Mixing
If $f$ has an unbound derivative - is $f$ necessarily non-Lipshitz?
$begingroup$
So we assume that $f$ is differentiable on some closed interval $[a,b]$ and has a derivative $f'$ which is not bounded.
Is there a way to show that $f$ is non-Lipshitz?
If not, is there some intuitive counter example (like of the form $x^psin{left(frac{1}{x}right)})$?
calculus lipschitz-functions
asked Jan 13 at 16:28
Ran KiriRan Kiri
14215
$endgroup$
add a comment |
$begingroup$
So we assume that $f$ is differentiable on some closed interval $[a,b]$ and has a derivative $f'$ which is not bounded.
Is there a way to show that $f$ is non-Lipshitz?
If not, is there some intuitive counter example (like of the form $x^psin{left(frac{1}{x}right)})$?
calculus lipschitz-functions
asked Jan 13 at 16:28
Ran KiriRan Kiri
14215
$endgroup$
$begingroup$
Try it the other way around: If $f$ is Lipschitz continuous (and differentiable) then the derivative is bounded.
$endgroup$
– Martin R
Jan 13 at 16:31
$begingroup$
This is immediate from the definition of $f'$. Hint: If $lim_{hto0}phi(h)>M$ then there exists $hne0$ such that $phi(h)>M$...
$endgroup$
– David C. Ullrich
Jan 13 at 16:40
add a comment |
$begingroup$
So we assume that $f$ is differentiable on some closed interval $[a,b]$ and has a derivative $f'$ which is not bounded.
Is there a way to show that $f$ is non-Lipshitz?
If not, is there some intuitive counter example (like of the form $x^psin{left(frac{1}{x}right)})$?
calculus lipschitz-functions
asked Jan 13 at 16:28
Ran KiriRan Kiri
14215
$endgroup$
So we assume that $f$ is differentiable on some closed interval $[a,b]$ and has a derivative $f'$ which is not bounded.
Is there a way to show that $f$ is non-Lipshitz?
If not, is there some intuitive counter example (like of the form $x^psin{left(frac{1}{x}right)})$?
calculus lipschitz-functions
calculus lipschitz-functions
asked Jan 13 at 16:28
Ran KiriRan Kiri
14215
asked Jan 13 at 16:28
Ran KiriRan Kiri
14215
asked Jan 13 at 16:28
Ran KiriRan Kiri
14215
asked Jan 13 at 16:28
Ran KiriRan Kiri
14215
asked Jan 13 at 16:28
Ran KiriRan Kiri
14215
14215
$begingroup$
Try it the other way around: If $f$ is Lipschitz continuous (and differentiable) then the derivative is bounded.
$endgroup$
– Martin R
Jan 13 at 16:31
$begingroup$
This is immediate from the definition of $f'$. Hint: If $lim_{hto0}phi(h)>M$ then there exists $hne0$ such that $phi(h)>M$...
$endgroup$
– David C. Ullrich
Jan 13 at 16:40
add a comment |
$begingroup$
Try it the other way around: If $f$ is Lipschitz continuous (and differentiable) then the derivative is bounded.
$endgroup$
– Martin R
Jan 13 at 16:31
$begingroup$
This is immediate from the definition of $f'$. Hint: If $lim_{hto0}phi(h)>M$ then there exists $hne0$ such that $phi(h)>M$...
$endgroup$
– David C. Ullrich
Jan 13 at 16:40
$begingroup$
Try it the other way around: If $f$ is Lipschitz continuous (and differentiable) then the derivative is bounded.
$endgroup$
– Martin R
Jan 13 at 16:31
$begingroup$
Try it the other way around: If $f$ is Lipschitz continuous (and differentiable) then the derivative is bounded.
$endgroup$
– Martin R
Jan 13 at 16:31
$begingroup$
This is immediate from the definition of $f'$. Hint: If $lim_{hto0}phi(h)>M$ then there exists $hne0$ such that $phi(h)>M$...
$endgroup$
– David C. Ullrich
Jan 13 at 16:40
$begingroup$
This is immediate from the definition of $f'$. Hint: If $lim_{hto0}phi(h)>M$ then there exists $hne0$ such that $phi(h)>M$...
