Proving the Lp norm is a norm.












0












$begingroup$


I want to prove the $L_p$ norm on continuous functions is in fact a norm. I have proven definitiveness and homogeneity but am struggling with the triangle inequality. I am using the fact that if the closed unit ball is convex then the triangle inequality holds.



Attempt so far:



Let $f$,$g$ be two functions in the closed unit ball, let $h=af+(1-a)g$, for some $a in [0,1]$.



Then the $L_p$ norm of $h$ is



$$left(int|af+(1-a)g|^pright)^{1/p}$$



I think i want to try and decompose this into 2 components, one containing the integral of $f$, one containing the integral of $g$ but am unsure how to do this given the power of $p$.
Thanks in advance
Dan






real-analysis norm normed-spaces







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  • $begingroup$
    Please use MathJax.
    $endgroup$
    – Morgan Rodgers
    Jan 16 at 18:22






  • 1




    $begingroup$
    How did such a poorly written post get upvoted in a matter of seconds ? Please, use MathJax to formulate your questions.
    $endgroup$
    – Rebellos
    Jan 16 at 18:24










  • $begingroup$
    what is mathjax?
    $endgroup$
    – user601175
    Jan 16 at 18:24










  • $begingroup$
    @user601175 I've edited MathJax into your post: you might want to take a look at the code for how it works.
    $endgroup$
    – user3482749
    Jan 16 at 18:25
















0












$begingroup$


I want to prove the $L_p$ norm on continuous functions is in fact a norm. I have proven definitiveness and homogeneity but am struggling with the triangle inequality. I am using the fact that if the closed unit ball is convex then the triangle inequality holds.



Attempt so far:



Let $f$,$g$ be two functions in the closed unit ball, let $h=af+(1-a)g$, for some $a in [0,1]$.



Then the $L_p$ norm of $h$ is



$$left(int|af+(1-a)g|^pright)^{1/p}$$



I think i want to try and decompose this into 2 components, one containing the integral of $f$, one containing the integral of $g$ but am unsure how to do this given the power of $p$.
Thanks in advance
Dan






real-analysis norm normed-spaces







share|cite|improve this question











$endgroup$












  • $begingroup$
    Please use MathJax.
    $endgroup$
    – Morgan Rodgers
    Jan 16 at 18:22






  • 1




    $begingroup$
    How did such a poorly written post get upvoted in a matter of seconds ? Please, use MathJax to formulate your questions.
    $endgroup$
    – Rebellos
    Jan 16 at 18:24










  • $begingroup$
    what is mathjax?
    $endgroup$
    – user601175
    Jan 16 at 18:24










  • $begingroup$
    @user601175 I've edited MathJax into your post: you might want to take a look at the code for how it works.
    $endgroup$
    – user3482749
    Jan 16 at 18:25














0












0








0





$begingroup$


I want to prove the $L_p$ norm on continuous functions is in fact a norm. I have proven definitiveness and homogeneity but am struggling with the triangle inequality. I am using the fact that if the closed unit ball is convex then the triangle inequality holds.



Attempt so far:



Let $f$,$g$ be two functions in the closed unit ball, let $h=af+(1-a)g$, for some $a in [0,1]$.



Then the $L_p$ norm of $h$ is



$$left(int|af+(1-a)g|^pright)^{1/p}$$



I think i want to try and decompose this into 2 components, one containing the integral of $f$, one containing the integral of $g$ but am unsure how to do this given the power of $p$.
Thanks in advance
Dan






real-analysis norm normed-spaces







share|cite|improve this question











$endgroup$




I want to prove the $L_p$ norm on continuous functions is in fact a norm. I have proven definitiveness and homogeneity but am struggling with the triangle inequality. I am using the fact that if the closed unit ball is convex then the triangle inequality holds.



Attempt so far:



Let $f$,$g$ be two functions in the closed unit ball, let $h=af+(1-a)g$, for some $a in [0,1]$.



