An inequality involving integrals and square root












2












$begingroup$


Could someone help me with this inequality?




$$left( int_a^b sqrt{sum_{j=1}^n f_j^2(x)} dx right)^2 geq sum_{j=1}^n left(int_a^b f_j(x) dx right)^2$$




I tried to used the concavity of square root and then Cauchy Schwarz, but the latter is in the wrong direction. I appreciate any help!






real-analysis calculus integration analysis inequality







share|cite|improve this question









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  • 2




    $begingroup$
    This is a special case of Mi kowski's Integral Inequality. See en.wikipedia.org/wiki/Minkowski_inequality
    $endgroup$
    – Kavi Rama Murthy
    Jan 1 at 10:17










  • $begingroup$
    Wow! I didn't know this integral inequality. Interested in how to show that.
    $endgroup$
    – MikeTeX
    Jan 1 at 10:41
















2












$begingroup$


Could someone help me with this inequality?




$$left( int_a^b sqrt{sum_{j=1}^n f_j^2(x)} dx right)^2 geq sum_{j=1}^n left(int_a^b f_j(x) dx right)^2$$




I tried to used the concavity of square root and then Cauchy Schwarz, but the latter is in the wrong direction. I appreciate any help!






real-analysis calculus integration analysis inequality







share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    This is a special case of Mi kowski's Integral Inequality. See en.wikipedia.org/wiki/Minkowski_inequality
    $endgroup$
    – Kavi Rama Murthy
    Jan 1 at 10:17










  • $begingroup$
    Wow! I didn't know this integral inequality. Interested in how to show that.
    $endgroup$
    – MikeTeX
    Jan 1 at 10:41














2












2








2





$begingroup$


Could someone help me with this inequality?




$$left( int_a^b sqrt{sum_{j=1}^n f_j^2(x)} dx right)^2 geq sum_{j=1}^n left(int_a^b f_j(x) dx right)^2$$




I tried to used the concavity of square root and then Cauchy Schwarz, but the latter is in the wrong direction. I appreciate any help!






real-analysis calculus integration analysis inequality







share|cite|improve this question









$endgroup$




Could someone help me with this inequality?




$$left( int_a^b sqrt{sum_{j=1}^n f_j^2(x)} dx right)^2 geq sum_{j=1}^n left(int_a^b f_j(x) dx right)^2$$




I tried to used the concavity of square root and then Cauchy Schwarz, but the latter is in the wrong direction. I appreciate any help!






real-analysis calculus integration analysis inequality




real-analysis calculus integration analysis inequality






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 1 at 9:59









JiuJiu

515113




515113








  • 2




    $begingroup$
    This is a special case of Mi kowski's Integral Inequality. See en.wikipedia.org/wiki/Minkowski_inequality
    $endgroup$
    – Kavi Rama Murthy
    Jan 1 at 10:17










  • $begingroup$
    Wow! I didn't know this integral inequality. Interested in how to show that.
    $endgroup$
    – MikeTeX
    Jan 1 at 10:41














  • 2




    $begingroup$
    This is a special case of Mi kowski's Integral Inequality. See en.wikipedia.org/wiki/Minkowski_inequality
    $endgroup$
    – Kavi Rama Murthy
    Jan 1 at 10:17










  • $begingroup$
    Wow! I didn't know this integral inequality. Interested in how to show that.
    $endgroup$
    – MikeTeX
    Jan 1 at 10:41








2




2




$begingroup$
This is a special case of Mi kowski's Integral Inequality. See en.wikipedia.org/wiki/Minkowski_inequality
$endgroup$
– Kavi Rama Murthy
Jan 1 at 10:17




$begingroup$
This is a special case of Mi kowski's Integral Inequality. See en.wikipedia.org/wiki/Minkowski_inequality
$endgroup$
– Kavi Rama Murthy
Jan 1 at 10:17












$begingroup$
Wow! I didn't know this integral inequality. Interested in how to show that.
$endgroup$
– MikeTeX
Jan 1 at 10:41




$begingroup$
Wow! I didn't know this integral inequality. Interested in how to show that.
$endgroup$
– MikeTeX
Jan 1 at 10:41










2 Answers
2






active

oldest

votes


















4












$begingroup$

Let $(a_1,a_2,cdots,a_n)$ be a unit vector in $mathbb R^{n}$. Then $sum_j a_j int_a^{b} f_j (x) dx=int_a^{b}sum_j a_j f_j (x) dx leq int_a^{b} sqrt {sum f_j^{2}}$ by C-S inequlaity. By taking sup over all u nit vectors $(a_1,a_2,cdots,a_n)$ we get the desire dinequality.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    This sup is nothing else than $int _a^b sqrt{sum f_j^2}$, so I don't see how you get the inequality.
    $endgroup$
    – MikeTeX
    Jan 1 at 10:58










