If $f'$ is bounded on $mathbb{R}$, then is $f$ uniformly continuous?
$begingroup$
Say $f : mathbb{R} to mathbb{R}$ is differentiable.
Show if $f'$ is bounded, then $f$ is uniformly continuous.
Try
For any $[a, b] subset mathbb{R}$$(a<b)$, I can show $f$ is uniformly continuous, as follows.
Say $x_0 in [a,b]$ is given. Fix $epsilon >0$.
Since $exists M >0$ s.t. $|f'(x)| le M$, we show $|y - x_0| < epsilon/M Rightarrow |f(y) - f(x_0)| < epsilon$.
Assume $exists y_0 in [a,b]$ s.t. $|y_0 - x_0| < epsilon/M$ and $|f(y_0) - f(x_0)| ge epsilon$.
Since $frac{f(y_0) - f(x_0)}{y_0 - x_0} > epsilon times M/epsilon = M$, by MVT $exists c in left[min{x_0, y_0} max{x_0, y_0}right]$ s.t. $f'(c) = frac{f(y_0) - f(x_0)}{y_0 - x_0}$. Observe $|f'(c)| > M$, which is a contradiction to $|f'(x)| le M$.
But I'm not sure I can generalize this fact on $[a,b]$ to $mathbb{R}$.
real-analysis
asked Nov 29 '18 at 21:17
MoreblueMoreblue
8741216
$endgroup$
add a comment |
$begingroup$
Say $f : mathbb{R} to mathbb{R}$ is differentiable.
Show if $f'$ is bounded, then $f$ is uniformly continuous.
Try
For any $[a, b] subset mathbb{R}$$(a<b)$, I can show $f$ is uniformly continuous, as follows.
Say $x_0 in [a,b]$ is given. Fix $epsilon >0$.
Since $exists M >0$ s.t. $|f'(x)| le M$, we show $|y - x_0| < epsilon/M Rightarrow |f(y) - f(x_0)| < epsilon$.
Assume $exists y_0 in [a,b]$ s.t. $|y_0 - x_0| < epsilon/M$ and $|f(y_0) - f(x_0)| ge epsilon$.
Since $frac{f(y_0) - f(x_0)}{y_0 - x_0} > epsilon times M/epsilon = M$, by MVT $exists c in left[min{x_0, y_0} max{x_0, y_0}right]$ s.t. $f'(c) = frac{f(y_0) - f(x_0)}{y_0 - x_0}$. Observe $|f'(c)| > M$, which is a contradiction to $|f'(x)| le M$.
But I'm not sure I can generalize this fact on $[a,b]$ to $mathbb{R}$.
real-analysis
asked Nov 29 '18 at 21:17
MoreblueMoreblue
8741216
$endgroup$
$begingroup$
Moreblue.f' bounded implies f Lipschitz continuous implies f uniformly continuous.en.m.wikipedia.org/wiki/Lipschitz_continuity
$endgroup$
– Peter Szilas
Nov 29 '18 at 21:49
add a comment |
$begingroup$
Say $f : mathbb{R} to mathbb{R}$ is differentiable.
Show if $f'$ is bounded, then $f$ is uniformly continuous.
Try
For any $[a, b] subset mathbb{R}$$(a<b)$, I can show $f$ is uniformly continuous, as follows.
Say $x_0 in [a,b]$ is given. Fix $epsilon >0$.
Since $exists M >0$ s.t. $|f'(x)| le M$, we show $|y - x_0| < epsilon/M Rightarrow |f(y) - f(x_0)| < epsilon$.
Assume $exists y_0 in [a,b]$ s.t. $|y_0 - x_0| < epsilon/M$ and $|f(y_0) - f(x_0)| ge epsilon$.
Since $frac{f(y_0) - f(x_0)}{y_0 - x_0} > epsilon times M/epsilon = M$, by MVT $exists c in left[min{x_0, y_0} max{x_0, y_0}right]$ s.t. $f'(c) = frac{f(y_0) - f(x_0)}{y_0 - x_0}$. Observe $|f'(c)| > M$, which is a contradiction to $|f'(x)| le M$.
But I'm not sure I can generalize this fact on $[a,b]$ to $mathbb{R}$.
real-analysis
asked Nov 29 '18 at 21:17
MoreblueMoreblue
8741216
$endgroup$
Say $f : mathbb{R} to mathbb{R}$ is differentiable.
Show if $f'$ is bounded, then $f$ is uniformly continuous.
Try
For any $[a, b] subset mathbb{R}$$(a<b)$, I can show $f$ is uniformly continuous, as follows.