Then the $L_p$ norm of $h$ is



$$left(int|af+(1-a)g|^pright)^{1/p}$$



I think i want to try and decompose this into 2 components, one containing the integral of $f$, one containing the integral of $g$ but am unsure how to do this given the power of $p$.
Thanks in advance
Dan






real-analysis norm normed-spaces




real-analysis norm normed-spaces






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 16 at 18:25









user3482749

4,3291119




4,3291119










asked Jan 16 at 18:20









user601175user601175

114




114












  • $begingroup$
    Please use MathJax.
    $endgroup$
    – Morgan Rodgers
    Jan 16 at 18:22






  • 1




    $begingroup$
    How did such a poorly written post get upvoted in a matter of seconds ? Please, use MathJax to formulate your questions.
    $endgroup$
    – Rebellos
    Jan 16 at 18:24










  • $begingroup$
    what is mathjax?
    $endgroup$
    – user601175
    Jan 16 at 18:24










  • $begingroup$
    @user601175 I've edited MathJax into your post: you might want to take a look at the code for how it works.
    $endgroup$
    – user3482749
    Jan 16 at 18:25


















  • $begingroup$
    Please use MathJax.
    $endgroup$
    – Morgan Rodgers
    Jan 16 at 18:22






  • 1




    $begingroup$
    How did such a poorly written post get upvoted in a matter of seconds ? Please, use MathJax to formulate your questions.
    $endgroup$
    – Rebellos
    Jan 16 at 18:24










  • $begingroup$
    what is mathjax?
    $endgroup$
    – user601175
    Jan 16 at 18:24










  • $begingroup$
    @user601175 I've edited MathJax into your post: you might want to take a look at the code for how it works.
    $endgroup$
    – user3482749
    Jan 16 at 18:25
















$begingroup$
Please use MathJax.
$endgroup$
– Morgan Rodgers
Jan 16 at 18:22




$begingroup$
Please use MathJax.
$endgroup$
– Morgan Rodgers
Jan 16 at 18:22




1




1




$begingroup$
How did such a poorly written post get upvoted in a matter of seconds ? Please, use MathJax to formulate your questions.
$endgroup$
– Rebellos
Jan 16 at 18:24




$begingroup$
How did such a poorly written post get upvoted in a matter of seconds ? Please, use MathJax to formulate your questions.
$endgroup$
– Rebellos
Jan 16 at 18:24












$begingroup$
what is mathjax?
$endgroup$
– user601175
Jan 16 at 18:24




$begingroup$
what is mathjax?
$endgroup$
– user601175
Jan 16 at 18:24












$begingroup$
@user601175 I've edited MathJax into your post: you might want to take a look at the code for how it works.
$endgroup$
– user3482749
Jan 16 at 18:25




$begingroup$
@user601175 I've edited MathJax into your post: you might want to take a look at the code for how it works.
$endgroup$
– user3482749
Jan 16 at 18:25










2 Answers
2






active

oldest

votes


















1












$begingroup$

There's a few ways to proceed. Probably the easiest is to not worry about the unit ball, and just prove the triangle inequality directly (that proof is available everywhere, for example on Wikipedia, if you want to look), but you asked for a version via convexity, so I'll give it a go:



First, note that $|af+(1-a)g|_{L_p} leq 1$ if and only if $|af+(1-a)g|_{L_p}^p leq 1$, so we'll deal with the latter, since it gets rid of that awkward power of $1/p$.



At this point, things are relatively easy: first, we note that $x mapsto x^p$ is convex on the non-negative real half-line (feel free to insert a proof of this here if you don't have one already), so (combining this with the triangle inequality for $|cdot|$) we have $|af(x)+(1-a)g(x)|^p leq left(a|f(x)| + (1-a)|g(x)| right)^pleq a|f(x)|^p + (1-a)|g(x)|^p$.



Thus, by standard properties of integrals, the definition of $|cdot|_{L_p}$, and the fact that $f$ and $g$ lie in the closed unit ball, we have



begin{align*}|af+(1-a)g|_{L_p}^p &= int |af(x)+(1-a)g(x)|^pmathrm{d}x \&leq int a|f(x)|^p+(1-a)|g(x)|^pmathrm{d}x
\&= aint|f(x)|^pmathrm{d}x + (1-a)int|g(x)|^pmathrm{d}x
\&= a|f|_{L_p}^p + (1-a)|g|_{L_p}^p
\&leq a + (1 -a)
\&= 1end{align*}



Thus, $|af+(1-a)g|_{L_p} leq 1$ also, so our open unit ball is convex, hence we have the triangle inequality.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you very much, i didnt spot that x->x^p is convex
    $endgroup$
    – user601175
    Jan 16 at 18:49



















