Mixing
$f$ is continuous function on $[0, infty)$, and the limit $lim_{n to infty} frac{f(x)}{x}= a in mathbb{R}$...
$begingroup$
I'm not sure how to prove or give a counter example to this. I wasn't able to prove it but I couldn't think of any counter example. Here's what I tried: By definition there exists $M>0$ such that for every $x>0$, $(a-1)x<f(x)<(a+1)x$. $f$ is uniformly continuous in $[0,M]$ (Cantor), but I'm not sure how to prove this for $(M, infty)$. Can someone please help?
real-analysis calculus continuity uniform-continuity
asked Jan 22 at 10:24
OmerOmer
3969
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add a comment |
$begingroup$
I'm not sure how to prove or give a counter example to this. I wasn't able to prove it but I couldn't think of any counter example. Here's what I tried: By definition there exists $M>0$ such that for every $x>0$, $(a-1)x<f(x)<(a+1)x$. $f$ is uniformly continuous in $[0,M]$ (Cantor), but I'm not sure how to prove this for $(M, infty)$. Can someone please help?
real-analysis calculus continuity uniform-continuity
asked Jan 22 at 10:24
OmerOmer
3969
$endgroup$
$begingroup$
@Surb but the derivative of ln(x+1) is 1/(x+1) which is bounded in [0,infinity) so ln(x+1) is uniformly continuous.
$endgroup$
– Omer
Jan 22 at 10:28
1
$begingroup$
How about $f(x)=sin{e^{x^2}}$?
$endgroup$
– Mindlack
Jan 22 at 10:29
2
$begingroup$
@Mindlack actually, I think sin(x^2) is enough. because if we take $x_n=sqrt{2 pi n}$ and $y_n= sqrt{2 pi n + pi/2}$ then $x_n-y_n$ tends to 0 but f(x_n)-f(y_n) doesn't.
$endgroup$
– Omer
Jan 22 at 10:33
$begingroup$
@DavideGiraudo Yes, sorry, I edited the title.
$endgroup$
– Omer
Jan 22 at 10:35
$begingroup$
Intuitively, for nice enough functions (in this case I would say $C^1$), uniform continuity is equivalent to boundedness of the derivative, and the limit condition to boundedness of the function. So unless the counterexample is "pathological", you should be looking for a bounded function with unbounded derivative, and $sin(x^2)$ is a fine example (or more generally $sin$ of any fast growing function).
$endgroup$
– Mees de Vries
Jan 22 at 10:37
add a comment |
$begingroup$
I'm not sure how to prove or give a counter example to this. I wasn't able to prove it but I couldn't think of any counter example. Here's what I tried: By definition there exists $M>0$ such that for every $x>0$, $(a-1)x<f(x)<(a+1)x$. $f$ is uniformly continuous in $[0,M]$ (Cantor), but I'm not sure how to prove this for $(M, infty)$. Can someone please help?
real-analysis calculus continuity uniform-continuity
asked Jan 22 at 10:24
OmerOmer
3969
$endgroup$
I'm not sure how to prove or give a counter example to this. I wasn't able to prove it but I couldn't think of any counter example. Here's what I tried: By definition there exists $M>0$ such that for every $x>0$, $(a-1)x<f(x)<(a+1)x$. $f$ is uniformly continuous in $[0,M]$ (Cantor), but I'm not sure how to prove this for $(M, infty)$. Can someone please help?
real-analysis calculus continuity uniform-continuity
real-analysis calculus continuity uniform-continuity
asked Jan 22 at 10:24
OmerOmer
3969
asked Jan 22 at 10:24
OmerOmer
3969
edited Jan 22 at 10:34
Omer
asked Jan 22 at 10:24
OmerOmer
3969
asked Jan 22 at 10:24
OmerOmer
3969
asked Jan 22 at 10:24
OmerOmer
3969
3969
$begingroup$
@Surb but the derivative of ln(x+1) is 1/(x+1) which is bounded in [0,infinity) so ln(x+1) is uniformly continuous.
$endgroup$
– Omer
Jan 22 at 10:28
1
$begingroup$
How about $f(x)=sin{e^{x^2}}$?