$endgroup$
– David C. Ullrich
Jan 13 at 16:40
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Yes, that conclusion holds. A proof by contraposition would be to show that (for a differentiable function $f$ on $[a, b]$)
$$
text{$f$ is Lipschitz} implies text{$f'$ is bounded.}
$$
So assume that
$f$ is Lipschitz continuous with some Lipschitz constant $L$. Then
$$
leftvert frac{f(x)-f(x_0)}{x-x_0} right vert le L
$$
for all $x, x_0 in [a, b]$ with $x ne x_0$. Taking the limit $x to x_0$ it follows that
$$
|f'(x_0)| le L
$$
for all $x_0 in [a, b]$, i.e. $f'$ is bounded.
answered Jan 13 at 16:46
Martin RMartin R
28.8k33356
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Yes, that conclusion holds. A proof by contraposition would be to show that (for a differentiable function $f$ on $[a, b]$)
$$
text{$f$ is Lipschitz} implies text{$f'$ is bounded.}
$$
So assume that
$f$ is Lipschitz continuous with some Lipschitz constant $L$. Then
$$
leftvert frac{f(x)-f(x_0)}{x-x_0} right vert le L
$$
for all $x, x_0 in [a, b]$ with $x ne x_0$. Taking the limit $x to x_0$ it follows that
$$
|f'(x_0)| le L
$$
for all $x_0 in [a, b]$, i.e. $f'$ is bounded.
answered Jan 13 at 16:46
Martin RMartin R
28.8k33356
$endgroup$
add a comment |
$begingroup$
Yes, that conclusion holds. A proof by contraposition would be to show that (for a differentiable function $f$ on $[a, b]$)
$$
text{$f$ is Lipschitz} implies text{$f'$ is bounded.}
$$
So assume that
$f$ is Lipschitz continuous with some Lipschitz constant $L$. Then
$$
leftvert frac{f(x)-f(x_0)}{x-x_0} right vert le L
$$
for all $x, x_0 in [a, b]$ with $x ne x_0$. Taking the limit $x to x_0$ it follows that
$$
|f'(x_0)| le L
$$
for all $x_0 in [a, b]$, i.e. $f'$ is bounded.
answered Jan 13 at 16:46
Martin RMartin R
28.8k33356
$endgroup$
add a comment |
$begingroup$
Yes, that conclusion holds. A proof by contraposition would be to show that (for a differentiable function $f$ on $[a, b]$)
$$
text{$f$ is Lipschitz} implies text{$f'$ is bounded.}
$$
So assume that
$f$ is Lipschitz continuous with some Lipschitz constant $L$. Then
$$
leftvert frac{f(x)-f(x_0)}{x-x_0} right vert le L
$$
for all $x, x_0 in [a, b]$ with $x ne x_0$. Taking the limit $x to x_0$ it follows that
$$
|f'(x_0)| le L
$$
for all $x_0 in [a, b]$, i.e. $f'$ is bounded.
answered Jan 13 at 16:46
Martin RMartin R
28.8k33356
$endgroup$
Yes, that conclusion holds. A proof by contraposition would be to show that (for a differentiable function $f$ on $[a, b]$)
$$
text{$f$ is Lipschitz} implies text{$f'$ is bounded.}
$$
So assume that
$f$ is Lipschitz continuous with some Lipschitz constant $L$. Then
$$
leftvert frac{f(x)-f(x_0)}{x-x_0} right vert le L
$$
for all $x, x_0 in [a, b]$ with $x ne x_0$. Taking the limit $x to x_0$ it follows that
$$
|f'(x_0)| le L
$$
for all $x_0 in [a, b]$, i.e. $f'$ is bounded.
answered Jan 13 at 16:46
Martin RMartin R
28.8k33356
edited Jan 13 at 16:54
answered Jan 13 at 16:46
Martin RMartin R
28.8k33356
answered Jan 13 at 16:46
Martin RMartin R
28.8k33356
answered Jan 13 at 16:46
Martin RMartin R
28.8k33356
28.8k33356
add a comment |
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$begingroup$
Try it the other way around: If $f$ is Lipschitz continuous (and differentiable) then the derivative is bounded.
$endgroup$
– Martin R
Jan 13 at 16:31
$begingroup$
This is immediate from the definition of $f'$. Hint: If $lim_{hto0}phi(h)>M$ then there exists $hne0$ such that $phi(h)>M$...
$endgroup$
– David C. Ullrich
Jan 13 at 16:40