  • $begingroup$
    No,the sup is square root of the RHS.
    $endgroup$
    – Kavi Rama Murthy
    Jan 1 at 11:52










  • $begingroup$
    Denote by $b_j$ the integral of $f_j$. What is sup of $sum a_j b_j$ over all unit vectors $a$?.
    $endgroup$
    – Kavi Rama Murthy
    Jan 1 at 11:56










  • $begingroup$
    It's OK, but your "sup" is misleading and unnecessarily complicates the matter. Simply define a priori $b_j$ to be the integral of $f_j$, $B$ to be the vector $(b_1, ldots, b_n)$, $F$ to be $(f_1(x), ldots, f_n(x))$ and $A$ to be the vector $B/||B||_2$ to get the result with Cauchy-Shwartz, as you did: $||B||_2 = Acdot B = int Acdot F leq int ||F||_2$. (note: I've upvoted for your nice answer anyway).
    $endgroup$
    – MikeTeX
    Jan 1 at 19:43












  • $begingroup$
    You are right. @mike tex
    $endgroup$
    – Kavi Rama Murthy
    Jan 2 at 0:25



















2












$begingroup$

Hint. Show the inequality for $n=2$, then the induction step is easy:
$$
sum_{j=1}^{n+1}left(int_{a}^{b}f_{j}(x)dxright)^{2}
=sum_{j=1}^{n}left(int_{a}^{b}f_{j}(x)dxright)^{2}+left(int_{a}^{b}f_{n+1}(x)dxright)^{2}\
leq left(int_{a}^{b} sqrt{sum_{j=1}^{n}f_{j}(x)^{2}}dxright)^{2}+left(int_{a}^{b}f_{n+1}(x)dxright)^{2}leq
left(int_{a}^{b} sqrt{sum_{j=1}^{n+1}f_{j}(x)^{2}}dxright)^{2}.$$



Hint for $n=2$. Note that
$$|f_{1}(x)|=sqrt{sqrt{f_{1}(x)^{2}+f_{2}(x)^{2}}+f_{2}(x)}cdot sqrt{sqrt{f_{1}(x)^{2}+f_{2}(x)^{2}}-f_{2}(x)}$$
then apply the Cauchy–Schwarz inequality






share|cite|improve this answer











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    2 Answers
    2






    active

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    2 Answers
    2






    active

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    active

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    active

    oldest

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    4












    $begingroup$

    Let $(a_1,a_2,cdots,a_n)$ be a unit vector in $mathbb R^{n}$. Then $sum_j a_j int_a^{b} f_j (x) dx=int_a^{b}sum_j a_j f_j (x) dx leq int_a^{b} sqrt {sum f_j^{2}}$ by C-S inequlaity. By taking sup over all u nit vectors $(a_1,a_2,cdots,a_n)$ we get the desire dinequality.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      This sup is nothing else than $int _a^b sqrt{sum f_j^2}$, so I don't see how you get the inequality.
      $endgroup$
      – MikeTeX
      Jan 1 at 10:58










    • $begingroup$
      No,the sup is square root of the RHS.
      $endgroup$
      – Kavi Rama Murthy
      Jan 1 at 11:52










    • $begingroup$
      Denote by $b_j$ the integral of $f_j$. What is sup of $sum a_j b_j$ over all unit vectors $a$?.
      $endgroup$
      – Kavi Rama Murthy
      Jan 1 at 11:56










    • $begingroup$
      It's OK, but your "sup" is misleading and unnecessarily complicates the matter. Simply define a priori $b_j$ to be the integral of $f_j$, $B$ to be the vector $(b_1, ldots, b_n)$, $F$ to be $(f_1(x), ldots, f_n(x))$ and $A$ to be the vector $B/||B||_2$ to get the result with Cauchy-Shwartz, as you did: $||B||_2 = Acdot B = int Acdot F leq int ||F||_2$. (note: I've upvoted for your nice answer anyway).
      $endgroup$
      – MikeTeX
      Jan 1 at 19:43












    • $begingroup$
      You are right. @mike tex
      $endgroup$
      – Kavi Rama Murthy
      Jan 2 at 0:25
















