$C^1$ Lipschitz function linear growth The Next CEO of Stack OverflowApproximate continous function with linear growth condition by Lipschitz functionWhy is this function not locally Lipschitz?Classify the growth of functions and find a more general growth functionLipschitz/Hölder continuity of cosine functionLinear functions $mathbbC^nlongrightarrowmathbbC^m$ are Lipschitz continousLipschitz Continuity For a Vector Function.Some questions about function's growth.Linear function: relation between linearity and continuityDifferentiable Manifolds, Metric Space Structure, Lipschitz ContinuityDifference between “Rate of growth of a function” and “Order of growth of a function”
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$C^1$ Lipschitz function linear growth
The Next CEO of Stack OverflowApproximate continous function with linear growth condition by Lipschitz functionWhy is this function not locally Lipschitz?Classify the growth of functions and find a more general growth functionLipschitz/Hölder continuity of cosine functionLinear functions $mathbbC^nlongrightarrowmathbbC^m$ are Lipschitz continousLipschitz Continuity For a Vector Function.Some questions about function's growth.Linear function: relation between linearity and continuityDifferentiable Manifolds, Metric Space Structure, Lipschitz ContinuityDifference between “Rate of growth of a function” and “Order of growth of a function”
$begingroup$
I would like to know how to prove that a $C^1$ Lipschitz function has linear growth. (Actually I don't even know if it is true, it is a question in my exam – it says prove that, so it means it is true).
Thanks in advance.
analysis
edited Mar 27 at 19:42
Bernard
123k741117
asked Mar 27 at 19:40
Minkowski YaacovMinkowski Yaacov
113
$endgroup$
add a comment |
$begingroup$
I would like to know how to prove that a $C^1$ Lipschitz function has linear growth. (Actually I don't even know if it is true, it is a question in my exam – it says prove that, so it means it is true).
Thanks in advance.
analysis
edited Mar 27 at 19:42
Bernard
123k741117
asked Mar 27 at 19:40
Minkowski YaacovMinkowski Yaacov
113
$endgroup$
$begingroup$
By "linear growth" do you mean that $tfracf(x)$ is bounded for $|x|>1$?
$endgroup$
– amsmath
Mar 27 at 20:40
$begingroup$
yes, more precisely |f(x)|<=k(1+|x|)
$endgroup$
– Minkowski Yaacov
Mar 27 at 20:45
add a comment |
$begingroup$
I would like to know how to prove that a $C^1$ Lipschitz function has linear growth. (Actually I don't even know if it is true, it is a question in my exam – it says prove that, so it means it is true).
Thanks in advance.
analysis
edited Mar 27 at 19:42
Bernard
123k741117
asked Mar 27 at 19:40
Minkowski YaacovMinkowski Yaacov
113
$endgroup$
I would like to know how to prove that a $C^1$ Lipschitz function has linear growth. (Actually I don't even know if it is true, it is a question in my exam – it says prove that, so it means it is true).
Thanks in advance.
analysis
analysis
edited Mar 27 at 19:42
Bernard
123k741117
asked Mar 27 at 19:40
Minkowski YaacovMinkowski Yaacov
113
edited Mar 27 at 19:42
Bernard
123k741117
asked Mar 27 at 19:40
Minkowski YaacovMinkowski Yaacov
113
edited Mar 27 at 19:42
Bernard
123k741117
edited Mar 27 at 19:42
Bernard
123k741117
edited Mar 27 at 19:42
Bernard
123k741117
123k741117
asked Mar 27 at 19:40
Minkowski YaacovMinkowski Yaacov
113
asked Mar 27 at 19:40
Minkowski YaacovMinkowski Yaacov
113
asked Mar 27 at 19:40
Minkowski YaacovMinkowski Yaacov
113
113
$begingroup$
By "linear growth" do you mean that $tfracf(x)$ is bounded for $|x|>1$?
$endgroup$
– amsmath
Mar 27 at 20:40
$begingroup$
yes, more precisely |f(x)|<=k(1+|x|)
$endgroup$
– Minkowski Yaacov
Mar 27 at 20:45
add a comment |
$begingroup$
By "linear growth" do you mean that $tfracf(x)$ is bounded for $|x|>1$?
$endgroup$
– amsmath
Mar 27 at 20:40
$begingroup$
yes, more precisely |f(x)|<=k(1+|x|)
$endgroup$
– Minkowski Yaacov
Mar 27 at 20:45
$begingroup$
By "linear growth" do you mean that $tfracf(x)$ is bounded for $|x|>1$?
$endgroup$
– amsmath
Mar 27 at 20:40
$begingroup$
By "linear growth" do you mean that $tfracf(x)$ is bounded for $|x|>1$?
$endgroup$
– amsmath
Mar 27 at 20:40
$begingroup$
yes, more precisely |f(x)|<=k(1+|x|)
$endgroup$
– Minkowski Yaacov
Mar 27 at 20:45
$begingroup$
yes, more precisely |f(x)|<=k(1+|x|)
$endgroup$
– Minkowski Yaacov
Mar 27 at 20:45
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
No need for $C^1$:
$$fracf(x)lefracf(0)lefrac) + le L+|f(0)|.$$
answered Mar 27 at 20:47
amsmathamsmath
3,278420
$endgroup$
$begingroup$
thanks but i just realize that i wanted to proof the opposite: that a function with linear growth is C^1 and Lipschitz. I'm sorry, my bad.
$endgroup$
– Minkowski Yaacov
Mar 27 at 20:51
1
$begingroup$
Then ask a new question, upvote and check my answer.
