If $f$ is continuous and with a limit at infinity then $f$ is uniformly continuous
$begingroup$
Let $f : [a, +infty) rightarrow mathbb{R}$ be continuous, such that
$lim_{x rightarrow +infty} f(x) = l in mathbb{R}$. Show that f is
uniformly continuous.
I want to split the domain - one in which we have continuity, and the other in which the values of f are approaching $l$. This means there are theree possibilities for our $x$ and $y$ using the uniformly continuous definition: x and y are in the first part, x and y are in the second, x is in the first and y in the second.
How would I do this proof?
real-analysis limits continuity uniform-continuity
edited Nov 7 '16 at 20:16
Did
248k23224462
asked Nov 7 '16 at 19:01
user381164user381164
724
$endgroup$
add a comment |
$begingroup$
Let $f : [a, +infty) rightarrow mathbb{R}$ be continuous, such that
$lim_{x rightarrow +infty} f(x) = l in mathbb{R}$. Show that f is
uniformly continuous.
I want to split the domain - one in which we have continuity, and the other in which the values of f are approaching $l$. This means there are theree possibilities for our $x$ and $y$ using the uniformly continuous definition: x and y are in the first part, x and y are in the second, x is in the first and y in the second.
How would I do this proof?
real-analysis limits continuity uniform-continuity
edited Nov 7 '16 at 20:16
Did
248k23224462
asked Nov 7 '16 at 19:01
user381164user381164
724
$endgroup$
2
$begingroup$
To avoid splitting into three cases, you could allow the pieces to overlap and make sure to choose $delta$ less than the width of the overlap. This avoids the case where $x$ is in one piece and $y$ is in the other.
$endgroup$
– J. Loreaux
Nov 7 '16 at 19:06
$begingroup$
@GregordeCillia A uniformly continuous function can have unbounded derivative
$endgroup$
– zhw.
Nov 7 '16 at 19:12
$begingroup$
Yes. That's why I deleted my comment after a few mjnutes. I was thinking anout lipschitz continuouity and wanted to give an osciallator-like counterexample
$endgroup$
– Gregor de Cillia
Nov 7 '16 at 19:21
$begingroup$
Fix epsilon. Let delta(x) be the maximum delta used in the definition of continuity at x. Show delta(x) has a nonzero limit given the conditions of the problem.
$endgroup$
– Alephnull
Nov 7 '16 at 20:35
add a comment |
$begingroup$
Let $f : [a, +infty) rightarrow mathbb{R}$ be continuous, such that
$lim_{x rightarrow +infty} f(x) = l in mathbb{R}$. Show that f is
uniformly continuous.
I want to split the domain - one in which we have continuity, and the other in which the values of f are approaching $l$. This means there are theree possibilities for our $x$ and $y$ using the uniformly continuous definition: x and y are in the first part, x and y are in the second, x is in the first and y in the second.
How would I do this proof?
real-analysis limits continuity uniform-continuity
edited Nov 7 '16 at 20:16
Did
248k23224462
asked Nov 7 '16 at 19:01
user381164user381164
724
$endgroup$
Let $f : [a, +infty) rightarrow mathbb{R}$ be continuous, such that
$lim_{x rightarrow +infty} f(x) = l in mathbb{R}$. Show that f is
uniformly continuous.
I want to split the domain - one in which we have continuity, and the other in which the values of f are approaching $l$. This means there are theree possibilities for our $x$ and $y$ using the uniformly continuous definition: x and y are in the first part, x and y are in the second, x is in the first and y in the second.
How would I do this proof?
real-analysis limits continuity uniform-continuity
real-analysis limits continuity uniform-continuity
edited Nov 7 '16 at 20:16
Did
248k23224462
asked Nov 7 '16 at 19:01
user381164user381164
724
edited Nov 7 '16 at 20:16
Did
248k23224462
asked Nov 7 '16 at 19:01
user381164user381164
724
edited Nov 7 '16 at 20:16
Did
248k23224462
edited Nov 7 '16 at 20:16
Did
248k23224462
edited Nov 7 '16 at 20:16
Did
248k23224462
248k23224462
asked Nov 7 '16 at 19:01
user381164user381164
724
asked Nov 7 '16 at 19:01
user381164user381164
724
asked Nov 7 '16 at 19:01
user381164user381164
724
724
2
$begingroup$
To avoid splitting into three cases, you could allow the pieces to overlap and make sure to choose $delta$ less than the width of the overlap. This avoids the case where $x$ is in one piece and $y$ is in the other.