$endgroup$
– Mindlack
Jan 22 at 10:29
2
$begingroup$
@Mindlack actually, I think sin(x^2) is enough. because if we take $x_n=sqrt{2 pi n}$ and $y_n= sqrt{2 pi n + pi/2}$ then $x_n-y_n$ tends to 0 but f(x_n)-f(y_n) doesn't.
$endgroup$
– Omer
Jan 22 at 10:33
$begingroup$
@DavideGiraudo Yes, sorry, I edited the title.
$endgroup$
– Omer
Jan 22 at 10:35
$begingroup$
Intuitively, for nice enough functions (in this case I would say $C^1$), uniform continuity is equivalent to boundedness of the derivative, and the limit condition to boundedness of the function. So unless the counterexample is "pathological", you should be looking for a bounded function with unbounded derivative, and $sin(x^2)$ is a fine example (or more generally $sin$ of any fast growing function).
$endgroup$
– Mees de Vries
Jan 22 at 10:37
add a comment |
$begingroup$
@Surb but the derivative of ln(x+1) is 1/(x+1) which is bounded in [0,infinity) so ln(x+1) is uniformly continuous.
$endgroup$
– Omer
Jan 22 at 10:28
1
$begingroup$
How about $f(x)=sin{e^{x^2}}$?
$endgroup$
– Mindlack
Jan 22 at 10:29
2
$begingroup$
@Mindlack actually, I think sin(x^2) is enough. because if we take $x_n=sqrt{2 pi n}$ and $y_n= sqrt{2 pi n + pi/2}$ then $x_n-y_n$ tends to 0 but f(x_n)-f(y_n) doesn't.
$endgroup$
– Omer
Jan 22 at 10:33
$begingroup$
@DavideGiraudo Yes, sorry, I edited the title.
$endgroup$
– Omer
Jan 22 at 10:35
$begingroup$
Intuitively, for nice enough functions (in this case I would say $C^1$), uniform continuity is equivalent to boundedness of the derivative, and the limit condition to boundedness of the function. So unless the counterexample is "pathological", you should be looking for a bounded function with unbounded derivative, and $sin(x^2)$ is a fine example (or more generally $sin$ of any fast growing function).
$endgroup$
– Mees de Vries
Jan 22 at 10:37
$begingroup$
@Surb but the derivative of ln(x+1) is 1/(x+1) which is bounded in [0,infinity) so ln(x+1) is uniformly continuous.
$endgroup$
– Omer
Jan 22 at 10:28
$begingroup$
@Surb but the derivative of ln(x+1) is 1/(x+1) which is bounded in [0,infinity) so ln(x+1) is uniformly continuous.
$endgroup$
– Omer
Jan 22 at 10:28
1
1
$begingroup$
How about $f(x)=sin{e^{x^2}}$?
$endgroup$
– Mindlack
Jan 22 at 10:29
$begingroup$
How about $f(x)=sin{e^{x^2}}$?
$endgroup$
– Mindlack
Jan 22 at 10:29
2
2
$begingroup$
@Mindlack actually, I think sin(x^2) is enough. because if we take $x_n=sqrt{2 pi n}$ and $y_n= sqrt{2 pi n + pi/2}$ then $x_n-y_n$ tends to 0 but f(x_n)-f(y_n) doesn't.
$endgroup$
– Omer
Jan 22 at 10:33
$begingroup$
@Mindlack actually, I think sin(x^2) is enough. because if we take $x_n=sqrt{2 pi n}$ and $y_n= sqrt{2 pi n + pi/2}$ then $x_n-y_n$ tends to 0 but f(x_n)-f(y_n) doesn't.
$endgroup$
– Omer
Jan 22 at 10:33
$begingroup$
@DavideGiraudo Yes, sorry, I edited the title.
$endgroup$
– Omer
Jan 22 at 10:35
$begingroup$
@DavideGiraudo Yes, sorry, I edited the title.
$endgroup$
– Omer
Jan 22 at 10:35
$begingroup$
Intuitively, for nice enough functions (in this case I would say $C^1$), uniform continuity is equivalent to boundedness of the derivative, and the limit condition to boundedness of the function. So unless the counterexample is "pathological", you should be looking for a bounded function with unbounded derivative, and $sin(x^2)$ is a fine example (or more generally $sin$ of any fast growing function).
