Jensen's type inequality for inner product












1














Let $mathcal{H}$ be a real Hilbert space and $x,y,z in mathcal{H}$. If $x$ and $y$ are a convex combination of $z$, that is
$$ z := tx + (1-t)y , quad t in (0,1) , $$
then we have the following inequality
$$ leftlVert z rightrVert ^{2} leq t leftlVert x rightrVert ^{2} + (1-t)leftlVert y rightrVert ^{2} tag{1}label{1} . $$
Notice that eqref{1} is better than the Cauchy - Schwarz inequality in the sense that both coefficients are strictly less than $1$.




The question is when we only have
$$ leftlangle z , v rightrangle = leftlangle tx + (1-t)y , v rightrangle , quad t in (0,1) , tag{2}label{2} $$
and $v in mathcal{H}$ is arbitrary given, can we still derive the inequality eqref{1}.




A naive approach gives
$$ leftlVert z rightrVert ^{2} - leftlVert z-v rightrVert ^{2} = t leftlVert x rightrVert ^{2} + (1-t) leftlVert y rightrVert ^{2} - t (1-t) leftlVert x-y rightrVert ^{2} - leftlVert tx + (1-t)y-v rightrVert ^{2} $$



If we can not get eqref{1}, is there any alternative one in which I can avoid or at least weaken the term $- leftlVert z-v rightrVert ^{2}$, that is, to have $-c leftlVert z-v rightrVert ^{2}$ for some $c in (0,1)$?






inequality inner-product-space jensen-inequality







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    1














    Let $mathcal{H}$ be a real Hilbert space and $x,y,z in mathcal{H}$. If $x$ and $y$ are a convex combination of $z$, that is
    $$ z := tx + (1-t)y , quad t in (0,1) , $$
    then we have the following inequality
    $$ leftlVert z rightrVert ^{2} leq t leftlVert x rightrVert ^{2} + (1-t)leftlVert y rightrVert ^{2} tag{1}label{1} . $$
    Notice that eqref{1} is better than the Cauchy - Schwarz inequality in the sense that both coefficients are strictly less than $1$.




    The question is when we only have
    $$ leftlangle z , v rightrangle = leftlangle tx + (1-t)y , v rightrangle , quad t in (0,1) , tag{2}label{2} $$
    and $v in mathcal{H}$ is arbitrary given, can we still derive the inequality eqref{1}.




    A naive approach gives
    $$ leftlVert z rightrVert ^{2} - leftlVert z-v rightrVert ^{2} = t leftlVert x rightrVert ^{2} + (1-t) leftlVert y rightrVert ^{2} - t (1-t) leftlVert x-y rightrVert ^{2} - leftlVert tx + (1-t)y-v rightrVert ^{2} $$



    If we can not get eqref{1}, is there any alternative one in which I can avoid or at least weaken the term $- leftlVert z-v rightrVert ^{2}$, that is, to have $-c leftlVert z-v rightrVert ^{2}$ for some $c in (0,1)$?






    inequality inner-product-space jensen-inequality







    share|cite|improve this question

























      1












      1








      1







      Let $mathcal{H}$ be a real Hilbert space and $x,y,z in mathcal{H}$. If $x$ and $y$ are a convex combination of $z$, that is
      $$ z := tx + (1-t)y , quad t in (0,1) , $$
      then we have the following inequality
      $$ leftlVert z rightrVert ^{2} leq t leftlVert x rightrVert ^{2} + (1-t)leftlVert y rightrVert ^{2} tag{1}label{1} . $$
      Notice that eqref{1} is better than the Cauchy - Schwarz inequality in the sense that both coefficients are strictly less than $1$.




      The question is when we only have
      $$ leftlangle z , v rightrangle = leftlangle tx + (1-t)y , v rightrangle , quad t in (0,1) , tag{2}label{2} $$
      and $v in mathcal{H}$ is arbitrary given, can we still derive the inequality eqref{1}.




