Using Hölder's inequality to prove statement for a specific vector $u$












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$begingroup$



Suppose that $uinmathbb{R}^{ntimes 1}$ is such that $sum_{i=1}^n u_i=0$.



Prove that $|u^top v|leq||u||_ 1(frac{v_{max}-v_{min}}2)$, where $v_{max}=max |v_i|$ and $v_{min}=min|v_i|$ for any $v in mathbb{R}^{ntimes 1}$.




I tried to prove this similarly to how my professor did it, using the fact that $|u^top(v-alpha e)|=|u^top v|$ if $e=(1,cdots,1)$ and $alpha$ is any real number because the sum of the components of $u$ is $0$. Then I used Hölder's inequality to get $|u^top v|leq||u||_1||v-alpha e||_infty$, and this is where I got stuck.



My professor does not justify why $$inf ||v-alpha e||_infty =frac{v_{max}-v_{min}}2.$$ I guess that the average of the maximum and minimum values of $v$ could minimize $||cdot||_infty$, but I'm not really convinced without a proper explanation and I can't prove it myself.



I guess what we really need to find is $$lim_{ptoinfty} inf_alpha (sum_{i=1}^n |v_i - alpha|^p)^frac1p $$ (because the limit of the p-norms is $||cdot||_infty$), which does not sound so easy and straightforward, though maybe it actually is and I'm not seeing how.






real-analysis linear-algebra matrices







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$endgroup$

















    0












    $begingroup$



    Suppose that $uinmathbb{R}^{ntimes 1}$ is such that $sum_{i=1}^n u_i=0$.



    Prove that $|u^top v|leq||u||_ 1(frac{v_{max}-v_{min}}2)$, where $v_{max}=max |v_i|$ and $v_{min}=min|v_i|$ for any $v in mathbb{R}^{ntimes 1}$.




    I tried to prove this similarly to how my professor did it, using the fact that $|u^top(v-alpha e)|=|u^top v|$ if $e=(1,cdots,1)$ and $alpha$ is any real number because the sum of the components of $u$ is $0$. Then I used Hölder's inequality to get $|u^top v|leq||u||_1||v-alpha e||_infty$, and this is where I got stuck.



    My professor does not justify why $$inf ||v-alpha e||_infty =frac{v_{max}-v_{min}}2.$$ I guess that the average of the maximum and minimum values of $v$ could minimize $||cdot||_infty$, but I'm not really convinced without a proper explanation and I can't prove it myself.



    I guess what we really need to find is $$lim_{ptoinfty} inf_alpha (sum_{i=1}^n |v_i - alpha|^p)^frac1p $$ (because the limit of the p-norms is $||cdot||_infty$), which does not sound so easy and straightforward, though maybe it actually is and I'm not seeing how.






    real-analysis linear-algebra matrices







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$



      Suppose that $uinmathbb{R}^{ntimes 1}$ is such that $sum_{i=1}^n u_i=0$.



      Prove that $|u^top v|leq||u||_ 1(frac{v_{max}-v_{min}}2)$, where $v_{max}=max |v_i|$ and $v_{min}=min|v_i|$ for any $v in mathbb{R}^{ntimes 1}$.




      I tried to prove this similarly to how my professor did it, using the fact that $|u^top(v-alpha e)|=|u^top v|$ if $e=(1,cdots,1)$ and $alpha$ is any real number because the sum of the components of $u$ is $0$. Then I used Hölder's inequality to get $|u^top v|leq||u||_1||v-alpha e||_infty$, and this is where I got stuck.



      My professor does not justify why $$inf ||v-alpha e||_infty =frac{v_{max}-v_{min}}2.$$ I guess that the average of the maximum and minimum values of $v$ could minimize $||cdot||_infty$, but I'm not really convinced without a proper explanation and I can't prove it myself.



      I guess what we really need to find is $$lim_{ptoinfty} inf_alpha (sum_{i=1}^n |v_i - alpha|^p)^frac1p $$ (because the limit of the p-norms is $||cdot||_infty$), which does not sound so easy and straightforward, though maybe it actually is and I'm not seeing how.






      real-analysis linear-algebra matrices







      share|cite|improve this question











      $endgroup$





      Suppose that $uinmathbb{R}^{ntimes 1}$ is such that $sum_{i=1}^n u_i=0$.



      Prove that $|u^top v|leq||u||_ 1(frac{v_{max}-v_{min}}2)$, where $v_{max}=max |v_i|$ and $v_{min}=min|v_i|$ for any $v in mathbb{R}^{ntimes 1}$.




      I tried to prove this similarly to how my professor did it, using the fact that $|u^top(v-alpha e)|=|u^top v|$ if $e=(1,cdots,1)$ and $alpha$ is any real number because the sum of the components of $u$ is $0$. Then I used Hölder's inequality to get $|u^top v|leq||u||_1||v-alpha e||_infty$, and this is where I got stuck.



      My professor does not justify why $$inf ||v-alpha e||_infty =frac{v_{max}-v_{min}}2.$$ I guess that the average of the maximum and minimum values of $v$ could minimize $||cdot||_infty$, but I'm not really convinced without a proper explanation and I can't prove it myself.