0












$begingroup$

I can't understand the intuition of taking some functions that belong in the closed unit ball, as $mathcal{L}^p$ is generally the space :



$$mathcal{L}^p = left{f:[a,b] to mathbb R, ; text{measurable} : left(int_a^b |f(x) |^pmathrm{d}xright)^{1/p}< + infty right}$$



Hint :



Let $f,g in mathcal{L}^P$. Then, it is :



begin{align*}
|f+g |_p & = left(int_a^b |f(x) + g(x)|^pmathrm{d}xright)^{1/p} \
&leq left(int_a^b left(|f(x)| + |g(x)| right)^p mathrm{d}x right)^{1/p}
end{align*}






share|cite|improve this answer









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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    There's a few ways to proceed. Probably the easiest is to not worry about the unit ball, and just prove the triangle inequality directly (that proof is available everywhere, for example on Wikipedia, if you want to look), but you asked for a version via convexity, so I'll give it a go:



    First, note that $|af+(1-a)g|_{L_p} leq 1$ if and only if $|af+(1-a)g|_{L_p}^p leq 1$, so we'll deal with the latter, since it gets rid of that awkward power of $1/p$.



    At this point, things are relatively easy: first, we note that $x mapsto x^p$ is convex on the non-negative real half-line (feel free to insert a proof of this here if you don't have one already), so (combining this with the triangle inequality for $|cdot|$) we have $|af(x)+(1-a)g(x)|^p leq left(a|f(x)| + (1-a)|g(x)| right)^pleq a|f(x)|^p + (1-a)|g(x)|^p$.



    Thus, by standard properties of integrals, the definition of $|cdot|_{L_p}$, and the fact that $f$ and $g$ lie in the closed unit ball, we have



    begin{align*}|af+(1-a)g|_{L_p}^p &= int |af(x)+(1-a)g(x)|^pmathrm{d}x \&leq int a|f(x)|^p+(1-a)|g(x)|^pmathrm{d}x
    \&= aint|f(x)|^pmathrm{d}x + (1-a)int|g(x)|^pmathrm{d}x
    \&= a|f|_{L_p}^p + (1-a)|g|_{L_p}^p
    \&leq a + (1 -a)
    \&= 1end{align*}



    Thus, $|af+(1-a)g|_{L_p} leq 1$ also, so our open unit ball is convex, hence we have the triangle inequality.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Thank you very much, i didnt spot that x->x^p is convex
      $endgroup$
      – user601175
      Jan 16 at 18:49
















    1












    $begingroup$

    There's a few ways to proceed. Probably the easiest is to not worry about the unit ball, and just prove the triangle inequality directly (that proof is available everywhere, for example on Wikipedia, if you want to look), but you asked for a version via convexity, so I'll give it a go:



    First, note that $|af+(1-a)g|_{L_p} leq 1$ if and only if $|af+(1-a)g|_{L_p}^p leq 1$, so we'll deal with the latter, since it gets rid of that awkward power of $1/p$.



    At this point, things are relatively easy: first, we note that $x mapsto x^p$ is convex on the non-negative real half-line (feel free to insert a proof of this here if you don't have one already), so (combining this with the triangle inequality for $|cdot|$) we have $|af(x)+(1-a)g(x)|^p leq left(a|f(x)| + (1-a)|g(x)| right)^pleq a|f(x)|^p + (1-a)|g(x)|^p$.



    Thus, by standard properties of integrals, the definition of $|cdot|_{L_p}$, and the fact that $f$ and $g$ lie in the closed unit ball, we have



    begin{align*}|af+(1-a)g|_{L_p}^p &= int |af(x)+(1-a)g(x)|^pmathrm{d}x \&leq int a|f(x)|^p+(1-a)|g(x)|^pmathrm{d}x
    \&= aint|f(x)|^pmathrm{d}x + (1-a)int|g(x)|^pmathrm{d}x
    \&= a|f|_{L_p}^p + (1-a)|g|_{L_p}^p
    \&leq a + (1 -a)
    \&= 1end{align*}



    Thus, $|af+(1-a)g|_{L_p} leq 1$ also, so our open unit ball is convex, hence we have the triangle inequality.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Thank you very much, i didnt spot that x->x^p is convex
      $endgroup$
      – user601175
      Jan 16 at 18:49














    1












    1








    1





    $begingroup$

    There's a few ways to proceed. Probably the easiest is to not worry about the unit ball, and just prove the triangle inequality directly (that proof is available everywhere, for example on Wikipedia, if you want to look), but you asked for a version via convexity, so I'll give it a go:



    First, note that $|af+(1-a)g|_{L_p} leq 1$ if and only if $|af+(1-a)g|_{L_p}^p leq 1$, so we'll deal with the latter, since it gets rid of that awkward power of $1/p$.



