$L^1$ inequality between ordered real numbers implies $L^2$ norm inequality












2















Let $x_1,ldots,x_n$ be real numbers such that $|x_1|geq ldotsgeq |x_n|$ and $displaystyle sum_{i=k+1}^n |x_i| leq alpha sum_{i=1}^k |x_i|$ where $1leq k leq n-1$ and $alpha >0$.



Prove that $$sum_{i=k+1}^n x_i^2 leq alpha sum_{i=1}^k x_i^2$$




I have managed to prove a looser inequality: $$begin{align}
alpha sum_{i=1}^k x_i^2
&geq frac{alpha}{k}left(sum_{i=1}^k |x_i| right)^2 quad quad text{Cauchy-Schwarz}\
&geq frac 1k sum_{i=k+1}^n |x_i| sum_{i=1}^k |x_i|\
&geq frac 1k left(sum_{i=k+1}^n |x_i| right)^2\
&geq frac 1k sum_{i=k+1}^n x_i^2 quad quad text{since } |z|_1 geq |z|_2
end{align}$$



Since $frac 1k$ is not $geq 1$, I'm stuck...






real-analysis inequality cauchy-schwarz-inequality







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    Let $x_1,ldots,x_n$ be real numbers such that $|x_1|geq ldotsgeq |x_n|$ and $displaystyle sum_{i=k+1}^n |x_i| leq alpha sum_{i=1}^k |x_i|$ where $1leq k leq n-1$ and $alpha >0$.



    Prove that $$sum_{i=k+1}^n x_i^2 leq alpha sum_{i=1}^k x_i^2$$




    I have managed to prove a looser inequality: $$begin{align}
    alpha sum_{i=1}^k x_i^2
    &geq frac{alpha}{k}left(sum_{i=1}^k |x_i| right)^2 quad quad text{Cauchy-Schwarz}\
    &geq frac 1k sum_{i=k+1}^n |x_i| sum_{i=1}^k |x_i|\
    &geq frac 1k left(sum_{i=k+1}^n |x_i| right)^2\
    &geq frac 1k sum_{i=k+1}^n x_i^2 quad quad text{since } |z|_1 geq |z|_2
    end{align}$$



    Since $frac 1k$ is not $geq 1$, I'm stuck...






    real-analysis inequality cauchy-schwarz-inequality







    share|cite|improve this question



























      2












      2








      2


      1






      Let $x_1,ldots,x_n$ be real numbers such that $|x_1|geq ldotsgeq |x_n|$ and $displaystyle sum_{i=k+1}^n |x_i| leq alpha sum_{i=1}^k |x_i|$ where $1leq k leq n-1$ and $alpha >0$.



      Prove that $$sum_{i=k+1}^n x_i^2 leq alpha sum_{i=1}^k x_i^2$$




      I have managed to prove a looser inequality: $$begin{align}
      alpha sum_{i=1}^k x_i^2
      &geq frac{alpha}{k}left(sum_{i=1}^k |x_i| right)^2 quad quad text{Cauchy-Schwarz}\
      &geq frac 1k sum_{i=k+1}^n |x_i| sum_{i=1}^k |x_i|\
      &geq frac 1k left(sum_{i=k+1}^n |x_i| right)^2\
      &geq frac 1k sum_{i=k+1}^n x_i^2 quad quad text{since } |z|_1 geq |z|_2
      end{align}$$



      Since $frac 1k$ is not $geq 1$, I'm stuck...






      real-analysis inequality cauchy-schwarz-inequality







      share|cite|improve this question
















      Let $x_1,ldots,x_n$ be real numbers such that $|x_1|geq ldotsgeq |x_n|$ and $displaystyle sum_{i=k+1}^n |x_i| leq alpha sum_{i=1}^k |x_i|$ where $1leq k leq n-1$ and $alpha >0$.



      Prove that $$sum_{i=k+1}^n x_i^2 leq alpha sum_{i=1}^k x_i^2$$




      I have managed to prove a looser inequality: $$begin{align}
      alpha sum_{i=1}^k x_i^2
      &geq frac{alpha}{k}left(sum_{i=1}^k |x_i| right)^2 quad quad text{Cauchy-Schwarz}\
      &geq frac 1k sum_{i=k+1}^n |x_i| sum_{i=1}^k |x_i|\
      &geq frac 1k left(sum_{i=k+1}^n |x_i| right)^2\
      &geq frac 1k sum_{i=k+1}^n x_i^2 quad quad text{since } |z|_1 geq |z|_2
      end{align}$$



      Since $frac 1k$ is not $geq 1$, I'm stuck...






      real-analysis inequality cauchy-schwarz-inequality




      real-analysis inequality cauchy-schwarz-inequality






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      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Nov 21 '18 at 11:28









      Tianlalu

      3,09621038




      3,09621038










      asked Nov 21 '18 at 10:15









      Issou Chankla

      254




      254






















          1 Answer
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          It is quite simple: Note that $|x_i| le |x_k|$ for any $i in {k+1,ldots,n}$. Thus
          $$tag{1}sum_{i=k+1}^n x_i^2 le |x_k| sum_{i=k+1}^n |x_i| le alpha |x_k| sum_{i=1}^k |x_i|,$$
          where the given condition is used in the last step. Now $|x_k x_i| le x_i^2$ for an $i=1,ldots,k$, because $|x_k| le |x_i|$ for all $i=1,ldots,k$. So proceeding in (1) gives the claimed estimate
          $$sum_{i=k+1}^n x_i^2lealpha sum_{i=1}^k x_i^2.$$






          share|cite|improve this answer





















          • Nice, thank you. This line of thought did not occur to me.
            – Issou Chankla
            Nov 21 '18 at 11:17