$endgroup$
– Mees de Vries
Jan 22 at 10:37
$begingroup$
Intuitively, for nice enough functions (in this case I would say $C^1$), uniform continuity is equivalent to boundedness of the derivative, and the limit condition to boundedness of the function. So unless the counterexample is "pathological", you should be looking for a bounded function with unbounded derivative, and $sin(x^2)$ is a fine example (or more generally $sin$ of any fast growing function).
$endgroup$
– Mees de Vries
Jan 22 at 10:37
add a comment |
1 Answer
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No.
For a function to be uniformly continuous you can't have segments where the "rate of change" is arbitrarily large.
So for example, $f(x) = cos(x^2)$ is not uniformly continuous, because there are arbitrarily short segments where the function goes from $-1$ to $1$ ($[sqrt{(2n-1)pi},sqrt{2npi}]$ for each $n$).
However, this does not constitute a counterexample because clearly $f(x)$ has no limit at infinity.
Alright, we can fix this, $f(x)$ is bounded so multiplying it by a vanishing function should do the trick, and indeed $g(x) = frac{cos(x^2)}{x}$ vanishes at infinity.
This is no good either, because (as you can check) $g(x)$ has a bound derivative and therefore it is uniformly continuous.
The geometric meaning of this is essentially that multiplying by $1/x$ has tamed the rate of change so much that it is no longer arbitrarily fast. To overcome this, we need to make the rate of change even faster.
So we choose $h(x) = frac{cos(x^3)}{x}$ (or, alternatively, make the decline to $0$ softer by choosing $h(x) = frac{cos(x^2)}{sqrt{x}}$), which, as you can readily check, is not uniformly continuous.
answered Jan 22 at 10:38
Shai DesheShai Deshe
921513
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add a comment |
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$begingroup$
No.
For a function to be uniformly continuous you can't have segments where the "rate of change" is arbitrarily large.
So for example, $f(x) = cos(x^2)$ is not uniformly continuous, because there are arbitrarily short segments where the function goes from $-1$ to $1$ ($[sqrt{(2n-1)pi},sqrt{2npi}]$ for each $n$).
However, this does not constitute a counterexample because clearly $f(x)$ has no limit at infinity.
Alright, we can fix this, $f(x)$ is bounded so multiplying it by a vanishing function should do the trick, and indeed $g(x) = frac{cos(x^2)}{x}$ vanishes at infinity.
This is no good either, because (as you can check) $g(x)$ has a bound derivative and therefore it is uniformly continuous.
The geometric meaning of this is essentially that multiplying by $1/x$ has tamed the rate of change so much that it is no longer arbitrarily fast. To overcome this, we need to make the rate of change even faster.
So we choose $h(x) = frac{cos(x^3)}{x}$ (or, alternatively, make the decline to $0$ softer by choosing $h(x) = frac{cos(x^2)}{sqrt{x}}$), which, as you can readily check, is not uniformly continuous.
answered Jan 22 at 10:38
Shai DesheShai Deshe
921513
$endgroup$
add a comment |
$begingroup$
No.
For a function to be uniformly continuous you can't have segments where the "rate of change" is arbitrarily large.
So for example, $f(x) = cos(x^2)$ is not uniformly continuous, because there are arbitrarily short segments where the function goes from $-1$ to $1$ ($[sqrt{(2n-1)pi},sqrt{2npi}]$ for each $n$).
However, this does not constitute a counterexample because clearly $f(x)$ has no limit at infinity.
Alright, we can fix this, $f(x)$ is bounded so multiplying it by a vanishing function should do the trick, and indeed $g(x) = frac{cos(x^2)}{x}$ vanishes at infinity.
This is no good either, because (as you can check) $g(x)$ has a bound derivative and therefore it is uniformly continuous.
The geometric meaning of this is essentially that multiplying by $1/x$ has tamed the rate of change so much that it is no longer arbitrarily fast. To overcome this, we need to make the rate of change even faster.
