to also make $\{f_{n}\}$ a sequence of positive functions?

[Edit]: Additional question two: Is it possible to make $\{f_{n}\}$ an increasing sequence of positive functions?

Thanks in advance for your help!

  • $\begingroup$ Sure one can! Define $D = \{(x,y) : y < f(x)\}$ and consider $D_t = \{(x,y) : (x + t y, y + t x) \in D\}$. Then it is not very difficult to see that $D_t = \{(x, y) : y < f_t(x)\}$ for a continuous, strictly decreasing function $f_t$, and $f_t(x)$ increases to $f(x)$ as $t \to 0^+$. $\endgroup$ Jun 14, 2021 at 9:12
  • $\begingroup$ (I realise the above works for $f$ defined on all of $\mathbb R$, and needs an appropriate adjustment for $f : (0,\infty) \to (0,\infty)$.) $\endgroup$ Jun 14, 2021 at 9:17

1 Answer 1


Take $f_n(x) =n\int_{x}^{x+1/n} f(t) dt-x/n$

  • $\begingroup$ Thank you for your answer! On which domain is this function defined? It seems to be like a part of a composed function. $\endgroup$
    – Shaq155
    Jun 14, 2021 at 11:05
  • $\begingroup$ On $(0,\infty)$, the $-x/n$ part is just for strict monotonicity $\endgroup$ Jun 14, 2021 at 11:20
  • $\begingroup$ How can strictly monotonic function have compact support?! $\endgroup$ Jun 15, 2021 at 19:14
  • $\begingroup$ Is it possible to make $\{f_{n}\}$ a sequence of positive functions? $\endgroup$
    – Shaq155
    Jun 21, 2021 at 18:37
  • $\begingroup$ Of course, replace $-x/n$ to $+1/(nx+n)$ $\endgroup$ Jun 21, 2021 at 19:19

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