      I guess what we really need to find is $$lim_{ptoinfty} inf_alpha (sum_{i=1}^n |v_i - alpha|^p)^frac1p $$ (because the limit of the p-norms is $||cdot||_infty$), which does not sound so easy and straightforward, though maybe it actually is and I'm not seeing how.






      real-analysis linear-algebra matrices




      real-analysis linear-algebra matrices






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 2 at 20:27









      Namaste

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      asked Jan 2 at 19:28









      AstlyDichrarAstlyDichrar

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          $begingroup$

          Note that
          $$
          v_{min}-alpha leq v_i-alpha leq v_max-alpha,
          $$          and thus
          $$
          max_i |v_i -alpha|= max{|v_min -alpha|, |v_max - alpha|}.
          $$           The right-hand side is the maximum of $d(v_min, alpha)$ and $d(v_max, alpha)$. One can see that the minimizer of it is $$alpha =frac{v_min + v_max}{2}$$ and $$
          inf_{alphain mathbb{R}} |v - alpha e|_infty = frac{v_max - v_min}{2}
          $$           follows.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Oh, I get it now. We want to pick an $alpha$ that makes both terms we want to find the maximum of be the same will minimize the maximum. Thanks for the explanation!
            $endgroup$
            – AstlyDichrar
            Jan 2 at 22:19











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          $begingroup$

          Note that
          $$
          v_{min}-alpha leq v_i-alpha leq v_max-alpha,
          $$          and thus
          $$
          max_i |v_i -alpha|= max{|v_min -alpha|, |v_max - alpha|}.
          $$           The right-hand side is the maximum of $d(v_min, alpha)$ and $d(v_max, alpha)$. One can see that the minimizer of it is $$alpha =frac{v_min + v_max}{2}$$ and $$
          inf_{alphain mathbb{R}} |v - alpha e|_infty = frac{v_max - v_min}{2}
          $$           follows.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Oh, I get it now. We want to pick an $alpha$ that makes both terms we want to find the maximum of be the same will minimize the maximum. Thanks for the explanation!
            $endgroup$
            – AstlyDichrar
            Jan 2 at 22:19
















          2












          $begingroup$

          Note that
          $$
          v_{min}-alpha leq v_i-alpha leq v_max-alpha,
          $$          and thus
          $$
          max_i |v_i -alpha|= max{|v_min -alpha|, |v_max - alpha|}.
          $$           The right-hand side is the maximum of $d(v_min, alpha)$ and $d(v_max, alpha)$. One can see that the minimizer of it is $$alpha =frac{v_min + v_max}{2}$$ and $$
          inf_{alphain mathbb{R}} |v - alpha e|_infty = frac{v_max - v_min}{2}
          $$           follows.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Oh, I get it now. We want to pick an $alpha$ that makes both terms we want to find the maximum of be the same will minimize the maximum. Thanks for the explanation!
            $endgroup$
            – AstlyDichrar
            Jan 2 at 22:19














          2












          2








          2





          $begingroup$

          Note that
          $$
          v_{min}-alpha leq v_i-alpha leq v_max-alpha,
          $$          and thus
          $$
          max_i |v_i -alpha|= max{|v_min -alpha|, |v_max - alpha|}.
          $$           The right-hand side is the maximum of $d(v_min, alpha)$ and $d(v_max, alpha)$. One can see that the minimizer of it is $$alpha =frac{v_min + v_max}{2}$$ and $$
          inf_{alphain mathbb{R}} |v - alpha e|_infty = frac{v_max - v_min}{2}
          $$           follows.






          share|cite|improve this answer









          $endgroup$



          Note that
          $$
          v_{min}-alpha leq v_i-alpha leq v_max-alpha,
          $$          and thus
          $$
          max_i |v_i -alpha|= max{|v_min -alpha|, |v_max - alpha|}.
          $$           The right-hand side is the maximum of $d(v_min, alpha)$ and $d(v_max, alpha)$. One can see that the minimizer of it is $$alpha =frac{v_min + v_max}{2}$$ and $$
          inf_{alphain mathbb{R}} |v - alpha e|_infty = frac{v_max - v_min}{2}
          $$           follows.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 2 at 19:48









          SongSong

          16.6k11043




          16.6k11043












          • $begingroup$
            Oh, I get it now. We want to pick an $alpha$ that makes both terms we want to find the maximum of be the same will minimize the maximum. Thanks for the explanation!
            $endgroup$
            – AstlyDichrar
            Jan 2 at 22:19


















          • $begingroup$
            Oh, I get it now. We want to pick an $alpha$ that makes both terms we want to find the maximum of be the same will minimize the maximum. Thanks for the explanation!
            $endgroup$
            – AstlyDichrar
            Jan 2 at 22:19
















          $begingroup$
          Oh, I get it now. We want to pick an $alpha$ that makes both terms we want to find the maximum of be the same will minimize the maximum. Thanks for the explanation!
          $endgroup$
          – AstlyDichrar
          Jan 2 at 22:19




          $begingroup$
          Oh, I get it now. We want to pick an $alpha$ that makes both terms we want to find the maximum of be the same will minimize the maximum. Thanks for the explanation!
          $endgroup$
          – AstlyDichrar
          Jan 2 at 22:19


















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