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Proving that $|x|^p$ is convex for $pin (1,2)$ using second derivative
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I am doing an exercise in which I first proved that any $fin C^1[a,b]$ which is twice differentiable on (a,b) with $f''geq 0$ is convex, and now I need to deduce that $x mapsto |x|^p$ is convex for $p geq 1$.
It's clear for $p=1,2$ and for $p>2$ I've shown that the function is twice differentiable with nonnegative second derivative so convexity follows from the first part. However for $pin (1,2)$ the function is not twice differentiable at $0$ so I'm not sure what to do. I don't think the question assumes that $p$ is an integer so it's not clear how to apply the first part in this case. Of course it is twice differentiable with nonnegative second derivative everywhere else but when the two points have opposite signs it is unclear how to show convexity.
real-analysis convex-analysis
asked Jan 15 at 10:29
AlephNullAlephNull
3539
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add a comment |
$begingroup$
I am doing an exercise in which I first proved that any $fin C^1[a,b]$ which is twice differentiable on (a,b) with $f''geq 0$ is convex, and now I need to deduce that $x mapsto |x|^p$ is convex for $p geq 1$.
It's clear for $p=1,2$ and for $p>2$ I've shown that the function is twice differentiable with nonnegative second derivative so convexity follows from the first part. However for $pin (1,2)$ the function is not twice differentiable at $0$ so I'm not sure what to do. I don't think the question assumes that $p$ is an integer so it's not clear how to apply the first part in this case. Of course it is twice differentiable with nonnegative second derivative everywhere else but when the two points have opposite signs it is unclear how to show convexity.
real-analysis convex-analysis
asked Jan 15 at 10:29
AlephNullAlephNull
3539
$endgroup$
add a comment |
$begingroup$
I am doing an exercise in which I first proved that any $fin C^1[a,b]$ which is twice differentiable on (a,b) with $f''geq 0$ is convex, and now I need to deduce that $x mapsto |x|^p$ is convex for $p geq 1$.
It's clear for $p=1,2$ and for $p>2$ I've shown that the function is twice differentiable with nonnegative second derivative so convexity follows from the first part. However for $pin (1,2)$ the function is not twice differentiable at $0$ so I'm not sure what to do. I don't think the question assumes that $p$ is an integer so it's not clear how to apply the first part in this case. Of course it is twice differentiable with nonnegative second derivative everywhere else but when the two points have opposite signs it is unclear how to show convexity.
real-analysis convex-analysis
asked Jan 15 at 10:29
AlephNullAlephNull
3539
$endgroup$
I am doing an exercise in which I first proved that any $fin C^1[a,b]$ which is twice differentiable on (a,b) with $f''geq 0$ is convex, and now I need to deduce that $x mapsto |x|^p$ is convex for $p geq 1$.
It's clear for $p=1,2$ and for $p>2$ I've shown that the function is twice differentiable with nonnegative second derivative so convexity follows from the first part. However for $pin (1,2)$ the function is not twice differentiable at $0$ so I'm not sure what to do. I don't think the question assumes that $p$ is an integer so it's not clear how to apply the first part in this case. Of course it is twice differentiable with nonnegative second derivative everywhere else but when the two points have opposite signs it is unclear how to show convexity.
real-analysis convex-analysis
real-analysis convex-analysis
asked Jan 15 at 10:29
AlephNullAlephNull
3539
asked Jan 15 at 10:29
AlephNullAlephNull
3539
asked Jan 15 at 10:29
AlephNullAlephNull
3539
asked Jan 15 at 10:29
AlephNullAlephNull
3539
asked Jan 15 at 10:29
AlephNullAlephNull
3539
3539
add a comment |
add a comment |
1 Answer
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$begingroup$
If $f'$ is monotone, then $f$ is convex. (Moreover, if $f'$ is strictly monontone increasing, then $f$ is strictly convex.) This criterion can be used if the second derivate doesn't exist as in this case.
