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This is the question that I posed myself and set out to solve before I came to MSE:

Suppose that $\{f_n\}$ is a sequence of real-valued functions defined on $[a,b]$ such that each function has the intermediate value property. Suppose there is another function $f:[a,b]\rightarrow\mathbb{R}$ such that the sequence $\{f_n\}$ approaches $f$ uniformly.

Does it follow that $f$ has the intermediate value property?

I then came to MSE and found this question and answer which is easily adapted to answer my problem in the negative---$f$ need not have the IVP.

However, before I came here I thought I came up with a proof for the opposite; that is, I thought I proved that $f$ must have the IVP. But, I cannot find the error in my reasoning. Could someone please find it?

Suppose that $x$ and $y$ are two elements of $[a,b]$ such that $f(x)<f(y)$. Let $c$ be a real number that satisfies $f(x)<c<f(y)$.

Now choose $N$ from $\mathbb{N}$ so large such that $f_n(x)<c<f_n(y)$ for every $n\geq N$.

Since every function in the sequence $\{f_n\}$ has the IVP, there is a sequence $\{c_n\}$ that begins its indexing at $N$ such that $$x<c_n<y\quad\text{and}\quad f_n(c_n)=c$$ for every $n\geq N$.

By Bolzano-Weierstrass there is a subsequence $\{c_{n_k}\}$ and an element $z$ of $[x,y]$ such that $c_{n_k}\rightarrow z$ as $k$ tends to infinity.

Clearly, the subsequence $\{f_{n_k}(c_{n_k})\}$ approaches $c$. However, the subsequence $\{f_{n_k}(c_{n_k})\}$ also approaches $f(z)$ since the convergence is uniform.

Therefore $f(z)=c$.

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  • $\begingroup$ Until now, I was unaware that fake-proofs is a tag...perhaps Donald Trump invented it -_- $\endgroup$ – Matt A Pelto Jul 5 '18 at 7:46
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You seem to be invoking a property different from uniform convergence that also happens to imply that $f$ is continuous. Some problem I did for fun: enter image description here

Since none of the functions are assumed or otherwise known to be continuous for all we know we could just as well have $f_n(z)=f(y)+1>f(y)$ for all $n \in \mathbb{N}$. In this hypothetical scenario, we would of course have that $f(z)=f(y)+1 \neq c$ ( $f_n \to f$ uniformly).


I can't think of a counterexample right now but I will post in here if I come up with one.

So every counterexample that I have tried to cook up somehow fails. If there is a counterexample, then I want to say it is not obvious. The statement could always be undecidable which is not to say that I have considered trying to even prove the claim that you mention but merely just throwing this out there as a possibility.

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I'm pretty sure it is not necessary that $$f_{n_k}(c_{n_k})\to f(z),$$ since each function in your sequence is not continuous. Everything else is true. I can't find a counter example here though.

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  • $\begingroup$ This seems to be it. But a counter-example would finish this question. $\endgroup$ – Robert Wolfe Jul 4 '18 at 21:42

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