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Let $f_n, f,g \colon \mathbb R \to \mathbb R$ be continuous. Assume that $f_n \to f$ uniformly. Does it follow that $g(f_n) \to g(f)$ uniformly? Here $g(f)$ stands for the composition of $g$ and $f.$

I would expect the answer to be negative as in order to get the uniform convergence of $g(f_n) \to g(f)$ I would expect to require $g$ to be uniformly continuous.

carlos85
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  • When you write $g(f_n)$, do you mean composition of the two functions? – Keen-ameteur Feb 23 '23 at 16:44
  • yes I mean the composition – carlos85 Feb 23 '23 at 16:47
  • Possible duplicate: https://math.stackexchange.com/questions/1018409/composition-of-uniformly-convergence-sequence-with-continuous-function-is-unifo – legionwhale Feb 23 '23 at 17:10
  • @legionwhale the order of composition is reversed, hence I think this question is different from the one in your link! – Ssay Feb 23 '23 at 17:12
  • @Seansong The question given here is also answered in the accepted answer. – legionwhale Feb 23 '23 at 17:12
  • @carlos85 Why not? If not, you should suggest an edit for the answer. – legionwhale Feb 23 '23 at 17:26
  • @legionwhale The answer there uses that the $f_n$'s are defined on a compact domain. Otherwise their image would not necessarily be bounded in some $[-M,M]$. – Keen-ameteur Feb 23 '23 at 17:51
  • @Keen-ameteur It is not difficult to construct a counterexample for the non-compact case based on the idea of the comment where one user asks "Does this hold if the compactness assumption is removed?" and the other says "No. Here's why...". Indeed, you have given the one I was about to write. So I still maintain that all relevant information is available on that page. – legionwhale Feb 23 '23 at 17:57
  • @legionwhale I'm not sure what you mean. It seemed to me like you said that the answers there solves the issue here. This is not the case. Also, there are no hints to build a counter examples from the answer on that thread. – Keen-ameteur Feb 24 '23 at 11:13
  • @Keen-ameteur I don't understand how you're not sure what I mean, but it's not important. – legionwhale Feb 24 '23 at 15:04

2 Answers2

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I would suggest a different solution than that of @Sean song since I think there is an easier counterexample.

Take $g(x)=e^x$, $f_n(x)= x+\frac{1}{n}$ and $f(x)=x$. Then $\Vert f_n-f\Vert_\infty=\frac{1}{n}\to 0$.

Since $g'(x)=g(x)$, we know by the mean value theorem that for any $a<b$, there exists a $c\in (a,b)$ satisfying

$$ e^c(b-a)= e^a(1-e^{b-a}) . $$

Hence, for any sequence $\{ x_n \}$ we have

$$ \Big\vert e^{c_n}\cdot \frac{1}{n} \Big\vert= \big\vert e^{x_n} \cdot(1-e^{\frac{1}{n}}) \Big\vert=\vert g\circ f(x_n)-g\circ f(x_n)\vert$$

for some $x<c_n<x+\frac{1}{n}$. So long as $\frac{e^{c_n}}{n} \geq \frac{e^{x_n}}{n} \geq 1$ for all $n\in \mathbb{N}$, this will show that there is no uniform convergence. Choosing $x_n=n$, will do the trick since

$$ \frac{e^x}{x}\to \infty \quad \text{as } \; x\to \infty. $$

Keen-ameteur
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EDIT : As pointed out in @keen-ameteur and my own comments below, there is a mistake in the answer - to find a counter example I cannot use $g$ with domain $x\geq2,$ for then since $f_n\to0$ we cannot assess at all how $g(f_n)$ behaves. Thank you very much for pointing this out.