The function $f(x) = |x|^p$ is differentiable for $p>1$ and we have
$$f'(x) = mathrm{sign}(x) |x|^{p-1}.$$
Thus, we see that $f'$ is strictly increasing and therefore $f$ is strictly convex.
answered Jan 15 at 12:34
p4schp4sch
5,305317
$endgroup$
1
$begingroup$
This is a nice and simple solution because in fact the proof of the second derivative method only uses the fact that $f'$ is increasing.
$endgroup$
– AlephNull
Jan 15 at 12:46
add a comment |
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1 Answer
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active
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1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
If $f'$ is monotone, then $f$ is convex. (Moreover, if $f'$ is strictly monontone increasing, then $f$ is strictly convex.) This criterion can be used if the second derivate doesn't exist as in this case.
The function $f(x) = |x|^p$ is differentiable for $p>1$ and we have
$$f'(x) = mathrm{sign}(x) |x|^{p-1}.$$
Thus, we see that $f'$ is strictly increasing and therefore $f$ is strictly convex.
answered Jan 15 at 12:34
p4schp4sch
5,305317
$endgroup$
1
$begingroup$
This is a nice and simple solution because in fact the proof of the second derivative method only uses the fact that $f'$ is increasing.
$endgroup$
– AlephNull
Jan 15 at 12:46
add a comment |
$begingroup$
If $f'$ is monotone, then $f$ is convex. (Moreover, if $f'$ is strictly monontone increasing, then $f$ is strictly convex.) This criterion can be used if the second derivate doesn't exist as in this case.
The function $f(x) = |x|^p$ is differentiable for $p>1$ and we have
$$f'(x) = mathrm{sign}(x) |x|^{p-1}.$$
Thus, we see that $f'$ is strictly increasing and therefore $f$ is strictly convex.
answered Jan 15 at 12:34
p4schp4sch
5,305317
$endgroup$
1
$begingroup$
This is a nice and simple solution because in fact the proof of the second derivative method only uses the fact that $f'$ is increasing.
$endgroup$
– AlephNull
Jan 15 at 12:46
add a comment |
$begingroup$
If $f'$ is monotone, then $f$ is convex. (Moreover, if $f'$ is strictly monontone increasing, then $f$ is strictly convex.) This criterion can be used if the second derivate doesn't exist as in this case.
The function $f(x) = |x|^p$ is differentiable for $p>1$ and we have
$$f'(x) = mathrm{sign}(x) |x|^{p-1}.$$
Thus, we see that $f'$ is strictly increasing and therefore $f$ is strictly convex.
answered Jan 15 at 12:34
p4schp4sch
5,305317
$endgroup$
If $f'$ is monotone, then $f$ is convex. (Moreover, if $f'$ is strictly monontone increasing, then $f$ is strictly convex.) This criterion can be used if the second derivate doesn't exist as in this case.
The function $f(x) = |x|^p$ is differentiable for $p>1$ and we have
$$f'(x) = mathrm{sign}(x) |x|^{p-1}.$$
Thus, we see that $f'$ is strictly increasing and therefore $f$ is strictly convex.
answered Jan 15 at 12:34
p4schp4sch
5,305317
answered Jan 15 at 12:34
p4schp4sch
5,305317
answered Jan 15 at 12:34
p4schp4sch
5,305317
answered Jan 15 at 12:34
p4schp4sch
5,305317
5,305317
1
$begingroup$
This is a nice and simple solution because in fact the proof of the second derivative method only uses the fact that $f'$ is increasing.
$endgroup$
– AlephNull
Jan 15 at 12:46
add a comment |
1
$begingroup$
This is a nice and simple solution because in fact the proof of the second derivative method only uses the fact that $f'$ is increasing.
$endgroup$
– AlephNull
Jan 15 at 12:46
1
1
$begingroup$
This is a nice and simple solution because in fact the proof of the second derivative method only uses the fact that $f'$ is increasing.
$endgroup$
– AlephNull
Jan 15 at 12:46
$begingroup$
This is a nice and simple solution because in fact the proof of the second derivative method only uses the fact that $f'$ is increasing.
$endgroup$
– AlephNull
Jan 15 at 12:46
add a comment |
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