$endgroup$
– J. Loreaux
Nov 7 '16 at 19:06
$begingroup$
@GregordeCillia A uniformly continuous function can have unbounded derivative
$endgroup$
– zhw.
Nov 7 '16 at 19:12
$begingroup$
Yes. That's why I deleted my comment after a few mjnutes. I was thinking anout lipschitz continuouity and wanted to give an osciallator-like counterexample
$endgroup$
– Gregor de Cillia
Nov 7 '16 at 19:21
$begingroup$
Fix epsilon. Let delta(x) be the maximum delta used in the definition of continuity at x. Show delta(x) has a nonzero limit given the conditions of the problem.
$endgroup$
– Alephnull
Nov 7 '16 at 20:35
add a comment |
2
$begingroup$
To avoid splitting into three cases, you could allow the pieces to overlap and make sure to choose $delta$ less than the width of the overlap. This avoids the case where $x$ is in one piece and $y$ is in the other.
$endgroup$
– J. Loreaux
Nov 7 '16 at 19:06
$begingroup$
@GregordeCillia A uniformly continuous function can have unbounded derivative
$endgroup$
– zhw.
Nov 7 '16 at 19:12
$begingroup$
Yes. That's why I deleted my comment after a few mjnutes. I was thinking anout lipschitz continuouity and wanted to give an osciallator-like counterexample
$endgroup$
– Gregor de Cillia
Nov 7 '16 at 19:21
$begingroup$
Fix epsilon. Let delta(x) be the maximum delta used in the definition of continuity at x. Show delta(x) has a nonzero limit given the conditions of the problem.
$endgroup$
– Alephnull
Nov 7 '16 at 20:35
2
2
$begingroup$
To avoid splitting into three cases, you could allow the pieces to overlap and make sure to choose $delta$ less than the width of the overlap. This avoids the case where $x$ is in one piece and $y$ is in the other.
$endgroup$
– J. Loreaux
Nov 7 '16 at 19:06
$begingroup$
To avoid splitting into three cases, you could allow the pieces to overlap and make sure to choose $delta$ less than the width of the overlap. This avoids the case where $x$ is in one piece and $y$ is in the other.
$endgroup$
– J. Loreaux
Nov 7 '16 at 19:06
$begingroup$
@GregordeCillia A uniformly continuous function can have unbounded derivative
$endgroup$
– zhw.
Nov 7 '16 at 19:12
$begingroup$
@GregordeCillia A uniformly continuous function can have unbounded derivative
$endgroup$
– zhw.
Nov 7 '16 at 19:12
$begingroup$
Yes. That's why I deleted my comment after a few mjnutes. I was thinking anout lipschitz continuouity and wanted to give an osciallator-like counterexample
$endgroup$
– Gregor de Cillia
Nov 7 '16 at 19:21
$begingroup$
Yes. That's why I deleted my comment after a few mjnutes. I was thinking anout lipschitz continuouity and wanted to give an osciallator-like counterexample
$endgroup$
– Gregor de Cillia
Nov 7 '16 at 19:21
$begingroup$
Fix epsilon. Let delta(x) be the maximum delta used in the definition of continuity at x. Show delta(x) has a nonzero limit given the conditions of the problem.
$endgroup$
– Alephnull
Nov 7 '16 at 20:35
$begingroup$
Fix epsilon. Let delta(x) be the maximum delta used in the definition of continuity at x. Show delta(x) has a nonzero limit given the conditions of the problem.
$endgroup$
– Alephnull
Nov 7 '16 at 20:35
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
It suffices you do the following. Given $epsilon>0$, take $N$ large so that, for $x>N$,
$$|f(x)- l|<varepsilon/2$$
It follows that if $x,y>N$, then $|f(x)-f(y)|<varepsilon$. Now consider the interval $[a,N+1]$, which is compact. Here $f$ is uniformly continuous, so there is some $delta >0$ such that if $|x-y|<delta$ then $|f(x)-f(y)|<varepsilon$. We can assume that $delta<1$.
Consider now arbitrary $x,ygeqslant a$ that are at most $delta$ units apart. If both $x,yleqslant N+1$, then you're done. Our choice of $delta$ is such that if $x> N+1$ then also $y>N$, and you're also done.
answered Nov 7 '16 at 19:13
Pedro Tamaroff♦Pedro Tamaroff
96.9k10153297
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add a comment |
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1 Answer
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active
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1 Answer
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active
oldest
votes
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oldest
votes
active
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votes
$begingroup$
It suffices you do the following. Given $epsilon>0$, take $N$ large so that, for $x>N$,
$$|f(x)- l|<varepsilon/2$$
It follows that if $x,y>N$, then $|f(x)-f(y)|<varepsilon$. Now consider the interval $[a,N+1]$, which is compact. Here $f$ is uniformly continuous, so there is some $delta >0$ such that if $|x-y|<delta$ then $|f(x)-f(y)|<varepsilon$. We can assume that $delta<1$.