    4












    $begingroup$

    Let $(a_1,a_2,cdots,a_n)$ be a unit vector in $mathbb R^{n}$. Then $sum_j a_j int_a^{b} f_j (x) dx=int_a^{b}sum_j a_j f_j (x) dx leq int_a^{b} sqrt {sum f_j^{2}}$ by C-S inequlaity. By taking sup over all u nit vectors $(a_1,a_2,cdots,a_n)$ we get the desire dinequality.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      This sup is nothing else than $int _a^b sqrt{sum f_j^2}$, so I don't see how you get the inequality.
      $endgroup$
      – MikeTeX
      Jan 1 at 10:58










    • $begingroup$
      No,the sup is square root of the RHS.
      $endgroup$
      – Kavi Rama Murthy
      Jan 1 at 11:52










    • $begingroup$
      Denote by $b_j$ the integral of $f_j$. What is sup of $sum a_j b_j$ over all unit vectors $a$?.
      $endgroup$
      – Kavi Rama Murthy
      Jan 1 at 11:56










    • $begingroup$
      It's OK, but your "sup" is misleading and unnecessarily complicates the matter. Simply define a priori $b_j$ to be the integral of $f_j$, $B$ to be the vector $(b_1, ldots, b_n)$, $F$ to be $(f_1(x), ldots, f_n(x))$ and $A$ to be the vector $B/||B||_2$ to get the result with Cauchy-Shwartz, as you did: $||B||_2 = Acdot B = int Acdot F leq int ||F||_2$. (note: I've upvoted for your nice answer anyway).
      $endgroup$
      – MikeTeX
      Jan 1 at 19:43












    • $begingroup$
      You are right. @mike tex
      $endgroup$
      – Kavi Rama Murthy
      Jan 2 at 0:25














    4












    4








    4





    $begingroup$

    Let $(a_1,a_2,cdots,a_n)$ be a unit vector in $mathbb R^{n}$. Then $sum_j a_j int_a^{b} f_j (x) dx=int_a^{b}sum_j a_j f_j (x) dx leq int_a^{b} sqrt {sum f_j^{2}}$ by C-S inequlaity. By taking sup over all u nit vectors $(a_1,a_2,cdots,a_n)$ we get the desire dinequality.






    share|cite|improve this answer









    $endgroup$



    Let $(a_1,a_2,cdots,a_n)$ be a unit vector in $mathbb R^{n}$. Then $sum_j a_j int_a^{b} f_j (x) dx=int_a^{b}sum_j a_j f_j (x) dx leq int_a^{b} sqrt {sum f_j^{2}}$ by C-S inequlaity. By taking sup over all u nit vectors $(a_1,a_2,cdots,a_n)$ we get the desire dinequality.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Jan 1 at 10:24









    Kavi Rama MurthyKavi Rama Murthy

    63.9k42464




    63.9k42464












    • $begingroup$
      This sup is nothing else than $int _a^b sqrt{sum f_j^2}$, so I don't see how you get the inequality.
      $endgroup$
      – MikeTeX
      Jan 1 at 10:58










    • $begingroup$
      No,the sup is square root of the RHS.
      $endgroup$
      – Kavi Rama Murthy
      Jan 1 at 11:52










    • $begingroup$
      Denote by $b_j$ the integral of $f_j$. What is sup of $sum a_j b_j$ over all unit vectors $a$?.
      $endgroup$
      – Kavi Rama Murthy
      Jan 1 at 11:56










    • $begingroup$
      It's OK, but your "sup" is misleading and unnecessarily complicates the matter. Simply define a priori $b_j$ to be the integral of $f_j$, $B$ to be the vector $(b_1, ldots, b_n)$, $F$ to be $(f_1(x), ldots, f_n(x))$ and $A$ to be the vector $B/||B||_2$ to get the result with Cauchy-Shwartz, as you did: $||B||_2 = Acdot B = int Acdot F leq int ||F||_2$. (note: I've upvoted for your nice answer anyway).
      $endgroup$
      – MikeTeX
      Jan 1 at 19:43












    • $begingroup$
      You are right. @mike tex
      $endgroup$
      – Kavi Rama Murthy
      Jan 2 at 0:25


















    • $begingroup$
      This sup is nothing else than $int _a^b sqrt{sum f_j^2}$, so I don't see how you get the inequality.
      $endgroup$
      – MikeTeX
      Jan 1 at 10:58










    • $begingroup$
      No,the sup is square root of the RHS.
      $endgroup$
      – Kavi Rama Murthy
      Jan 1 at 11:52










    • $begingroup$
      Denote by $b_j$ the integral of $f_j$. What is sup of $sum a_j b_j$ over all unit vectors $a$?.
      $endgroup$
      – Kavi Rama Murthy
      Jan 1 at 11:56










