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Definition:

Let $f: [0, 1] \rightarrow \mathbb{R}$ be a function. Say $f$ is Totally-Nonmonotonic iff for any $a, b$ in [$0, 1$], $f$ is NOT monotonic on [$a, b$].

Question:

Let $f: [0, 1] \rightarrow \mathbb{R}$ (both of which are equipped with the topology generated by the usual distance) be a continuous function and let $\epsilon$ be a real positive real number. Prove that there exists a continuous totally-nonmonotonic function $g: [0, 1] \rightarrow \mathbb{R}$ such that |$f(t) - g(t)$| $\leq \epsilon$ for every $t \in [0, 1]$.

I am tempted to apply Stone-Weirstrass Theorem here but then realize that although the set of all totally-nonmonotonic continuous functions does separate points in $C[X, \mathbb{R}]$, however, it is not a subalgebra because, assuming $h$ is a totally-nonmonotonic continuous function, both $h$ and $-h$ are in the set but $h + (-h)$ is not in the set. I think the purpose here is to prove that the set of all totally-nonmonotonic continuous function is dense in $C[X, \mathbb{R}]$. Could you anybody provide me some hints if you have any ideas?

Any responses will be appreciated.

twnly
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Sanae
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2 Answers2

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Build a continuous and nowhere monotonic $h$ on $[0,1]$ with $h(0)=0$,$h(1)=1$ and $|h|\le1$ as done here. For fixed $\epsilon>0$ there is some $n$ s.t. $|x-y|<1/n$ implies $|f(x)-f(y)|<\epsilon$. Let $I_k=[\frac kn,\frac{k+1}n]$, $k=0,...,n-1$ partition $[0,1]$ and set $h_k(x)=(f(\frac{k+1}n)-f(\frac kn))h(nx-k)+f(\frac kn)$. Observe $g(x)=\sum_k \chi_{I_k}h_k(x)$ is continuous and nowhere monotonic. Now any $x\in [0,1]$ belongs to some $I_k$ so $|f(x)-g(x)|\le|f(x)-f(k/n)|+|f(k/n)-g(x)|\le 2\epsilon$.

GuPe
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  • I suppose that the function $h$ your answer is the one appear in the first answer in your link. It is quite smart for you to "shrink" the whole $h$ into bunch of pieces by creating $h_k$. Thank you so much for your help! – Sanae Dec 27 '18 at 06:47
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A standard application of Baire Category Theorem says that the set of functions in $C[0,1]$ that are differntiable at at least one point is of first category. Hence its complement is dense. It follows that any $f \in C[0,1]$ can be approximated uniformly by nowhere differentiable functions. Any nowhere differentiable function is nowhere monotonic.

Kavi Rama Murthy
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  • First of all, thank you for your responses. You claim that in the metric space $(C[0, 1], d_{\infty})$, the set of all nowhere differentiable functions are dense, which is what confused me. I could not find any existing proofs online. Could you please provide me some hints about how to prove your claim? – Sanae Dec 27 '18 at 06:17
  • Theorem 17.8 of 'Real and Abstract Analysis' by Hewitt and Stromberg has a proof. @SanaeKochiya – Kavi Rama Murthy Dec 27 '18 at 06:30
  • Very appreciate your help. I got the pdf and will check it by myself. – Sanae Dec 27 '18 at 06:38