Show that the function $|t|^p$ is convex on $Bbb{R}$












1












$begingroup$


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







share|cite|improve this question











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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
















1












$begingroup$


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







share|cite|improve this question











$endgroup$








  • 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














1












1








1





$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Feb 11 at 17:44









Xander Henderson

14.6k103555




14.6k103555










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














  • 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










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.






share|cite|improve this 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.






    share|cite|improve this answer











    $endgroup$


















      2












      $begingroup$

      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.






      share|cite|improve this answer











      $endgroup$
















        2












        2








        2





        $begingroup$

        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.






        share|cite|improve this answer











        $endgroup$



        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.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 16 at 20:55

























        answered Jan 16 at 20:40









        BigbearZzzBigbearZzz

        8,75121652




        8,75121652






























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