Prove $ | sum_{k=1}^{n} frac{z_k}{k} |^2 le ( sum_{k=1}^{n} frac{1}{k^2} ) (sum_{k=1}^{n}| z_k|^2 ) $












2












$begingroup$


I want to prove that:
$ left| sum_{k=1}^{n} frac{z_k}{k} right|^2 le left( sum_{k=1}^{n} frac{1}{k^2} right) cdot left(sum_{k=1}^{n} left| z_kright|^2 right) $

where $z_k in mathbb C $


Probably I should use Schwarz inequality:
$$left| leftlangle x,y rightrangle right| le left| left| x right| right| cdot left| left| y right| right| = sqrt{leftlangle x,x rightrangle cdot leftlangle y,y rightrangle} $$
after square
$$ left| leftlangle x,y rightrangle right|^2 le leftlangle x,x rightrangle cdot leftlangle y,y rightrangle $$


I thought that I can:
$$ vec{x} = [frac{1}{1^2},frac{1}{2^2},...,frac{1}{n^2}]^T $$
$$ vec{y} = [left| z_1right|^2,...,left| z_nright|^2]^T $$
and put
$$ leftlangle x,yrightrangle = sum_{k=1}^{n} x_k cdot y_k $$
Right side I get instantly. The problem is with left side:
$$ left| sum_{k=1}^{n} frac{|z_k|^2}{k^2} right|^2 = left| sum_{k=1}^{n} left| frac{z_k^2}{k^2} right| right|^2 = left| sum_{k=1}^{n} frac{z_k^2}{k^2} right|^2 $$ but it is not$$ left| sum_{k=1}^{n} frac{z_k}{k} right|^2 $$
And I have 2 questions:

1. Can I define that: $ leftlangle x,yrightrangle = sum_{k=1}^{n} x_k cdot y_k $

2. Where is the mistake? / How can I finish that?






linear-algebra inequality proof-writing







share|cite|improve this question











$endgroup$

















    2












    $begingroup$


    I want to prove that:
    $ left| sum_{k=1}^{n} frac{z_k}{k} right|^2 le left( sum_{k=1}^{n} frac{1}{k^2} right) cdot left(sum_{k=1}^{n} left| z_kright|^2 right) $

    where $z_k in mathbb C $


    Probably I should use Schwarz inequality:
    $$left| leftlangle x,y rightrangle right| le left| left| x right| right| cdot left| left| y right| right| = sqrt{leftlangle x,x rightrangle cdot leftlangle y,y rightrangle} $$
    after square
    $$ left| leftlangle x,y rightrangle right|^2 le leftlangle x,x rightrangle cdot leftlangle y,y rightrangle $$


    I thought that I can:
    $$ vec{x} = [frac{1}{1^2},frac{1}{2^2},...,frac{1}{n^2}]^T $$
    $$ vec{y} = [left| z_1right|^2,...,left| z_nright|^2]^T $$
    and put
    $$ leftlangle x,yrightrangle = sum_{k=1}^{n} x_k cdot y_k $$
    Right side I get instantly. The problem is with left side:
    $$ left| sum_{k=1}^{n} frac{|z_k|^2}{k^2} right|^2 = left| sum_{k=1}^{n} left| frac{z_k^2}{k^2} right| right|^2 = left| sum_{k=1}^{n} frac{z_k^2}{k^2} right|^2 $$ but it is not$$ left| sum_{k=1}^{n} frac{z_k}{k} right|^2 $$
    And I have 2 questions:

    1. Can I define that: $ leftlangle x,yrightrangle = sum_{k=1}^{n} x_k cdot y_k $

    2. Where is the mistake? / How can I finish that?






    linear-algebra inequality proof-writing







    share|cite|improve this question











    $endgroup$















      2












      2








      2





      $begingroup$


      I want to prove that:
      $ left| sum_{k=1}^{n} frac{z_k}{k} right|^2 le left( sum_{k=1}^{n} frac{1}{k^2} right) cdot left(sum_{k=1}^{n} left| z_kright|^2 right) $

      where $z_k in mathbb C $


      Probably I should use Schwarz inequality:
      $$left| leftlangle x,y rightrangle right| le left| left| x right| right| cdot left| left| y right| right| = sqrt{leftlangle x,x rightrangle cdot leftlangle y,y rightrangle} $$
      after square
      $$ left| leftlangle x,y rightrangle right|^2 le leftlangle x,x rightrangle cdot leftlangle y,y rightrangle $$