Original answer :

Consider the functions $f_n(x) = 1/x^n$ defined on $(2,\infty)$. Now these functions converge uniformly to $f(x) = 0$. To see this, let $\epsilon$ be given. Note that $$\|f_n(x) - 0\| = \frac{1}{x^n} < \frac{1}{2^n},$$ And thus there exists $N$, independent of $x$, such that $n>N$ implies $\|f_n(x)-0\|\leq \epsilon$ for all $x\in (2,\infty).$

However, if we define $g(x) = 1/x\,(x\geq2),$, then we have $g(f_n(x)) = x^n$. It follows that your sequence $g(f_n)$ does not even converge.

EDIT: To extend the domain to $\mathbb{R}$, I suppose we may simply extend the functions to be symmetric about $x=2$.

I guess i'll elaborate on my answer. Let $$g(x) = \begin{cases} 1/x & x\geq 2 \\ 1/(4-x) & x<2.\end{cases}$$

Then $g$ is continuous everywhere on $\mathbb{R},$ and the argument does not break. It is easy to extend the $f_n$ in a similar fashion, define $$f_n(x) = f_n(4-x)$$ whenever $x\leq 2.$

Since the composition of symmetric functions is again symmetric, it follows that $g\circ f_n$ turns out to be the graph of $$y=x^n,\, x\geq 2$$ flipped around the axis $x=2.$ As $n\to \infty$, we have $g(f_n(x)) \to \infty$ for all $x\in \mathbb{R}$.

Ssay
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    I'm not sure that this works as a counter-example, because the OP specifies that all their functions are continuous and defined on all of $\mathbb{R}$. – Maths Matador Feb 23 '23 at 16:55
  • @MathsMatador I've made an edit, would appreciate you checking it out. – Ssay Feb 23 '23 at 16:56
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    Yes this does not apply as $1/x$ Is not continuous – carlos85 Feb 23 '23 at 17:02
  • @AlexOrtiz I've defined it on $(2,\infty)$. Then we "flip" it around $x=2.$ Its explicit form would be given by $g(x) = 1/(4-x)$ for $x<2$. – Ssay Feb 23 '23 at 17:03
  • @carlos85 It is continuous on $(2,\infty)$. Then you flip it around the axis $x=2$: consider $g(x) = 1/(4-x)$ for $x<2$ and $g(x)=1/x$ for $x\geq2.$ – Ssay Feb 23 '23 at 17:04
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    I see, it would probably be good for you to add some of these details to your answer – Alex Ortiz Feb 23 '23 at 17:05
  • I think it better to define your functions earlier in your text. (So not in a sentence at the very end) – 311411 Feb 23 '23 at 17:08
  • @AlexOrtiz well said. I've elaborated a bit. – Ssay Feb 23 '23 at 17:08
  • @311411 Yep, I thought there would be clarity but I believe that was selfish. I've made edits, but at the end, so the original answer would be available though. Hope things are clear now. – Ssay Feb 23 '23 at 17:10
  • @Seansong How would you extend $\frac{1}{x^n}$ symmetrically around $2$? You would either break the uniform convergence to $0$, or have that $g\circ f_n(x)\neq x^n$. – Keen-ameteur Feb 23 '23 at 17:11
  • @Keen-ameteur thanks for the comment. I thought i could keep its uniform convergence to 0, by defining it with $f_n(x) = 1/(4-x)^n$ for $x\leq2$. Then the composition $g \circ f_n$ wouldnt exactly be $y=x^n$, but it would be the graph of $y=x^n$ on $x\geq 2$ flipped around the axis $x=2$. Of course, I think what's important is that the sequence $g\circ f_n$ simply doesn't converge at all. Please inform me if there are any errors. – Ssay Feb 23 '23 at 17:19
  • @Seansong Something seems off, since it looks like you should be able to show point-wise convergence from his conditions. – Keen-ameteur Feb 23 '23 at 17:42
  • @Keen-ameteur ah, good point! Since we have $f_n \to 0$, setting the domain of $g$ as $(2,\infty)$ would be a stupid move; I have to specify first how $g$ acts around zero. Your counter example seems to work. – Ssay Feb 23 '23 at 17:48