    • $begingroup$
      It's OK, but your "sup" is misleading and unnecessarily complicates the matter. Simply define a priori $b_j$ to be the integral of $f_j$, $B$ to be the vector $(b_1, ldots, b_n)$, $F$ to be $(f_1(x), ldots, f_n(x))$ and $A$ to be the vector $B/||B||_2$ to get the result with Cauchy-Shwartz, as you did: $||B||_2 = Acdot B = int Acdot F leq int ||F||_2$. (note: I've upvoted for your nice answer anyway).
      $endgroup$
      – MikeTeX
      Jan 1 at 19:43












    • $begingroup$
      You are right. @mike tex
      $endgroup$
      – Kavi Rama Murthy
      Jan 2 at 0:25
















    $begingroup$
    This sup is nothing else than $int _a^b sqrt{sum f_j^2}$, so I don't see how you get the inequality.
    $endgroup$
    – MikeTeX
    Jan 1 at 10:58




    $begingroup$
    This sup is nothing else than $int _a^b sqrt{sum f_j^2}$, so I don't see how you get the inequality.
    $endgroup$
    – MikeTeX
    Jan 1 at 10:58












    $begingroup$
    No,the sup is square root of the RHS.
    $endgroup$
    – Kavi Rama Murthy
    Jan 1 at 11:52




    $begingroup$
    No,the sup is square root of the RHS.
    $endgroup$
    – Kavi Rama Murthy
    Jan 1 at 11:52












    $begingroup$
    Denote by $b_j$ the integral of $f_j$. What is sup of $sum a_j b_j$ over all unit vectors $a$?.
    $endgroup$
    – Kavi Rama Murthy
    Jan 1 at 11:56




    $begingroup$
    Denote by $b_j$ the integral of $f_j$. What is sup of $sum a_j b_j$ over all unit vectors $a$?.
    $endgroup$
    – Kavi Rama Murthy
    Jan 1 at 11:56












    $begingroup$
    It's OK, but your "sup" is misleading and unnecessarily complicates the matter. Simply define a priori $b_j$ to be the integral of $f_j$, $B$ to be the vector $(b_1, ldots, b_n)$, $F$ to be $(f_1(x), ldots, f_n(x))$ and $A$ to be the vector $B/||B||_2$ to get the result with Cauchy-Shwartz, as you did: $||B||_2 = Acdot B = int Acdot F leq int ||F||_2$. (note: I've upvoted for your nice answer anyway).
    $endgroup$
    – MikeTeX
    Jan 1 at 19:43






    $begingroup$
    It's OK, but your "sup" is misleading and unnecessarily complicates the matter. Simply define a priori $b_j$ to be the integral of $f_j$, $B$ to be the vector $(b_1, ldots, b_n)$, $F$ to be $(f_1(x), ldots, f_n(x))$ and $A$ to be the vector $B/||B||_2$ to get the result with Cauchy-Shwartz, as you did: $||B||_2 = Acdot B = int Acdot F leq int ||F||_2$. (note: I've upvoted for your nice answer anyway).
    $endgroup$
    – MikeTeX
    Jan 1 at 19:43














    $begingroup$
    You are right. @mike tex
    $endgroup$
    – Kavi Rama Murthy
    Jan 2 at 0:25




    $begingroup$
    You are right. @mike tex
    $endgroup$
    – Kavi Rama Murthy
    Jan 2 at 0:25











    2












    $begingroup$

    Hint. Show the inequality for $n=2$, then the induction step is easy:
    $$
    sum_{j=1}^{n+1}left(int_{a}^{b}f_{j}(x)dxright)^{2}
    =sum_{j=1}^{n}left(int_{a}^{b}f_{j}(x)dxright)^{2}+left(int_{a}^{b}f_{n+1}(x)dxright)^{2}\
    leq left(int_{a}^{b} sqrt{sum_{j=1}^{n}f_{j}(x)^{2}}dxright)^{2}+left(int_{a}^{b}f_{n+1}(x)dxright)^{2}leq
    left(int_{a}^{b} sqrt{sum_{j=1}^{n+1}f_{j}(x)^{2}}dxright)^{2}.$$