So we choose $h(x) = frac{cos(x^3)}{x}$ (or, alternatively, make the decline to $0$ softer by choosing $h(x) = frac{cos(x^2)}{sqrt{x}}$), which, as you can readily check, is not uniformly continuous.
answered Jan 22 at 10:38
Shai DesheShai Deshe
921513
$endgroup$
add a comment |
$begingroup$
No.
For a function to be uniformly continuous you can't have segments where the "rate of change" is arbitrarily large.
So for example, $f(x) = cos(x^2)$ is not uniformly continuous, because there are arbitrarily short segments where the function goes from $-1$ to $1$ ($[sqrt{(2n-1)pi},sqrt{2npi}]$ for each $n$).
However, this does not constitute a counterexample because clearly $f(x)$ has no limit at infinity.
Alright, we can fix this, $f(x)$ is bounded so multiplying it by a vanishing function should do the trick, and indeed $g(x) = frac{cos(x^2)}{x}$ vanishes at infinity.
This is no good either, because (as you can check) $g(x)$ has a bound derivative and therefore it is uniformly continuous.
The geometric meaning of this is essentially that multiplying by $1/x$ has tamed the rate of change so much that it is no longer arbitrarily fast. To overcome this, we need to make the rate of change even faster.
So we choose $h(x) = frac{cos(x^3)}{x}$ (or, alternatively, make the decline to $0$ softer by choosing $h(x) = frac{cos(x^2)}{sqrt{x}}$), which, as you can readily check, is not uniformly continuous.
answered Jan 22 at 10:38
Shai DesheShai Deshe
921513
$endgroup$
No.
For a function to be uniformly continuous you can't have segments where the "rate of change" is arbitrarily large.
So for example, $f(x) = cos(x^2)$ is not uniformly continuous, because there are arbitrarily short segments where the function goes from $-1$ to $1$ ($[sqrt{(2n-1)pi},sqrt{2npi}]$ for each $n$).
However, this does not constitute a counterexample because clearly $f(x)$ has no limit at infinity.
Alright, we can fix this, $f(x)$ is bounded so multiplying it by a vanishing function should do the trick, and indeed $g(x) = frac{cos(x^2)}{x}$ vanishes at infinity.
This is no good either, because (as you can check) $g(x)$ has a bound derivative and therefore it is uniformly continuous.
The geometric meaning of this is essentially that multiplying by $1/x$ has tamed the rate of change so much that it is no longer arbitrarily fast. To overcome this, we need to make the rate of change even faster.
So we choose $h(x) = frac{cos(x^3)}{x}$ (or, alternatively, make the decline to $0$ softer by choosing $h(x) = frac{cos(x^2)}{sqrt{x}}$), which, as you can readily check, is not uniformly continuous.
answered Jan 22 at 10:38
Shai DesheShai Deshe
921513
answered Jan 22 at 10:38
Shai DesheShai Deshe
921513
answered Jan 22 at 10:38
Shai DesheShai Deshe
921513
answered Jan 22 at 10:38
Shai DesheShai Deshe
921513
921513
add a comment |
add a comment |
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$begingroup$
@Surb but the derivative of ln(x+1) is 1/(x+1) which is bounded in [0,infinity) so ln(x+1) is uniformly continuous.
$endgroup$
– Omer
Jan 22 at 10:28
1
$begingroup$
How about $f(x)=sin{e^{x^2}}$?
$endgroup$
– Mindlack
Jan 22 at 10:29
2
$begingroup$
@Mindlack actually, I think sin(x^2) is enough. because if we take $x_n=sqrt{2 pi n}$ and $y_n= sqrt{2 pi n + pi/2}$ then $x_n-y_n$ tends to 0 but f(x_n)-f(y_n) doesn't.
$endgroup$
– Omer
Jan 22 at 10:33
$begingroup$
@DavideGiraudo Yes, sorry, I edited the title.
$endgroup$
– Omer
Jan 22 at 10:35
$begingroup$
Intuitively, for nice enough functions (in this case I would say $C^1$), uniform continuity is equivalent to boundedness of the derivative, and the limit condition to boundedness of the function. So unless the counterexample is "pathological", you should be looking for a bounded function with unbounded derivative, and $sin(x^2)$ is a fine example (or more generally $sin$ of any fast growing function).
$endgroup$
– Mees de Vries
Jan 22 at 10:37