Consider now arbitrary $x,ygeqslant a$ that are at most $delta$ units apart. If both $x,yleqslant N+1$, then you're done. Our choice of $delta$ is such that if $x> N+1$ then also $y>N$, and you're also done.
answered Nov 7 '16 at 19:13
Pedro Tamaroff♦Pedro Tamaroff
96.9k10153297
$endgroup$
add a comment |
$begingroup$
It suffices you do the following. Given $epsilon>0$, take $N$ large so that, for $x>N$,
$$|f(x)- l|<varepsilon/2$$
It follows that if $x,y>N$, then $|f(x)-f(y)|<varepsilon$. Now consider the interval $[a,N+1]$, which is compact. Here $f$ is uniformly continuous, so there is some $delta >0$ such that if $|x-y|<delta$ then $|f(x)-f(y)|<varepsilon$. We can assume that $delta<1$.
Consider now arbitrary $x,ygeqslant a$ that are at most $delta$ units apart. If both $x,yleqslant N+1$, then you're done. Our choice of $delta$ is such that if $x> N+1$ then also $y>N$, and you're also done.
answered Nov 7 '16 at 19:13
Pedro Tamaroff♦Pedro Tamaroff
96.9k10153297
$endgroup$
add a comment |
$begingroup$
It suffices you do the following. Given $epsilon>0$, take $N$ large so that, for $x>N$,
$$|f(x)- l|<varepsilon/2$$
It follows that if $x,y>N$, then $|f(x)-f(y)|<varepsilon$. Now consider the interval $[a,N+1]$, which is compact. Here $f$ is uniformly continuous, so there is some $delta >0$ such that if $|x-y|<delta$ then $|f(x)-f(y)|<varepsilon$. We can assume that $delta<1$.
Consider now arbitrary $x,ygeqslant a$ that are at most $delta$ units apart. If both $x,yleqslant N+1$, then you're done. Our choice of $delta$ is such that if $x> N+1$ then also $y>N$, and you're also done.
answered Nov 7 '16 at 19:13
Pedro Tamaroff♦Pedro Tamaroff
96.9k10153297
$endgroup$
It suffices you do the following. Given $epsilon>0$, take $N$ large so that, for $x>N$,
$$|f(x)- l|<varepsilon/2$$
It follows that if $x,y>N$, then $|f(x)-f(y)|<varepsilon$. Now consider the interval $[a,N+1]$, which is compact. Here $f$ is uniformly continuous, so there is some $delta >0$ such that if $|x-y|<delta$ then $|f(x)-f(y)|<varepsilon$. We can assume that $delta<1$.
Consider now arbitrary $x,ygeqslant a$ that are at most $delta$ units apart. If both $x,yleqslant N+1$, then you're done. Our choice of $delta$ is such that if $x> N+1$ then also $y>N$, and you're also done.
answered Nov 7 '16 at 19:13
Pedro Tamaroff♦Pedro Tamaroff
96.9k10153297
answered Nov 7 '16 at 19:13
Pedro Tamaroff♦Pedro Tamaroff
96.9k10153297
answered Nov 7 '16 at 19:13
Pedro Tamaroff♦Pedro Tamaroff
96.9k10153297
answered Nov 7 '16 at 19:13
Pedro Tamaroff♦Pedro Tamaroff
96.9k10153297
96.9k10153297
add a comment |
add a comment |
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$begingroup$
To avoid splitting into three cases, you could allow the pieces to overlap and make sure to choose $delta$ less than the width of the overlap. This avoids the case where $x$ is in one piece and $y$ is in the other.
$endgroup$
– J. Loreaux
Nov 7 '16 at 19:06
$begingroup$
@GregordeCillia A uniformly continuous function can have unbounded derivative
$endgroup$
– zhw.
Nov 7 '16 at 19:12
$begingroup$
Yes. That's why I deleted my comment after a few mjnutes. I was thinking anout lipschitz continuouity and wanted to give an osciallator-like counterexample
$endgroup$
– Gregor de Cillia
Nov 7 '16 at 19:21
$begingroup$
Fix epsilon. Let delta(x) be the maximum delta used in the definition of continuity at x. Show delta(x) has a nonzero limit given the conditions of the problem.
$endgroup$
– Alephnull
Nov 7 '16 at 20:35