          • In general you lose information, if you apply Cauchy-Schwarz. The constants between $|cdot|_1$ and $|cdot|_2$ are also strict. In fact, the key point here is the condition $|x_1| ge ldots ge |x_n|$. (The statement is wrong without this condition!)
            – p4sch
            Nov 21 '18 at 12:04











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          1 Answer
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          active

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          active

          oldest

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          1














          It is quite simple: Note that $|x_i| le |x_k|$ for any $i in {k+1,ldots,n}$. Thus
          $$tag{1}sum_{i=k+1}^n x_i^2 le |x_k| sum_{i=k+1}^n |x_i| le alpha |x_k| sum_{i=1}^k |x_i|,$$
          where the given condition is used in the last step. Now $|x_k x_i| le x_i^2$ for an $i=1,ldots,k$, because $|x_k| le |x_i|$ for all $i=1,ldots,k$. So proceeding in (1) gives the claimed estimate
          $$sum_{i=k+1}^n x_i^2lealpha sum_{i=1}^k x_i^2.$$






          share|cite|improve this answer





















          • Nice, thank you. This line of thought did not occur to me.
            – Issou Chankla
            Nov 21 '18 at 11:17










          • In general you lose information, if you apply Cauchy-Schwarz. The constants between $|cdot|_1$ and $|cdot|_2$ are also strict. In fact, the key point here is the condition $|x_1| ge ldots ge |x_n|$. (The statement is wrong without this condition!)
            – p4sch
            Nov 21 '18 at 12:04
















          1














          It is quite simple: Note that $|x_i| le |x_k|$ for any $i in {k+1,ldots,n}$. Thus
          $$tag{1}sum_{i=k+1}^n x_i^2 le |x_k| sum_{i=k+1}^n |x_i| le alpha |x_k| sum_{i=1}^k |x_i|,$$
          where the given condition is used in the last step. Now $|x_k x_i| le x_i^2$ for an $i=1,ldots,k$, because $|x_k| le |x_i|$ for all $i=1,ldots,k$. So proceeding in (1) gives the claimed estimate
          $$sum_{i=k+1}^n x_i^2lealpha sum_{i=1}^k x_i^2.$$






          share|cite|improve this answer





















          • Nice, thank you. This line of thought did not occur to me.
            – Issou Chankla
            Nov 21 '18 at 11:17










          • In general you lose information, if you apply Cauchy-Schwarz. The constants between $|cdot|_1$ and $|cdot|_2$ are also strict. In fact, the key point here is the condition $|x_1| ge ldots ge |x_n|$. (The statement is wrong without this condition!)
            – p4sch
            Nov 21 '18 at 12:04














          1












          1








          1






          It is quite simple: Note that $|x_i| le |x_k|$ for any $i in {k+1,ldots,n}$. Thus
          $$tag{1}sum_{i=k+1}^n x_i^2 le |x_k| sum_{i=k+1}^n |x_i| le alpha |x_k| sum_{i=1}^k |x_i|,$$
          where the given condition is used in the last step. Now $|x_k x_i| le x_i^2$ for an $i=1,ldots,k$, because $|x_k| le |x_i|$ for all $i=1,ldots,k$. So proceeding in (1) gives the claimed estimate
          $$sum_{i=k+1}^n x_i^2lealpha sum_{i=1}^k x_i^2.$$






          share|cite|improve this answer












          It is quite simple: Note that $|x_i| le |x_k|$ for any $i in {k+1,ldots,n}$. Thus
          $$tag{1}sum_{i=k+1}^n x_i^2 le |x_k| sum_{i=k+1}^n |x_i| le alpha |x_k| sum_{i=1}^k |x_i|,$$
          where the given condition is used in the last step. Now $|x_k x_i| le x_i^2$ for an $i=1,ldots,k$, because $|x_k| le |x_i|$ for all $i=1,ldots,k$. So proceeding in (1) gives the claimed estimate
          $$sum_{i=k+1}^n x_i^2lealpha sum_{i=1}^k x_i^2.$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 21 '18 at 11:06









          p4sch

          4,760217




          4,760217












          • Nice, thank you. This line of thought did not occur to me.
            – Issou Chankla
            Nov 21 '18 at 11:17










          • In general you lose information, if you apply Cauchy-Schwarz. The constants between $|cdot|_1$ and $|cdot|_2$ are also strict. In fact, the key point here is the condition $|x_1| ge ldots ge |x_n|$. (The statement is wrong without this condition!)
            – p4sch
            Nov 21 '18 at 12:04


















          • Nice, thank you. This line of thought did not occur to me.
            – Issou Chankla
            Nov 21 '18 at 11:17










          • In general you lose information, if you apply Cauchy-Schwarz. The constants between $|cdot|_1$ and $|cdot|_2$ are also strict. In fact, the key point here is the condition $|x_1| ge ldots ge |x_n|$. (The statement is wrong without this condition!)
            – p4sch
            Nov 21 '18 at 12:04
















          Nice, thank you. This line of thought did not occur to me.
          – Issou Chankla
          Nov 21 '18 at 11:17




          Nice, thank you. This line of thought did not occur to me.
          – Issou Chankla
          Nov 21 '18 at 11:17












          In general you lose information, if you apply Cauchy-Schwarz. The constants between $|cdot|_1$ and $|cdot|_2$ are also strict. In fact, the key point here is the condition $|x_1| ge ldots ge |x_n|$. (The statement is wrong without this condition!)
          – p4sch
          Nov 21 '18 at 12:04




          In general you lose information, if you apply Cauchy-Schwarz. The constants between $|cdot|_1$ and $|cdot|_2$ are also strict. In fact, the key point here is the condition $|x_1| ge ldots ge |x_n|$. (The statement is wrong without this condition!)
          – p4sch
          Nov 21 '18 at 12:04


















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