Is there a bounded rational function that is not uniformly continuous on $mathbb{R}$?
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I think the answer is no. I tried looking for one hour and it didn't work. Isn't every rational bounded function uniformly continuous ? There's nothing online about them.
real-analysis uniform-continuity
asked Nov 16 at 21:08
joseph
1068
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up vote
2
down vote
favorite
I think the answer is no. I tried looking for one hour and it didn't work. Isn't every rational bounded function uniformly continuous ? There's nothing online about them.
real-analysis uniform-continuity
asked Nov 16 at 21:08
joseph
1068
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I think the answer is no. I tried looking for one hour and it didn't work. Isn't every rational bounded function uniformly continuous ? There's nothing online about them.
real-analysis uniform-continuity
asked Nov 16 at 21:08
joseph
1068
I think the answer is no. I tried looking for one hour and it didn't work. Isn't every rational bounded function uniformly continuous ? There's nothing online about them.
real-analysis uniform-continuity
real-analysis uniform-continuity
asked Nov 16 at 21:08
joseph
1068
asked Nov 16 at 21:08
joseph
1068
asked Nov 16 at 21:08
joseph
1068
asked Nov 16 at 21:08
joseph
1068
asked Nov 16 at 21:08
joseph
1068
1068
add a comment |
add a comment |
1 Answer
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Suppose $f=p/q$ is a rational function with coefficients in $mathbb{R}$, written as a quotient of polynomials $p$ and $q$ in lowest terms. Then $f$ is bounded on $mathbb{R}$ iff $q$ has no real roots (so $f$ has no real poles) and $deg pleq deg q$ (so $f(x)$ stays bounded as $xtopminfty$).
Now suppose $f$ is bounded and consider the derivative $$f'=frac{p'q-pq'}{q^2}.$$ We see that the denominator of $f'$ also has no real roots and the degree of the denominator is at least that of the numerator (in fact, the degree of the denominator is now strictly greater). Thus $f'$ is also bounded on $mathbb{R}$. It follows that $f$ is Lipschitz and in particular uniformly continuous.
answered Nov 16 at 21:15
Eric Wofsey
175k12202326
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
6
down vote
accepted
Suppose $f=p/q$ is a rational function with coefficients in $mathbb{R}$, written as a quotient of polynomials $p$ and $q$ in lowest terms. Then $f$ is bounded on $mathbb{R}$ iff $q$ has no real roots (so $f$ has no real poles) and $deg pleq deg q$ (so $f(x)$ stays bounded as $xtopminfty$).
Now suppose $f$ is bounded and consider the derivative $$f'=frac{p'q-pq'}{q^2}.$$ We see that the denominator of $f'$ also has no real roots and the degree of the denominator is at least that of the numerator (in fact, the degree of the denominator is now strictly greater). Thus $f'$ is also bounded on $mathbb{R}$. It follows that $f$ is Lipschitz and in particular uniformly continuous.
answered Nov 16 at 21:15
Eric Wofsey
175k12202326
add a comment |
up vote
6
down vote
accepted
Suppose $f=p/q$ is a rational function with coefficients in $mathbb{R}$, written as a quotient of polynomials $p$ and $q$ in lowest terms. Then $f$ is bounded on $mathbb{R}$ iff $q$ has no real roots (so $f$ has no real poles) and $deg pleq deg q$ (so $f(x)$ stays bounded as $xtopminfty$).
Now suppose $f$ is bounded and consider the derivative $$f'=frac{p'q-pq'}{q^2}.$$ We see that the denominator of $f'$ also has no real roots and the degree of the denominator is at least that of the numerator (in fact, the degree of the denominator is now strictly greater). Thus $f'$ is also bounded on $mathbb{R}$. It follows that $f$ is Lipschitz and in particular uniformly continuous.
answered Nov 16 at 21:15
Eric Wofsey
175k12202326
add a comment |
up vote
6
down vote
accepted
up vote
6
down vote
accepted
Suppose $f=p/q$ is a rational function with coefficients in $mathbb{R}$, written as a quotient of polynomials $p$ and $q$ in lowest terms. Then $f$ is bounded on $mathbb{R}$ iff $q$ has no real roots (so $f$ has no real poles) and $deg pleq deg q$ (so $f(x)$ stays bounded as $xtopminfty$).
Now suppose $f$ is bounded and consider the derivative $$f'=frac{p'q-pq'}{q^2}.$$ We see that the denominator of $f'$ also has no real roots and the degree of the denominator is at least that of the numerator (in fact, the degree of the denominator is now strictly greater). Thus $f'$ is also bounded on $mathbb{R}$. It follows that $f$ is Lipschitz and in particular uniformly continuous.
answered Nov 16 at 21:15
Eric Wofsey
175k12202326
Suppose $f=p/q$ is a rational function with coefficients in $mathbb{R}$, written as a quotient of polynomials $p$ and $q$ in lowest terms. Then $f$ is bounded on $mathbb{R}$ iff $q$ has no real roots (so $f$ has no real poles) and $deg pleq deg q$ (so $f(x)$ stays bounded as $xtopminfty$).
Now suppose $f$ is bounded and consider the derivative $$f'=frac{p'q-pq'}{q^2}.$$ We see that the denominator of $f'$ also has no real roots and the degree of the denominator is at least that of the numerator (in fact, the degree of the denominator is now strictly greater). Thus $f'$ is also bounded on $mathbb{R}$. It follows that $f$ is Lipschitz and in particular uniformly continuous.
answered Nov 16 at 21:15
Eric Wofsey
175k12202326
answered Nov 16 at 21:15
Eric Wofsey
175k12202326
answered Nov 16 at 21:15
Eric Wofsey
175k12202326
answered Nov 16 at 21:15
Eric Wofsey
175k12202326
175k12202326
add a comment |
add a comment |
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