      I thought that I can:
      $$ vec{x} = [frac{1}{1^2},frac{1}{2^2},...,frac{1}{n^2}]^T $$
      $$ vec{y} = [left| z_1right|^2,...,left| z_nright|^2]^T $$
      and put
      $$ leftlangle x,yrightrangle = sum_{k=1}^{n} x_k cdot y_k $$
      Right side I get instantly. The problem is with left side:
      $$ left| sum_{k=1}^{n} frac{|z_k|^2}{k^2} right|^2 = left| sum_{k=1}^{n} left| frac{z_k^2}{k^2} right| right|^2 = left| sum_{k=1}^{n} frac{z_k^2}{k^2} right|^2 $$ but it is not$$ left| sum_{k=1}^{n} frac{z_k}{k} right|^2 $$
      And I have 2 questions:

      1. Can I define that: $ leftlangle x,yrightrangle = sum_{k=1}^{n} x_k cdot y_k $

      2. Where is the mistake? / How can I finish that?






      linear-algebra inequality proof-writing







      share|cite|improve this question











      $endgroup$




      I want to prove that:
      $ left| sum_{k=1}^{n} frac{z_k}{k} right|^2 le left( sum_{k=1}^{n} frac{1}{k^2} right) cdot left(sum_{k=1}^{n} left| z_kright|^2 right) $

      where $z_k in mathbb C $


      Probably I should use Schwarz inequality:
      $$left| leftlangle x,y rightrangle right| le left| left| x right| right| cdot left| left| y right| right| = sqrt{leftlangle x,x rightrangle cdot leftlangle y,y rightrangle} $$
      after square
      $$ left| leftlangle x,y rightrangle right|^2 le leftlangle x,x rightrangle cdot leftlangle y,y rightrangle $$


      I thought that I can:
      $$ vec{x} = [frac{1}{1^2},frac{1}{2^2},...,frac{1}{n^2}]^T $$
      $$ vec{y} = [left| z_1right|^2,...,left| z_nright|^2]^T $$
      and put
      $$ leftlangle x,yrightrangle = sum_{k=1}^{n} x_k cdot y_k $$
      Right side I get instantly. The problem is with left side:
      $$ left| sum_{k=1}^{n} frac{|z_k|^2}{k^2} right|^2 = left| sum_{k=1}^{n} left| frac{z_k^2}{k^2} right| right|^2 = left| sum_{k=1}^{n} frac{z_k^2}{k^2} right|^2 $$ but it is not$$ left| sum_{k=1}^{n} frac{z_k}{k} right|^2 $$
      And I have 2 questions:

      1. Can I define that: $ leftlangle x,yrightrangle = sum_{k=1}^{n} x_k cdot y_k $

      2. Where is the mistake? / How can I finish that?






      linear-algebra inequality proof-writing




      linear-algebra inequality proof-writing






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 2 at 13:55









      Martin R

      27.9k33255




      27.9k33255










      asked Jan 2 at 13:35









      VirtualUserVirtualUser

      69112




      69112






















          1 Answer
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          $begingroup$

          With your choice of $x$ and $y$ the right-hand side would be
          $$
          left( sum_{k=1}^{n} frac{1}{k^4} right) cdot left(sum_{k=1}^{n} left| z_kright|^4 right) , .
          $$



          But applying Cauchy-Schwarz to $vec x = (z_1, ldots, z_n)^T$
          and $vec y = (1, frac 12,ldots frac 1n)^T$ with the standard complex inner product
          $$
          leftlangle vec x, vec yrightrangle = sum_{k=1}^{n} x_k cdot bar y_k
          $$
          gives the intended result.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Ahh, ok, there is mistake, but how about defining scalar product? Can I do this or should I consider situation in the general definition of scalar product?
            $endgroup$
            – VirtualUser
            Jan 2 at 13:47










          • $begingroup$
            @VirtualUser: I have added the definition of the scalar product.
            $endgroup$
            – Martin R
            Jan 2 at 13:58










          • $begingroup$
            Ok, but if I understood well, I can choose the scalar product?
            $endgroup$
            – VirtualUser
            Jan 2 at 14:00










          • $begingroup$
            @VirtualUser: $| langle x,y rangle |^2 le langle x,x rangle cdot langle y,y rangle$ holds in any inner product space, you can define $langle x,y rangle $ as you like as long as it satisfies the rules for an inner product. – This particular choice (the “standard“ complex inner product) gives the intended result.
            $endgroup$
            – Martin R
            Jan 2 at 14:06