    Hint for $n=2$. Note that
    $$|f_{1}(x)|=sqrt{sqrt{f_{1}(x)^{2}+f_{2}(x)^{2}}+f_{2}(x)}cdot sqrt{sqrt{f_{1}(x)^{2}+f_{2}(x)^{2}}-f_{2}(x)}$$
    then apply the Cauchy–Schwarz inequality






    share|cite|improve this answer











    $endgroup$


















      2












      $begingroup$

      Hint. Show the inequality for $n=2$, then the induction step is easy:
      $$
      sum_{j=1}^{n+1}left(int_{a}^{b}f_{j}(x)dxright)^{2}
      =sum_{j=1}^{n}left(int_{a}^{b}f_{j}(x)dxright)^{2}+left(int_{a}^{b}f_{n+1}(x)dxright)^{2}\
      leq left(int_{a}^{b} sqrt{sum_{j=1}^{n}f_{j}(x)^{2}}dxright)^{2}+left(int_{a}^{b}f_{n+1}(x)dxright)^{2}leq
      left(int_{a}^{b} sqrt{sum_{j=1}^{n+1}f_{j}(x)^{2}}dxright)^{2}.$$



      Hint for $n=2$. Note that
      $$|f_{1}(x)|=sqrt{sqrt{f_{1}(x)^{2}+f_{2}(x)^{2}}+f_{2}(x)}cdot sqrt{sqrt{f_{1}(x)^{2}+f_{2}(x)^{2}}-f_{2}(x)}$$
      then apply the Cauchy–Schwarz inequality






      share|cite|improve this answer











      $endgroup$
















        2












        2








        2





        $begingroup$

        Hint. Show the inequality for $n=2$, then the induction step is easy:
        $$
        sum_{j=1}^{n+1}left(int_{a}^{b}f_{j}(x)dxright)^{2}
        =sum_{j=1}^{n}left(int_{a}^{b}f_{j}(x)dxright)^{2}+left(int_{a}^{b}f_{n+1}(x)dxright)^{2}\
        leq left(int_{a}^{b} sqrt{sum_{j=1}^{n}f_{j}(x)^{2}}dxright)^{2}+left(int_{a}^{b}f_{n+1}(x)dxright)^{2}leq
        left(int_{a}^{b} sqrt{sum_{j=1}^{n+1}f_{j}(x)^{2}}dxright)^{2}.$$



        Hint for $n=2$. Note that
        $$|f_{1}(x)|=sqrt{sqrt{f_{1}(x)^{2}+f_{2}(x)^{2}}+f_{2}(x)}cdot sqrt{sqrt{f_{1}(x)^{2}+f_{2}(x)^{2}}-f_{2}(x)}$$
        then apply the Cauchy–Schwarz inequality






        share|cite|improve this answer











        $endgroup$



        Hint. Show the inequality for $n=2$, then the induction step is easy:
        $$
        sum_{j=1}^{n+1}left(int_{a}^{b}f_{j}(x)dxright)^{2}
        =sum_{j=1}^{n}left(int_{a}^{b}f_{j}(x)dxright)^{2}+left(int_{a}^{b}f_{n+1}(x)dxright)^{2}\
        leq left(int_{a}^{b} sqrt{sum_{j=1}^{n}f_{j}(x)^{2}}dxright)^{2}+left(int_{a}^{b}f_{n+1}(x)dxright)^{2}leq
        left(int_{a}^{b} sqrt{sum_{j=1}^{n+1}f_{j}(x)^{2}}dxright)^{2}.$$



        Hint for $n=2$. Note that
        $$|f_{1}(x)|=sqrt{sqrt{f_{1}(x)^{2}+f_{2}(x)^{2}}+f_{2}(x)}cdot sqrt{sqrt{f_{1}(x)^{2}+f_{2}(x)^{2}}-f_{2}(x)}$$
        then apply the Cauchy–Schwarz inequality







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 1 at 10:34

























        answered Jan 1 at 10:24









        Robert ZRobert Z

        99k1068139




        99k1068139






























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            Cabo Verde

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              Nota: Cabo-verdiano redireciona para este artigo. Para a língua falada em Cabo Verde, veja crioulo cabo-verdiano. Para outros significados, veja Cabo Verde (desambiguação). Coordenadas: 14° 55' N 23° 31' O Cabo Verde República de Cabo Verde Bandeira Brasão de Armas Lema: Unidade, Trabalho, Progresso Hino nacional: Cântico da Liberdade Gentílico: cabo-verdiano [ 1 ] Capital Praia 14° 55' N 23° 31' O Cidade mais populosa Praia Língua oficial Português Governo República unitária semipresidencialista [ 2 ]  - Presidente Jorge Carlos Fonseca  - Primeiro-ministro Ulisses Correia e Silva História    - Descoberta 1460   - Independência 00 • de Portugal 5 de Julho de 1975   - Multipartidarismo 13 de Janeiro de 1990  Área    - Total 4 033 km² (146.º)  - Água (%) 0,08 População    - Estimativa para 2017 560 899 [ 3 ] hab. (173.º)
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            Gyllenstierna

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