Dashawn Clark

2022-05-03

Find the antiderivative of f

Let $S=\{0\}\cup \{\frac{1}{n}:n\in \mathbb{N}\}$ and define $f:\mathbb{R}\to \mathbb{R}$

a) Prove that $|f(x)-f(y)|\le |x-y|$ for any $x,y\in \mathbb{R}$.

b) Find the antiderivatives of f

I obtained the following form for the function:

$f(x)=\{\begin{array}{ll}-x,\phantom{\rule{1em}{0ex}}& x\le 0\\ x-\frac{1}{n+1},\phantom{\rule{1em}{0ex}}& \frac{1}{n+1}\le x\le \frac{1}{2}(\frac{1}{n+1}+\frac{1}{n})\\ \frac{1}{n}-x,\phantom{\rule{1em}{0ex}}& \frac{1}{2}(\frac{1}{n+1}+\frac{1}{n})<x<\frac{1}{n}\\ x-1,\phantom{\rule{1em}{0ex}}& x\ge 1\end{array}$

I then tried to prove the inequality from a) considering many cases for x and y, but there are many possibilities. For the antiderivatives, I tried to combine the antiderivatives for each of the 4 branches of the function, but again, the calculations are messy.

Let $S=\{0\}\cup \{\frac{1}{n}:n\in \mathbb{N}\}$ and define $f:\mathbb{R}\to \mathbb{R}$

a) Prove that $|f(x)-f(y)|\le |x-y|$ for any $x,y\in \mathbb{R}$.

b) Find the antiderivatives of f

I obtained the following form for the function:

$f(x)=\{\begin{array}{ll}-x,\phantom{\rule{1em}{0ex}}& x\le 0\\ x-\frac{1}{n+1},\phantom{\rule{1em}{0ex}}& \frac{1}{n+1}\le x\le \frac{1}{2}(\frac{1}{n+1}+\frac{1}{n})\\ \frac{1}{n}-x,\phantom{\rule{1em}{0ex}}& \frac{1}{2}(\frac{1}{n+1}+\frac{1}{n})<x<\frac{1}{n}\\ x-1,\phantom{\rule{1em}{0ex}}& x\ge 1\end{array}$

I then tried to prove the inequality from a) considering many cases for x and y, but there are many possibilities. For the antiderivatives, I tried to combine the antiderivatives for each of the 4 branches of the function, but again, the calculations are messy.

Leia Wiggins

Beginner2022-05-04Added 18 answers

Step 1

Let S be a subset of $\mathbb{R}$.

1. For any $\u03f5>0$ there is ${s}^{\prime}\in S$ such that $|x-{s}^{\prime}|\le f(x)+\u03f5$. We know that $f(y)\le |y-{s}^{\prime}|=|y-x+x-{s}^{\prime}|\le |y-x|+f(x)+\u03f5$.

Since this holds for any $\u03f5>0$ we have $f(y)\le f(x)+|y-x|$.

Step 2

2. For any $\delta >0$ there is ${t}^{\prime}\in S$ such that $|y-{t}^{\prime}|\le f(y)+\delta $. Analogously, $f(x)\le |x-{t}^{\prime}|=|x-y+y-{t}^{\prime}|\le |x-y|+f(y)+\delta $.

Since this holds for any $\delta >0$ we have $f(x)\le f(y)+|y-x|$.

Step 3

3. Thus we have $f(x)-f(y)\le |x-y|\phantom{\rule{0ex}{0ex}}f(y)-f(x)\le |y-x|.$

Thus $|f(y)-f(x)|\le |y-x|$.

No let S be defined as in your example and you are done.

Let S be a subset of $\mathbb{R}$.

1. For any $\u03f5>0$ there is ${s}^{\prime}\in S$ such that $|x-{s}^{\prime}|\le f(x)+\u03f5$. We know that $f(y)\le |y-{s}^{\prime}|=|y-x+x-{s}^{\prime}|\le |y-x|+f(x)+\u03f5$.

Since this holds for any $\u03f5>0$ we have $f(y)\le f(x)+|y-x|$.

Step 2

2. For any $\delta >0$ there is ${t}^{\prime}\in S$ such that $|y-{t}^{\prime}|\le f(y)+\delta $. Analogously, $f(x)\le |x-{t}^{\prime}|=|x-y+y-{t}^{\prime}|\le |x-y|+f(y)+\delta $.

Since this holds for any $\delta >0$ we have $f(x)\le f(y)+|y-x|$.

Step 3

3. Thus we have $f(x)-f(y)\le |x-y|\phantom{\rule{0ex}{0ex}}f(y)-f(x)\le |y-x|.$

Thus $|f(y)-f(x)|\le |y-x|$.

No let S be defined as in your example and you are done.

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