Mixing
Show that the function $|t|^p$ is convex on $Bbb{R}$
$begingroup$
Let $f:Bbb{R}toBbb R$ be a function defined as $f(t) = |t|^p$ with $pgeq1$. I want to show that $f$ is convex by showing its second derivative is non-negative. I have tried calculating the second derivative but this led to confusion, is there a simpler way?
real-analysis convex-analysis
edited Feb 11 at 17:44
Xander Henderson
14.6k103555
asked Jan 16 at 20:11
Sam.SSam.S
649
$endgroup$
|
show 3 more comments
$begingroup$
Let $f:Bbb{R}toBbb R$ be a function defined as $f(t) = |t|^p$ with $pgeq1$. I want to show that $f$ is convex by showing its second derivative is non-negative. I have tried calculating the second derivative but this led to confusion, is there a simpler way?
real-analysis convex-analysis
edited Feb 11 at 17:44
Xander Henderson
14.6k103555
asked Jan 16 at 20:11
Sam.SSam.S
649
$endgroup$
2
$begingroup$
Can you describe some more the confusion you had?
$endgroup$
– OldGodzilla
Jan 16 at 20:13
$begingroup$
Hi Sam. Please look at how I edited your post, to see how to type math. It will be helpful if you need to make changes.
$endgroup$
– Chickenmancer
Jan 16 at 20:14
$begingroup$
I don't really understand how to actually compute this derivative, mainly due to the |t| part. But also what happens at t=0? since I am to show this property for all of t on R.
$endgroup$
– Sam.S
Jan 16 at 20:23
$begingroup$
@Sam.S Use the chain rule.
$endgroup$
– Will M.
Jan 16 at 20:25
1
$begingroup$
@Sam.S I've linked to the proof that the composition of an increasing convex function and a convex function is convex. More specifically it is this post math.stackexchange.com/a/473922/231327 I hope you might find it helpful.
$endgroup$
– BigbearZzz
Jan 16 at 21:06
|
show 3 more comments
$begingroup$
Let $f:Bbb{R}toBbb R$ be a function defined as $f(t) = |t|^p$ with $pgeq1$. I want to show that $f$ is convex by showing its second derivative is non-negative. I have tried calculating the second derivative but this led to confusion, is there a simpler way?
real-analysis convex-analysis
edited Feb 11 at 17:44
Xander Henderson
14.6k103555
asked Jan 16 at 20:11
Sam.SSam.S
649
$endgroup$
Let $f:Bbb{R}toBbb R$ be a function defined as $f(t) = |t|^p$ with $pgeq1$. I want to show that $f$ is convex by showing its second derivative is non-negative. I have tried calculating the second derivative but this led to confusion, is there a simpler way?
real-analysis convex-analysis
real-analysis convex-analysis
edited Feb 11 at 17:44
Xander Henderson
14.6k103555
asked Jan 16 at 20:11
Sam.SSam.S
649
edited Feb 11 at 17:44
Xander Henderson
14.6k103555
asked Jan 16 at 20:11
Sam.SSam.S
649
edited Feb 11 at 17:44
Xander Henderson
14.6k103555
edited Feb 11 at 17:44
Xander Henderson
14.6k103555
edited Feb 11 at 17:44
Xander Henderson
14.6k103555
14.6k103555
asked Jan 16 at 20:11
Sam.SSam.S
649
asked Jan 16 at 20:11
Sam.SSam.S
649
asked Jan 16 at 20:11
Sam.SSam.S
649
649
2
$begingroup$
Can you describe some more the confusion you had?
$endgroup$
– OldGodzilla
Jan 16 at 20:13
$begingroup$
Hi Sam. Please look at how I edited your post, to see how to type math. It will be helpful if you need to make changes.
$endgroup$
– Chickenmancer
Jan 16 at 20:14
$begingroup$
I don't really understand how to actually compute this derivative, mainly due to the |t| part. But also what happens at t=0? since I am to show this property for all of t on R.
$endgroup$
– Sam.S
Jan 16 at 20:23
$begingroup$
@Sam.S Use the chain rule.
$endgroup$
– Will M.
Jan 16 at 20:25
1
$begingroup$
@Sam.S I've linked to the proof that the composition of an increasing convex function and a convex function is convex. More specifically it is this post math.stackexchange.com/a/473922/231327 I hope you might find it helpful.