    At this point, things are relatively easy: first, we note that $x mapsto x^p$ is convex on the non-negative real half-line (feel free to insert a proof of this here if you don't have one already), so (combining this with the triangle inequality for $|cdot|$) we have $|af(x)+(1-a)g(x)|^p leq left(a|f(x)| + (1-a)|g(x)| right)^pleq a|f(x)|^p + (1-a)|g(x)|^p$.



    Thus, by standard properties of integrals, the definition of $|cdot|_{L_p}$, and the fact that $f$ and $g$ lie in the closed unit ball, we have



    begin{align*}|af+(1-a)g|_{L_p}^p &= int |af(x)+(1-a)g(x)|^pmathrm{d}x \&leq int a|f(x)|^p+(1-a)|g(x)|^pmathrm{d}x
    \&= aint|f(x)|^pmathrm{d}x + (1-a)int|g(x)|^pmathrm{d}x
    \&= a|f|_{L_p}^p + (1-a)|g|_{L_p}^p
    \&leq a + (1 -a)
    \&= 1end{align*}



    Thus, $|af+(1-a)g|_{L_p} leq 1$ also, so our open unit ball is convex, hence we have the triangle inequality.






    share|cite|improve this answer









    $endgroup$



    There's a few ways to proceed. Probably the easiest is to not worry about the unit ball, and just prove the triangle inequality directly (that proof is available everywhere, for example on Wikipedia, if you want to look), but you asked for a version via convexity, so I'll give it a go:



    First, note that $|af+(1-a)g|_{L_p} leq 1$ if and only if $|af+(1-a)g|_{L_p}^p leq 1$, so we'll deal with the latter, since it gets rid of that awkward power of $1/p$.



    At this point, things are relatively easy: first, we note that $x mapsto x^p$ is convex on the non-negative real half-line (feel free to insert a proof of this here if you don't have one already), so (combining this with the triangle inequality for $|cdot|$) we have $|af(x)+(1-a)g(x)|^p leq left(a|f(x)| + (1-a)|g(x)| right)^pleq a|f(x)|^p + (1-a)|g(x)|^p$.



    Thus, by standard properties of integrals, the definition of $|cdot|_{L_p}$, and the fact that $f$ and $g$ lie in the closed unit ball, we have



    begin{align*}|af+(1-a)g|_{L_p}^p &= int |af(x)+(1-a)g(x)|^pmathrm{d}x \&leq int a|f(x)|^p+(1-a)|g(x)|^pmathrm{d}x
    \&= aint|f(x)|^pmathrm{d}x + (1-a)int|g(x)|^pmathrm{d}x
    \&= a|f|_{L_p}^p + (1-a)|g|_{L_p}^p
    \&leq a + (1 -a)
    \&= 1end{align*}



    Thus, $|af+(1-a)g|_{L_p} leq 1$ also, so our open unit ball is convex, hence we have the triangle inequality.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Jan 16 at 18:46









    user3482749user3482749

    4,3291119




    4,3291119












    • $begingroup$
      Thank you very much, i didnt spot that x->x^p is convex
      $endgroup$
      – user601175
      Jan 16 at 18:49


















    • $begingroup$
      Thank you very much, i didnt spot that x->x^p is convex
      $endgroup$
      – user601175
      Jan 16 at 18:49
















    $begingroup$
    Thank you very much, i didnt spot that x->x^p is convex
    $endgroup$
    – user601175
    Jan 16 at 18:49




    $begingroup$
    Thank you very much, i didnt spot that x->x^p is convex
    $endgroup$
    – user601175
    Jan 16 at 18:49











    0












    $begingroup$

    I can't understand the intuition of taking some functions that belong in the closed unit ball, as $mathcal{L}^p$ is generally the space :



    $$mathcal{L}^p = left{f:[a,b] to mathbb R, ; text{measurable} : left(int_a^b |f(x) |^pmathrm{d}xright)^{1/p}< + infty right}$$