$endgroup$
– amsmath
Mar 27 at 20:54
$begingroup$
But I can answer that question right away: no. For a very simple reason: $|f(x)|le k(1+|x|)$ just means that the graph of $f$ is contained in some sector. But it can be anything there. Even completely non-continuous.
$endgroup$
– amsmath
Mar 27 at 20:59
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
No need for $C^1$:
$$fracf(x)lefracf(0)lefrac) + le L+|f(0)|.$$
answered Mar 27 at 20:47
amsmathamsmath
3,278420
$endgroup$
$begingroup$
thanks but i just realize that i wanted to proof the opposite: that a function with linear growth is C^1 and Lipschitz. I'm sorry, my bad.
$endgroup$
– Minkowski Yaacov
Mar 27 at 20:51
1
$begingroup$
Then ask a new question, upvote and check my answer.
$endgroup$
– amsmath
Mar 27 at 20:54
$begingroup$
But I can answer that question right away: no. For a very simple reason: $|f(x)|le k(1+|x|)$ just means that the graph of $f$ is contained in some sector. But it can be anything there. Even completely non-continuous.
$endgroup$
– amsmath
Mar 27 at 20:59
add a comment |
$begingroup$
No need for $C^1$:
$$fracf(x)lefracf(0)lefrac) + le L+|f(0)|.$$
answered Mar 27 at 20:47
amsmathamsmath
3,278420
$endgroup$
$begingroup$
thanks but i just realize that i wanted to proof the opposite: that a function with linear growth is C^1 and Lipschitz. I'm sorry, my bad.
$endgroup$
– Minkowski Yaacov
Mar 27 at 20:51
1
$begingroup$
Then ask a new question, upvote and check my answer.
$endgroup$
– amsmath
Mar 27 at 20:54
$begingroup$
But I can answer that question right away: no. For a very simple reason: $|f(x)|le k(1+|x|)$ just means that the graph of $f$ is contained in some sector. But it can be anything there. Even completely non-continuous.
$endgroup$
– amsmath
Mar 27 at 20:59
add a comment |
$begingroup$
No need for $C^1$:
$$fracf(x)lefracf(0)lefrac) + le L+|f(0)|.$$
answered Mar 27 at 20:47
amsmathamsmath
3,278420
$endgroup$
No need for $C^1$:
$$fracf(x)lefracf(0)lefrac) + le L+|f(0)|.$$
answered Mar 27 at 20:47
amsmathamsmath
3,278420
answered Mar 27 at 20:47
amsmathamsmath
3,278420
answered Mar 27 at 20:47
amsmathamsmath
3,278420
answered Mar 27 at 20:47
amsmathamsmath
3,278420
3,278420
$begingroup$
thanks but i just realize that i wanted to proof the opposite: that a function with linear growth is C^1 and Lipschitz. I'm sorry, my bad.
$endgroup$
– Minkowski Yaacov
Mar 27 at 20:51
1
$begingroup$
Then ask a new question, upvote and check my answer.
$endgroup$
– amsmath
Mar 27 at 20:54
$begingroup$
But I can answer that question right away: no. For a very simple reason: $|f(x)|le k(1+|x|)$ just means that the graph of $f$ is contained in some sector. But it can be anything there. Even completely non-continuous.
$endgroup$
– amsmath
Mar 27 at 20:59
add a comment |
$begingroup$
thanks but i just realize that i wanted to proof the opposite: that a function with linear growth is C^1 and Lipschitz. I'm sorry, my bad.
$endgroup$
– Minkowski Yaacov
Mar 27 at 20:51
1
$begingroup$
Then ask a new question, upvote and check my answer.
$endgroup$
– amsmath
Mar 27 at 20:54
$begingroup$
But I can answer that question right away: no. For a very simple reason: $|f(x)|le k(1+|x|)$ just means that the graph of $f$ is contained in some sector. But it can be anything there. Even completely non-continuous.
$endgroup$
– amsmath
Mar 27 at 20:59
$begingroup$
thanks but i just realize that i wanted to proof the opposite: that a function with linear growth is C^1 and Lipschitz. I'm sorry, my bad.
$endgroup$
– Minkowski Yaacov
Mar 27 at 20:51
$begingroup$
thanks but i just realize that i wanted to proof the opposite: that a function with linear growth is C^1 and Lipschitz. I'm sorry, my bad.
$endgroup$
– Minkowski Yaacov
Mar 27 at 20:51
1
1
$begingroup$
Then ask a new question, upvote and check my answer.
$endgroup$
– amsmath
Mar 27 at 20:54
$begingroup$
Then ask a new question, upvote and check my answer.
$endgroup$
– amsmath
Mar 27 at 20:54
$begingroup$
But I can answer that question right away: no. For a very simple reason: $|f(x)|le k(1+|x|)$ just means that the graph of $f$ is contained in some sector. But it can be anything there. Even completely non-continuous.
$endgroup$
– amsmath
Mar 27 at 20:59
$begingroup$
But I can answer that question right away: no. For a very simple reason: $|f(x)|le k(1+|x|)$ just means that the graph of $f$ is contained in some sector. But it can be anything there. Even completely non-continuous.
$endgroup$
– amsmath
Mar 27 at 20:59
add a comment |
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$begingroup$
By "linear growth" do you mean that $tfracf(x)$ is bounded for $|x|>1$?
$endgroup$
– amsmath
Mar 27 at 20:40
$begingroup$
yes, more precisely |f(x)|<=k(1+|x|)
$endgroup$
– Minkowski Yaacov
Mar 27 at 20:45