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Let $(f_n)$ be a sequence of monotone functions on a compact interval $I$ which converges pointwise to a continuous function $f$. Show that $f$ is monotone and that $(f_n)$ converges uniformly to $f$.

I am unable to show that $f$ is monotone, considering that each $f_n$ can be either increasing or decreasing and I don't know if this is even true.

I know that if all of them are increasing/decreasing then it is true.

macton
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2 Answers2

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It is true. If $f$ is constant then there is nothing to show. Otherwise, all but finitely many $f_n$ have the same monotony direction:

  • If $f(a) < f(b)$ for some $a<b$ then the pointwise convergence of $f_n$ to $f$ implies that $f_n(a) < f_n(b)$ for all sufficiently large $n$, so that all but finitely $f_n$ are (weakly) increasing, which implies that $f$ is (weakly) increasing.

  • Similarly if $f(a) > f(b)$ for some $a < b$ then all but finitely $f_n$ are (weakly) decreasing, which implies that $f$ is (weakly) decreasing.

Martin R
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  • that is exactly the opposite that I wanted to prove... – macton Jan 04 '21 at 15:38
  • @macton: Is it really? It shows that the limit function $f$ is constant, increasing or decreasing, i.e. it is monotone. Is that not what you were asking for? – Martin R Jan 04 '21 at 15:40
  • I thought you are directly assuming that f is already monotone, which is not because I didn't see carefully. That was my fault :( – macton Jan 04 '21 at 15:44
  • @macton: No problem, perhaps I did not formulate it clearly enough. The idea was that if $f$ is not constant then eventually the $f_n$ are all increasing or all decreasing, which implies that $f$ is monotone. – Martin R Jan 04 '21 at 16:02
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You must have a subsequence of $f_n$ that consists entirely of either monotonically increasing functions or of monotonically decreasing functions.

Arin Chaudhuri
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