$endgroup$
– BigbearZzz
Jan 16 at 21:06
|
show 3 more comments
2
$begingroup$
Can you describe some more the confusion you had?
$endgroup$
– OldGodzilla
Jan 16 at 20:13
$begingroup$
Hi Sam. Please look at how I edited your post, to see how to type math. It will be helpful if you need to make changes.
$endgroup$
– Chickenmancer
Jan 16 at 20:14
$begingroup$
I don't really understand how to actually compute this derivative, mainly due to the |t| part. But also what happens at t=0? since I am to show this property for all of t on R.
$endgroup$
– Sam.S
Jan 16 at 20:23
$begingroup$
@Sam.S Use the chain rule.
$endgroup$
– Will M.
Jan 16 at 20:25
1
$begingroup$
@Sam.S I've linked to the proof that the composition of an increasing convex function and a convex function is convex. More specifically it is this post math.stackexchange.com/a/473922/231327 I hope you might find it helpful.
$endgroup$
– BigbearZzz
Jan 16 at 21:06
2
2
$begingroup$
Can you describe some more the confusion you had?
$endgroup$
– OldGodzilla
Jan 16 at 20:13
$begingroup$
Can you describe some more the confusion you had?
$endgroup$
– OldGodzilla
Jan 16 at 20:13
$begingroup$
Hi Sam. Please look at how I edited your post, to see how to type math. It will be helpful if you need to make changes.
$endgroup$
– Chickenmancer
Jan 16 at 20:14
$begingroup$
Hi Sam. Please look at how I edited your post, to see how to type math. It will be helpful if you need to make changes.
$endgroup$
– Chickenmancer
Jan 16 at 20:14
$begingroup$
I don't really understand how to actually compute this derivative, mainly due to the |t| part. But also what happens at t=0? since I am to show this property for all of t on R.
$endgroup$
– Sam.S
Jan 16 at 20:23
$begingroup$
I don't really understand how to actually compute this derivative, mainly due to the |t| part. But also what happens at t=0? since I am to show this property for all of t on R.
$endgroup$
– Sam.S
Jan 16 at 20:23
$begingroup$
@Sam.S Use the chain rule.
$endgroup$
– Will M.
Jan 16 at 20:25
$begingroup$
@Sam.S Use the chain rule.
$endgroup$
– Will M.
Jan 16 at 20:25
1
1
$begingroup$
@Sam.S I've linked to the proof that the composition of an increasing convex function and a convex function is convex. More specifically it is this post math.stackexchange.com/a/473922/231327 I hope you might find it helpful.
$endgroup$
– BigbearZzz
Jan 16 at 21:06
$begingroup$
@Sam.S I've linked to the proof that the composition of an increasing convex function and a convex function is convex. More specifically it is this post math.stackexchange.com/a/473922/231327 I hope you might find it helpful.
$endgroup$
– BigbearZzz
Jan 16 at 21:06
|
show 3 more comments
1 Answer
1
active
oldest
votes
$begingroup$
The function $f:Bbb Rto[0,infty)$ defined by $$f(x)=|x|$$ is convex. Also, the function $g:[0,infty)to [0,infty)$ defined by
$$
g(t)=t^p
$$ is convex and increasing for $pge 1$. The composition of two such functions is convex, hence
$$
|t|^p = gcirc f(t)
$$
is a convex function.
You can prove that $g$ is convex using derivative test as you want.
Showing that $f$ is convex: Take any $x,yinBbb R$, we have
$$
|(1-lambda)x + lambda y| le |1-lambda||x| + |lambda| |y| = (1-lambda)|x| + lambda |y|
$$
for any $lambdain[0,1]$, which show that $f(x)=|x|$ is convex.
answered Jan 16 at 20:40
BigbearZzzBigbearZzz
8,75121652
$endgroup$
add a comment |
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1 Answer
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$begingroup$
The function $f:Bbb Rto[0,infty)$ defined by $$f(x)=|x|$$ is convex. Also, the function $g:[0,infty)to [0,infty)$ defined by
$$
g(t)=t^p
$$ is convex and increasing for $pge 1$. The composition of two such functions is convex, hence
$$
|t|^p = gcirc f(t)
$$
is a convex function.