    Hint :



    Let $f,g in mathcal{L}^P$. Then, it is :



    begin{align*}
    |f+g |_p & = left(int_a^b |f(x) + g(x)|^pmathrm{d}xright)^{1/p} \
    &leq left(int_a^b left(|f(x)| + |g(x)| right)^p mathrm{d}x right)^{1/p}
    end{align*}






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      I can't understand the intuition of taking some functions that belong in the closed unit ball, as $mathcal{L}^p$ is generally the space :



      $$mathcal{L}^p = left{f:[a,b] to mathbb R, ; text{measurable} : left(int_a^b |f(x) |^pmathrm{d}xright)^{1/p}< + infty right}$$



      Hint :



      Let $f,g in mathcal{L}^P$. Then, it is :



      begin{align*}
      |f+g |_p & = left(int_a^b |f(x) + g(x)|^pmathrm{d}xright)^{1/p} \
      &leq left(int_a^b left(|f(x)| + |g(x)| right)^p mathrm{d}x right)^{1/p}
      end{align*}






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        I can't understand the intuition of taking some functions that belong in the closed unit ball, as $mathcal{L}^p$ is generally the space :



        $$mathcal{L}^p = left{f:[a,b] to mathbb R, ; text{measurable} : left(int_a^b |f(x) |^pmathrm{d}xright)^{1/p}< + infty right}$$



        Hint :



        Let $f,g in mathcal{L}^P$. Then, it is :



        begin{align*}
        |f+g |_p & = left(int_a^b |f(x) + g(x)|^pmathrm{d}xright)^{1/p} \
        &leq left(int_a^b left(|f(x)| + |g(x)| right)^p mathrm{d}x right)^{1/p}
        end{align*}






        share|cite|improve this answer









        $endgroup$



        I can't understand the intuition of taking some functions that belong in the closed unit ball, as $mathcal{L}^p$ is generally the space :



        $$mathcal{L}^p = left{f:[a,b] to mathbb R, ; text{measurable} : left(int_a^b |f(x) |^pmathrm{d}xright)^{1/p}< + infty right}$$



        Hint :



        Let $f,g in mathcal{L}^P$. Then, it is :



        begin{align*}
        |f+g |_p & = left(int_a^b |f(x) + g(x)|^pmathrm{d}xright)^{1/p} \
        &leq left(int_a^b left(|f(x)| + |g(x)| right)^p mathrm{d}x right)^{1/p}
        end{align*}







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 16 at 18:32









        RebellosRebellos

        15.7k31250




        15.7k31250






























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            0 1 What exactly does the error message beginning with E: mean? I assume it has to do with the file structure on Ubuntu-on-Windows, but what exactly? The following package was automatically installed and is no longer required: os-prober Use 'apt-get autoremove' to remove it. The following extra packages will be installed: libxslt1.1 The following NEW packages will be installed: libxslt1.1 xmlstarlet 0 upgraded, 2 newly installed, 0 to remove and 0 not upgraded. Need to get 435 kB of archives. After this operation, 1023 kB of additional disk space will be used. Get:1 http://archive.ubuntu.com/ubuntu/ trusty/main libxslt1.1 amd64 1.1.28-2build1 [145 kB] Get:2 http://archive.ubuntu.com/ubuntu/ trusty/universe xmlstarlet amd64 1.5.0-1 [290 kB]
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            File:DeusFollowingSea.jpg

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            File:DeusFollowingSea.jpg From Wikipedia, the free encyclopedia Jump to navigation Jump to search File File history File usage No higher resolution available. DeusFollowingSea.jpg ‎ (238 × 211 pixels, file size: 12 KB, MIME type: image/jpeg ) Summary [ edit ] Media data and Non-free use rationale Description Cover art of the Deus CD Following Sea Author or copyright owner Deus Source (WP:NFCC#4) rocksucker.co.uk Date of publication 2012 Use in article (WP:NFCC#7) Following Sea Purpose of use in article (WP:NFCC#8) to serve as the primary means of visual identification at the top of the article dedicated to the work in question. Not replaceable with free media because (WP:NFCC#1) n.a. Minimal use (WP:NFCC#3) Low res image used in infobox of the article on the album Respect for commercial opportunities (WP:NFCC#2) n.a. Fair use Fair use of co
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