Say $x_0 in [a,b]$ is given. Fix $epsilon >0$.
Since $exists M >0$ s.t. $|f'(x)| le M$, we show $|y - x_0| < epsilon/M Rightarrow |f(y) - f(x_0)| < epsilon$.
Assume $exists y_0 in [a,b]$ s.t. $|y_0 - x_0| < epsilon/M$ and $|f(y_0) - f(x_0)| ge epsilon$.
Since $frac{f(y_0) - f(x_0)}{y_0 - x_0} > epsilon times M/epsilon = M$, by MVT $exists c in left[min{x_0, y_0} max{x_0, y_0}right]$ s.t. $f'(c) = frac{f(y_0) - f(x_0)}{y_0 - x_0}$. Observe $|f'(c)| > M$, which is a contradiction to $|f'(x)| le M$.
But I'm not sure I can generalize this fact on $[a,b]$ to $mathbb{R}$.
real-analysis
real-analysis
asked Nov 29 '18 at 21:17
MoreblueMoreblue
8741216
asked Nov 29 '18 at 21:17
MoreblueMoreblue
8741216
asked Nov 29 '18 at 21:17
MoreblueMoreblue
8741216
asked Nov 29 '18 at 21:17
MoreblueMoreblue
8741216
asked Nov 29 '18 at 21:17
MoreblueMoreblue
8741216
8741216
$begingroup$
Moreblue.f' bounded implies f Lipschitz continuous implies f uniformly continuous.en.m.wikipedia.org/wiki/Lipschitz_continuity
$endgroup$
– Peter Szilas
Nov 29 '18 at 21:49
add a comment |
$begingroup$
Moreblue.f' bounded implies f Lipschitz continuous implies f uniformly continuous.en.m.wikipedia.org/wiki/Lipschitz_continuity
$endgroup$
– Peter Szilas
Nov 29 '18 at 21:49
$begingroup$
Moreblue.f' bounded implies f Lipschitz continuous implies f uniformly continuous.en.m.wikipedia.org/wiki/Lipschitz_continuity
$endgroup$
– Peter Szilas
Nov 29 '18 at 21:49
$begingroup$
Moreblue.f' bounded implies f Lipschitz continuous implies f uniformly continuous.en.m.wikipedia.org/wiki/Lipschitz_continuity
$endgroup$
– Peter Szilas
Nov 29 '18 at 21:49
add a comment |
2 Answers
2
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oldest
votes
$begingroup$
hint
$f$ is differentiable at $Bbb R$ thus by MVT, for all $x,y$, there exists $c$ in $Bbb R$ such that
$$f(x)-f(y)=(x-y)f'(c)$$
but $f'$ is bounded, so
there exist $K> 0$ :
$$|f(x)-f(y)|le K|x-y|.$$
you can take $eta=frac{epsilon}{K}$.
answered Nov 29 '18 at 21:24
hamam_Abdallahhamam_Abdallah
38k21634
$endgroup$
add a comment |
$begingroup$
I don't think there is any trouble to generalize.
If $x,y in mathbb R$ the mean-value theorem gives you a point $z$ in between $x$ and $y$ satisfying $$f(x) - f(y) = f'(z)(x-y)$$ so that
$$|f(x) - f(y)| = |f'(z)| |x-y| le M |x-y|.$$
This holds for all $x$ and $y$.
Thus if $epsilon > 0$ and $|x-y| < epsilon /M$, then $|f(x) - f(y)| < epsilon$.
answered Nov 29 '18 at 21:22
Umberto P.Umberto P.
38.8k13064
$endgroup$
add a comment |
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2 Answers
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2 Answers
2
active
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$begingroup$
hint
$f$ is differentiable at $Bbb R$ thus by MVT, for all $x,y$, there exists $c$ in $Bbb R$ such that
$$f(x)-f(y)=(x-y)f'(c)$$
but $f'$ is bounded, so
there exist $K> 0$ :
$$|f(x)-f(y)|le K|x-y|.$$
you can take $eta=frac{epsilon}{K}$.
answered Nov 29 '18 at 21:24
hamam_Abdallahhamam_Abdallah
38k21634
$endgroup$
add a comment |
$begingroup$
hint
$f$ is differentiable at $Bbb R$ thus by MVT, for all $x,y$, there exists $c$ in $Bbb R$ such that
$$f(x)-f(y)=(x-y)f'(c)$$
but $f'$ is bounded, so
there exist $K> 0$ :
$$|f(x)-f(y)|le K|x-y|.$$
you can take $eta=frac{epsilon}{K}$.