You can prove that $g$ is convex using derivative test as you want.
Showing that $f$ is convex: Take any $x,yinBbb R$, we have
$$
|(1-lambda)x + lambda y| le |1-lambda||x| + |lambda| |y| = (1-lambda)|x| + lambda |y|
$$
for any $lambdain[0,1]$, which show that $f(x)=|x|$ is convex.
answered Jan 16 at 20:40
BigbearZzzBigbearZzz
8,75121652
$endgroup$
add a comment |
$begingroup$
The function $f:Bbb Rto[0,infty)$ defined by $$f(x)=|x|$$ is convex. Also, the function $g:[0,infty)to [0,infty)$ defined by
$$
g(t)=t^p
$$ is convex and increasing for $pge 1$. The composition of two such functions is convex, hence
$$
|t|^p = gcirc f(t)
$$
is a convex function.
You can prove that $g$ is convex using derivative test as you want.
Showing that $f$ is convex: Take any $x,yinBbb R$, we have
$$
|(1-lambda)x + lambda y| le |1-lambda||x| + |lambda| |y| = (1-lambda)|x| + lambda |y|
$$
for any $lambdain[0,1]$, which show that $f(x)=|x|$ is convex.
answered Jan 16 at 20:40
BigbearZzzBigbearZzz
8,75121652
$endgroup$
add a comment |
$begingroup$
The function $f:Bbb Rto[0,infty)$ defined by $$f(x)=|x|$$ is convex. Also, the function $g:[0,infty)to [0,infty)$ defined by
$$
g(t)=t^p
$$ is convex and increasing for $pge 1$. The composition of two such functions is convex, hence
$$
|t|^p = gcirc f(t)
$$
is a convex function.
You can prove that $g$ is convex using derivative test as you want.
Showing that $f$ is convex: Take any $x,yinBbb R$, we have
$$
|(1-lambda)x + lambda y| le |1-lambda||x| + |lambda| |y| = (1-lambda)|x| + lambda |y|
$$
for any $lambdain[0,1]$, which show that $f(x)=|x|$ is convex.
answered Jan 16 at 20:40
BigbearZzzBigbearZzz
8,75121652
$endgroup$
The function $f:Bbb Rto[0,infty)$ defined by $$f(x)=|x|$$ is convex. Also, the function $g:[0,infty)to [0,infty)$ defined by
$$
g(t)=t^p
$$ is convex and increasing for $pge 1$. The composition of two such functions is convex, hence
$$
|t|^p = gcirc f(t)
$$
is a convex function.
You can prove that $g$ is convex using derivative test as you want.
Showing that $f$ is convex: Take any $x,yinBbb R$, we have
$$
|(1-lambda)x + lambda y| le |1-lambda||x| + |lambda| |y| = (1-lambda)|x| + lambda |y|
$$
for any $lambdain[0,1]$, which show that $f(x)=|x|$ is convex.
answered Jan 16 at 20:40
BigbearZzzBigbearZzz
8,75121652
edited Jan 16 at 20:55
answered Jan 16 at 20:40
BigbearZzzBigbearZzz
8,75121652
answered Jan 16 at 20:40
BigbearZzzBigbearZzz
8,75121652
answered Jan 16 at 20:40
BigbearZzzBigbearZzz
8,75121652
8,75121652
add a comment |
add a comment |
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2
$begingroup$
Can you describe some more the confusion you had?
$endgroup$
– OldGodzilla
Jan 16 at 20:13
$begingroup$
Hi Sam. Please look at how I edited your post, to see how to type math. It will be helpful if you need to make changes.
$endgroup$
– Chickenmancer
Jan 16 at 20:14
$begingroup$
I don't really understand how to actually compute this derivative, mainly due to the |t| part. But also what happens at t=0? since I am to show this property for all of t on R.
$endgroup$
– Sam.S
Jan 16 at 20:23
$begingroup$
@Sam.S Use the chain rule.
$endgroup$
– Will M.
Jan 16 at 20:25
1
$begingroup$
@Sam.S I've linked to the proof that the composition of an increasing convex function and a convex function is convex. More specifically it is this post math.stackexchange.com/a/473922/231327 I hope you might find it helpful.
$endgroup$
– BigbearZzz
Jan 16 at 21:06