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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            張江高科駅

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            張江高科駅 ちょうこうこうか Zhangjiang Hi-Tech Park 所在地 中国上海市浦東新区 所属事業者 上海地鉄運営有限公司（上海軌道交通・駅詳細） 上海浦東現代有軌交通有限公司（駅詳細） テンプレートを表示 張江高科駅 各種表記 繁体字： 張江高科站 簡体字： 张江高科站 拼音 ： Zhàngjiāng Gāokē Zhàn 発音： ヂャンジャン ガオクー ヂャン 英文： Zhangjiang Hi-Tech Park Station テンプレートを表示 張江高科駅 （ちょうこうこうか-えき）は中華人民共和国上海市浦東新区に位置する、上海軌道交通2号線の駅である。 なお、張江高科駅前にある張江軌道電車の 張江地鉄駅駅 についてもここで述べる。 目次 1 利用可能な鉄道路線 2 沿革 3 駅構造 3.1 上海軌道交通 3.2 張江軌道電車 4 駅周辺 5 隣の駅 6 注釈 7 関連項目 利用可能な鉄道路線 上海軌道交通 ■ 2号線 上海浦東現代有軌交通有限公司 ■ 張江軌道電車 沿革 高架駅時代の姿 2000年12月27日 - 2号線の駅が高架駅として開業。 2010年 1月1日 - 張江軌道電車の開業に伴い、張江地鉄駅駅（停留所）を設置。 2月13日 - 駅地下化に伴い高架駅の運用停止。 2月24日 - 10日間の閉鎖を経て地下新駅の開業。 2017年2月9日 - リニューアル工事を経て再開業。乗降客の増加に対応し、ホームへのエレベーター、エスカレーターを増設、券売機エリアを拡張した [1] 。 駅構造 上海軌道交通 上海軌道交通 張江高科駅 5号口 ちょうこうこうか Zhangjiang Hi-Tech Park ◄ 竜陽路 (2.60km) (1.61km) 金科路 ► 所在地 中国上海市浦東新区松涛路 祖冲之路 （中国語版） 北緯31度12分11秒 東経121度34分50秒  /  北緯31.2...
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