          • $begingroup$
            ok, I understand, thanks for your time @Martin_R
            $endgroup$
            – VirtualUser
            Jan 2 at 14:13











          Your Answer





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          1 Answer
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          1 Answer
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          active

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          active

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          2












          $begingroup$

          With your choice of $x$ and $y$ the right-hand side would be
          $$
          left( sum_{k=1}^{n} frac{1}{k^4} right) cdot left(sum_{k=1}^{n} left| z_kright|^4 right) , .
          $$



          But applying Cauchy-Schwarz to $vec x = (z_1, ldots, z_n)^T$
          and $vec y = (1, frac 12,ldots frac 1n)^T$ with the standard complex inner product
          $$
          leftlangle vec x, vec yrightrangle = sum_{k=1}^{n} x_k cdot bar y_k
          $$
          gives the intended result.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Ahh, ok, there is mistake, but how about defining scalar product? Can I do this or should I consider situation in the general definition of scalar product?
            $endgroup$
            – VirtualUser
            Jan 2 at 13:47










          • $begingroup$
            @VirtualUser: I have added the definition of the scalar product.
            $endgroup$
            – Martin R
            Jan 2 at 13:58










          • $begingroup$
            Ok, but if I understood well, I can choose the scalar product?
            $endgroup$
            – VirtualUser
            Jan 2 at 14:00










          • $begingroup$
            @VirtualUser: $| langle x,y rangle |^2 le langle x,x rangle cdot langle y,y rangle$ holds in any inner product space, you can define $langle x,y rangle $ as you like as long as it satisfies the rules for an inner product. – This particular choice (the “standard“ complex inner product) gives the intended result.
            $endgroup$
            – Martin R
            Jan 2 at 14:06










          • $begingroup$
            ok, I understand, thanks for your time @Martin_R
            $endgroup$
            – VirtualUser
            Jan 2 at 14:13
















          2












          $begingroup$

          With your choice of $x$ and $y$ the right-hand side would be
          $$
          left( sum_{k=1}^{n} frac{1}{k^4} right) cdot left(sum_{k=1}^{n} left| z_kright|^4 right) , .
          $$



          But applying Cauchy-Schwarz to $vec x = (z_1, ldots, z_n)^T$
          and $vec y = (1, frac 12,ldots frac 1n)^T$ with the standard complex inner product
          $$
          leftlangle vec x, vec yrightrangle = sum_{k=1}^{n} x_k cdot bar y_k
          $$
          gives the intended result.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Ahh, ok, there is mistake, but how about defining scalar product? Can I do this or should I consider situation in the general definition of scalar product?
            $endgroup$
            – VirtualUser
            Jan 2 at 13:47










          • $begingroup$
            @VirtualUser: I have added the definition of the scalar product.
            $endgroup$
            – Martin R
            Jan 2 at 13:58










          • $begingroup$
            Ok, but if I understood well, I can choose the scalar product?
            $endgroup$
            – VirtualUser
            Jan 2 at 14:00










          • $begingroup$
            @VirtualUser: $| langle x,y rangle |^2 le langle x,x rangle cdot langle y,y rangle$ holds in any inner product space, you can define $langle x,y rangle $ as you like as long as it satisfies the rules for an inner product. – This particular choice (the “standard“ complex inner product) gives the intended result.
            $endgroup$
            – Martin R
            Jan 2 at 14:06










          • $begingroup$
            ok, I understand, thanks for your time @Martin_R
            $endgroup$
            – VirtualUser
            Jan 2 at 14:13














          2












          2








          2





          $begingroup$

          With your choice of $x$ and $y$ the right-hand side would be
          $$
          left( sum_{k=1}^{n} frac{1}{k^4} right) cdot left(sum_{k=1}^{n} left| z_kright|^4 right) , .
          $$



          But applying Cauchy-Schwarz to $vec x = (z_1, ldots, z_n)^T$
          and $vec y = (1, frac 12,ldots frac 1n)^T$ with the standard complex inner product
          $$
          leftlangle vec x, vec yrightrangle = sum_{k=1}^{n} x_k cdot bar y_k
          $$
          gives the intended result.






          share|cite|improve this answer











          $endgroup$



          With your choice of $x$ and $y$ the right-hand side would be
          $$
          left( sum_{k=1}^{n} frac{1}{k^4} right) cdot left(sum_{k=1}^{n} left| z_kright|^4 right) , .
          $$