answered Nov 29 '18 at 21:24
hamam_Abdallahhamam_Abdallah
38k21634
$endgroup$
add a comment |
$begingroup$
hint
$f$ is differentiable at $Bbb R$ thus by MVT, for all $x,y$, there exists $c$ in $Bbb R$ such that
$$f(x)-f(y)=(x-y)f'(c)$$
but $f'$ is bounded, so
there exist $K> 0$ :
$$|f(x)-f(y)|le K|x-y|.$$
you can take $eta=frac{epsilon}{K}$.
answered Nov 29 '18 at 21:24
hamam_Abdallahhamam_Abdallah
38k21634
$endgroup$
hint
$f$ is differentiable at $Bbb R$ thus by MVT, for all $x,y$, there exists $c$ in $Bbb R$ such that
$$f(x)-f(y)=(x-y)f'(c)$$
but $f'$ is bounded, so
there exist $K> 0$ :
$$|f(x)-f(y)|le K|x-y|.$$
you can take $eta=frac{epsilon}{K}$.
answered Nov 29 '18 at 21:24
hamam_Abdallahhamam_Abdallah
38k21634
answered Nov 29 '18 at 21:24
hamam_Abdallahhamam_Abdallah
38k21634
answered Nov 29 '18 at 21:24
hamam_Abdallahhamam_Abdallah
38k21634
answered Nov 29 '18 at 21:24
hamam_Abdallahhamam_Abdallah
38k21634
38k21634
add a comment |
add a comment |
$begingroup$
I don't think there is any trouble to generalize.
If $x,y in mathbb R$ the mean-value theorem gives you a point $z$ in between $x$ and $y$ satisfying $$f(x) - f(y) = f'(z)(x-y)$$ so that
$$|f(x) - f(y)| = |f'(z)| |x-y| le M |x-y|.$$
This holds for all $x$ and $y$.
Thus if $epsilon > 0$ and $|x-y| < epsilon /M$, then $|f(x) - f(y)| < epsilon$.
answered Nov 29 '18 at 21:22
Umberto P.Umberto P.
38.8k13064
$endgroup$
add a comment |
$begingroup$
I don't think there is any trouble to generalize.
If $x,y in mathbb R$ the mean-value theorem gives you a point $z$ in between $x$ and $y$ satisfying $$f(x) - f(y) = f'(z)(x-y)$$ so that
$$|f(x) - f(y)| = |f'(z)| |x-y| le M |x-y|.$$
This holds for all $x$ and $y$.
Thus if $epsilon > 0$ and $|x-y| < epsilon /M$, then $|f(x) - f(y)| < epsilon$.
answered Nov 29 '18 at 21:22
Umberto P.Umberto P.
38.8k13064
$endgroup$
add a comment |
$begingroup$
I don't think there is any trouble to generalize.
If $x,y in mathbb R$ the mean-value theorem gives you a point $z$ in between $x$ and $y$ satisfying $$f(x) - f(y) = f'(z)(x-y)$$ so that
$$|f(x) - f(y)| = |f'(z)| |x-y| le M |x-y|.$$
This holds for all $x$ and $y$.
Thus if $epsilon > 0$ and $|x-y| < epsilon /M$, then $|f(x) - f(y)| < epsilon$.
answered Nov 29 '18 at 21:22
Umberto P.Umberto P.
38.8k13064
$endgroup$
I don't think there is any trouble to generalize.
If $x,y in mathbb R$ the mean-value theorem gives you a point $z$ in between $x$ and $y$ satisfying $$f(x) - f(y) = f'(z)(x-y)$$ so that
$$|f(x) - f(y)| = |f'(z)| |x-y| le M |x-y|.$$
This holds for all $x$ and $y$.
Thus if $epsilon > 0$ and $|x-y| < epsilon /M$, then $|f(x) - f(y)| < epsilon$.
answered Nov 29 '18 at 21:22
Umberto P.Umberto P.
38.8k13064
answered Nov 29 '18 at 21:22
Umberto P.Umberto P.
38.8k13064
answered Nov 29 '18 at 21:22
Umberto P.Umberto P.
38.8k13064
answered Nov 29 '18 at 21:22
Umberto P.Umberto P.
38.8k13064
38.8k13064
add a comment |
add a comment |
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$begingroup$
Moreblue.f' bounded implies f Lipschitz continuous implies f uniformly continuous.en.m.wikipedia.org/wiki/Lipschitz_continuity
$endgroup$
– Peter Szilas
Nov 29 '18 at 21:49