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2022-07-22

Proving the existence of a non-monotone continuous function defined on [0,1]

Let $({I}_{n}{)}_{n\in \mathbb{N}}$ be the sequence of intervals of [0,1] with rational endpoints, and for every $n\in \mathbb{N}\text{}$ let ${E}_{n}=\{f\in C[0,1]:f\phantom{\rule{mediummathspace}{0ex}}\text{is monotone in}\phantom{\rule{mediummathspace}{0ex}}{I}_{n}\}$. Prove that for every $n\in \mathbb{N}$, ${E}_{n}$ is closed and nowhere dense in $(C[0,1],{d}_{\mathrm{\infty}})$. Deduce that there are continuous functions in the interval [0,1] which aren't monotone in any subinterval.

For a given n, ${E}_{n}$ can be expressed as ${E}_{n}={E}_{n\nearrow}\cup {E}_{n\swarrow}$ where ${E}_{n\nearrow}$ and ${E}_{n\swarrow}$ are the sets of monotonically increasing functions and monotonically decreasing functions in ${E}_{n}$ respectively. I am having problems trying to prove that these two sets are closed. I mean, take $f\in (C[0,1],{d}_{\mathrm{\infty}})$ such that there is $\{{f}_{k}{\}}_{k\in \mathbb{N}}\subset {E}_{n\nearrow}$ with ${f}_{k}\to f$. How can I prove $f\in {E}_{n\nearrow}$?. Suppose I could prove this, then I have to show that ${\overline{{E}_{n}}}^{\circ}={E}_{n}^{\circ}=\mathrm{\varnothing}$. This means that for every $f\in {E}_{n}$ and every $r>0$, there is $g\in B(f,r)$ such that g is not monotone. Again, I am stuck. If I could solve this two points, it's not difficult to check the hypothesis and apply the Baire category theorem to prove the last statement.

Let $({I}_{n}{)}_{n\in \mathbb{N}}$ be the sequence of intervals of [0,1] with rational endpoints, and for every $n\in \mathbb{N}\text{}$ let ${E}_{n}=\{f\in C[0,1]:f\phantom{\rule{mediummathspace}{0ex}}\text{is monotone in}\phantom{\rule{mediummathspace}{0ex}}{I}_{n}\}$. Prove that for every $n\in \mathbb{N}$, ${E}_{n}$ is closed and nowhere dense in $(C[0,1],{d}_{\mathrm{\infty}})$. Deduce that there are continuous functions in the interval [0,1] which aren't monotone in any subinterval.

For a given n, ${E}_{n}$ can be expressed as ${E}_{n}={E}_{n\nearrow}\cup {E}_{n\swarrow}$ where ${E}_{n\nearrow}$ and ${E}_{n\swarrow}$ are the sets of monotonically increasing functions and monotonically decreasing functions in ${E}_{n}$ respectively. I am having problems trying to prove that these two sets are closed. I mean, take $f\in (C[0,1],{d}_{\mathrm{\infty}})$ such that there is $\{{f}_{k}{\}}_{k\in \mathbb{N}}\subset {E}_{n\nearrow}$ with ${f}_{k}\to f$. How can I prove $f\in {E}_{n\nearrow}$?. Suppose I could prove this, then I have to show that ${\overline{{E}_{n}}}^{\circ}={E}_{n}^{\circ}=\mathrm{\varnothing}$. This means that for every $f\in {E}_{n}$ and every $r>0$, there is $g\in B(f,r)$ such that g is not monotone. Again, I am stuck. If I could solve this two points, it's not difficult to check the hypothesis and apply the Baire category theorem to prove the last statement.

Kali Galloway

Beginner2022-07-23Added 16 answers

Step 1

It is easy to show that ${E}_{n}$ is close. Just like what you did, let ${E}_{n}^{1}$ be the subset of ${E}_{n}$ containing all functions increasing on ${I}_{n}$. Let $\{{f}_{k}\}$ be a sequence in ${E}_{n}$ converging to $f\in C[0,1]$. Pick $x,y\in {I}_{n}$, $x<y$. Then ${f}_{n}(x)\le {f}_{n}(y)$ for all n. Take $n\to \mathrm{\infty}$ gives $f(x)\le f(y)$. Thus ${E}_{n}^{1}$ is closed. Now ${E}_{n}$ is the union of two close set, so it is close.

To show that ${E}_{n}$ has empty interior, let $f\in {E}_{n}$. Assume that f is inceasing (the case when f is decreasing can be done similarly). We perturb f a little bit to get a function h which is not monotone. Let Write ${I}_{n}=[a,b]$. Let $\u03f5>0$ be arbitrary. Let ${c}_{1},{c}_{2}\in (a,b)$ such that ${c}_{1}<{c}_{2},\text{and}\text{}f(b)-\u03f5f({c}_{2})\le f(b).$

${c}_{1},{c}_{2}$ can be found as f is continuous. Let g be a continuous function on [0,1] with the properties:

$|g|\le \u03f5,\text{}g(a)=g(b)=-\u03f5,\text{}g({c}_{1})=g({c}_{2})=0,\text{}.$

Step 2

Then $h=f-g$ satisfies $d(f,h{)}_{\mathrm{\infty}}\le \u03f5$ and $h(a)=f(a)-\u03f5<f(a)\le f({c}_{1})=h({c}_{1})$ and $h({c}_{2})=f({c}_{2})>f(b)-\u03f5=h(b)$. Thus h is not monotone. As f, $\u03f5>0$ are arbitrary, ${E}_{n}$ contains no open set. Hence ${E}_{n}$ are nowhere dense. (The conclusion is quite interesting).

It is easy to show that ${E}_{n}$ is close. Just like what you did, let ${E}_{n}^{1}$ be the subset of ${E}_{n}$ containing all functions increasing on ${I}_{n}$. Let $\{{f}_{k}\}$ be a sequence in ${E}_{n}$ converging to $f\in C[0,1]$. Pick $x,y\in {I}_{n}$, $x<y$. Then ${f}_{n}(x)\le {f}_{n}(y)$ for all n. Take $n\to \mathrm{\infty}$ gives $f(x)\le f(y)$. Thus ${E}_{n}^{1}$ is closed. Now ${E}_{n}$ is the union of two close set, so it is close.

To show that ${E}_{n}$ has empty interior, let $f\in {E}_{n}$. Assume that f is inceasing (the case when f is decreasing can be done similarly). We perturb f a little bit to get a function h which is not monotone. Let Write ${I}_{n}=[a,b]$. Let $\u03f5>0$ be arbitrary. Let ${c}_{1},{c}_{2}\in (a,b)$ such that ${c}_{1}<{c}_{2},\text{and}\text{}f(b)-\u03f5f({c}_{2})\le f(b).$

${c}_{1},{c}_{2}$ can be found as f is continuous. Let g be a continuous function on [0,1] with the properties:

$|g|\le \u03f5,\text{}g(a)=g(b)=-\u03f5,\text{}g({c}_{1})=g({c}_{2})=0,\text{}.$

Step 2

Then $h=f-g$ satisfies $d(f,h{)}_{\mathrm{\infty}}\le \u03f5$ and $h(a)=f(a)-\u03f5<f(a)\le f({c}_{1})=h({c}_{1})$ and $h({c}_{2})=f({c}_{2})>f(b)-\u03f5=h(b)$. Thus h is not monotone. As f, $\u03f5>0$ are arbitrary, ${E}_{n}$ contains no open set. Hence ${E}_{n}$ are nowhere dense. (The conclusion is quite interesting).

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