          But applying Cauchy-Schwarz to $vec x = (z_1, ldots, z_n)^T$
          and $vec y = (1, frac 12,ldots frac 1n)^T$ with the standard complex inner product
          $$
          leftlangle vec x, vec yrightrangle = sum_{k=1}^{n} x_k cdot bar y_k
          $$
          gives the intended result.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 2 at 13:52

























          answered Jan 2 at 13:45









          Martin RMartin R

          27.9k33255




          27.9k33255












          • $begingroup$
            Ahh, ok, there is mistake, but how about defining scalar product? Can I do this or should I consider situation in the general definition of scalar product?
            $endgroup$
            – VirtualUser
            Jan 2 at 13:47










          • $begingroup$
            @VirtualUser: I have added the definition of the scalar product.
            $endgroup$
            – Martin R
            Jan 2 at 13:58










          • $begingroup$
            Ok, but if I understood well, I can choose the scalar product?
            $endgroup$
            – VirtualUser
            Jan 2 at 14:00










          • $begingroup$
            @VirtualUser: $| langle x,y rangle |^2 le langle x,x rangle cdot langle y,y rangle$ holds in any inner product space, you can define $langle x,y rangle $ as you like as long as it satisfies the rules for an inner product. – This particular choice (the “standard“ complex inner product) gives the intended result.
            $endgroup$
            – Martin R
            Jan 2 at 14:06










          • $begingroup$
            ok, I understand, thanks for your time @Martin_R
            $endgroup$
            – VirtualUser
            Jan 2 at 14:13


















          • $begingroup$
            Ahh, ok, there is mistake, but how about defining scalar product? Can I do this or should I consider situation in the general definition of scalar product?
            $endgroup$
            – VirtualUser
            Jan 2 at 13:47










          • $begingroup$
            @VirtualUser: I have added the definition of the scalar product.
            $endgroup$
            – Martin R
            Jan 2 at 13:58










          • $begingroup$
            Ok, but if I understood well, I can choose the scalar product?
            $endgroup$
            – VirtualUser
            Jan 2 at 14:00










          • $begingroup$
            @VirtualUser: $| langle x,y rangle |^2 le langle x,x rangle cdot langle y,y rangle$ holds in any inner product space, you can define $langle x,y rangle $ as you like as long as it satisfies the rules for an inner product. – This particular choice (the “standard“ complex inner product) gives the intended result.
            $endgroup$
            – Martin R
            Jan 2 at 14:06










          • $begingroup$
            ok, I understand, thanks for your time @Martin_R
            $endgroup$
            – VirtualUser
            Jan 2 at 14:13
















          $begingroup$
          Ahh, ok, there is mistake, but how about defining scalar product? Can I do this or should I consider situation in the general definition of scalar product?
          $endgroup$
          – VirtualUser
          Jan 2 at 13:47




          $begingroup$
          Ahh, ok, there is mistake, but how about defining scalar product? Can I do this or should I consider situation in the general definition of scalar product?
          $endgroup$
          – VirtualUser
          Jan 2 at 13:47












          $begingroup$
          @VirtualUser: I have added the definition of the scalar product.
          $endgroup$
          – Martin R
          Jan 2 at 13:58




          $begingroup$
          @VirtualUser: I have added the definition of the scalar product.
          $endgroup$
          – Martin R
          Jan 2 at 13:58












          $begingroup$
          Ok, but if I understood well, I can choose the scalar product?
          $endgroup$
          – VirtualUser
          Jan 2 at 14:00




          $begingroup$
          Ok, but if I understood well, I can choose the scalar product?
          $endgroup$
          – VirtualUser
          Jan 2 at 14:00












          $begingroup$
          @VirtualUser: $| langle x,y rangle |^2 le langle x,x rangle cdot langle y,y rangle$ holds in any inner product space, you can define $langle x,y rangle $ as you like as long as it satisfies the rules for an inner product. – This particular choice (the “standard“ complex inner product) gives the intended result.
          $endgroup$
          – Martin R
          Jan 2 at 14:06




          $begingroup$
          @VirtualUser: $| langle x,y rangle |^2 le langle x,x rangle cdot langle y,y rangle$ holds in any inner product space, you can define $langle x,y rangle $ as you like as long as it satisfies the rules for an inner product. – This particular choice (the “standard“ complex inner product) gives the intended result.
          $endgroup$
          – Martin R
          Jan 2 at 14:06












          $begingroup$
          ok, I understand, thanks for your time @Martin_R
          $endgroup$
          – VirtualUser
          Jan 2 at 14:13




          $begingroup$
          ok, I understand, thanks for your time @Martin_R
          $endgroup$
          – VirtualUser
          Jan 2 at 14:13


















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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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