      A naive approach gives
      $$ leftlVert z rightrVert ^{2} - leftlVert z-v rightrVert ^{2} = t leftlVert x rightrVert ^{2} + (1-t) leftlVert y rightrVert ^{2} - t (1-t) leftlVert x-y rightrVert ^{2} - leftlVert tx + (1-t)y-v rightrVert ^{2} $$



      If we can not get eqref{1}, is there any alternative one in which I can avoid or at least weaken the term $- leftlVert z-v rightrVert ^{2}$, that is, to have $-c leftlVert z-v rightrVert ^{2}$ for some $c in (0,1)$?






      inequality inner-product-space jensen-inequality







      share|cite|improve this question













      Let $mathcal{H}$ be a real Hilbert space and $x,y,z in mathcal{H}$. If $x$ and $y$ are a convex combination of $z$, that is
      $$ z := tx + (1-t)y , quad t in (0,1) , $$
      then we have the following inequality
      $$ leftlVert z rightrVert ^{2} leq t leftlVert x rightrVert ^{2} + (1-t)leftlVert y rightrVert ^{2} tag{1}label{1} . $$
      Notice that eqref{1} is better than the Cauchy - Schwarz inequality in the sense that both coefficients are strictly less than $1$.




      The question is when we only have
      $$ leftlangle z , v rightrangle = leftlangle tx + (1-t)y , v rightrangle , quad t in (0,1) , tag{2}label{2} $$
      and $v in mathcal{H}$ is arbitrary given, can we still derive the inequality eqref{1}.




      A naive approach gives
      $$ leftlVert z rightrVert ^{2} - leftlVert z-v rightrVert ^{2} = t leftlVert x rightrVert ^{2} + (1-t) leftlVert y rightrVert ^{2} - t (1-t) leftlVert x-y rightrVert ^{2} - leftlVert tx + (1-t)y-v rightrVert ^{2} $$



      If we can not get eqref{1}, is there any alternative one in which I can avoid or at least weaken the term $- leftlVert z-v rightrVert ^{2}$, that is, to have $-c leftlVert z-v rightrVert ^{2}$ for some $c in (0,1)$?






      inequality inner-product-space jensen-inequality




      inequality inner-product-space jensen-inequality






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Nov 29 '18 at 22:01









      mortalmortal

      397414




      397414






















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          It fails in $mathcal{H}=mathbb{R}^3$ with the usual inner product. Let's consider the following particular case: $y=(0,0,0)$, $x=(2,0,0)$, $z=(lambda,0,0)$ with $lambda^2>2$, $v=(0,1,0)$ and $t=1/2$. Then $langle z,v rangle =0 =langle tx+(1-t)y,vrangle $ but $||z||^2=lambda^2>2=t||x||^2$.






          share|cite|improve this answer





















          • thanks for that. but what if $leftlangle z , v rightrangle neq 0$? since I don't think that it would be my case..
            – mortal
            Nov 29 '18 at 22:29








          • 1




            I think there is no chance because, in general, your hypothesis is very weak. For example, modify the example above by $z=(lambda,1,0)$ with $lambda^2>3$, $x=(2,2,0)$.
            – Dante Grevino
            Nov 29 '18 at 22:35










          • if then could you please have a look on the case with $$ z = a - t cdot c , x = b - t cdot d , z = - t cdot d , v = a - b . $$ That is the case I'm looking for. I hope it would help.
            – mortal
            Nov 29 '18 at 22:39






          • 1




            No problem, but you should post that question as a separate question and giving the precise hypothesis.
            – Dante Grevino
            Nov 29 '18 at 22:43










          • thank you! I'll do it.
            – mortal
            Nov 29 '18 at 22:45











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          1 Answer
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          active

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          2














          It fails in $mathcal{H}=mathbb{R}^3$ with the usual inner product. Let's consider the following particular case: $y=(0,0,0)$, $x=(2,0,0)$, $z=(lambda,0,0)$ with $lambda^2>2$, $v=(0,1,0)$ and $t=1/2$. Then $langle z,v rangle =0 =langle tx+(1-t)y,vrangle $ but $||z||^2=lambda^2>2=t||x||^2$.






          share|cite|improve this answer





















          • thanks for that. but what if $leftlangle z , v rightrangle neq 0$? since I don't think that it would be my case..
            – mortal
            Nov 29 '18 at 22:29








          • 1




            I think there is no chance because, in general, your hypothesis is very weak. For example, modify the example above by $z=(lambda,1,0)$ with $lambda^2>3$, $x=(2,2,0)$.
            – Dante Grevino
            Nov 29 '18 at 22:35










          • if then could you please have a look on the case with $$ z = a - t cdot c , x = b - t cdot d , z = - t cdot d , v = a - b . $$ That is the case I'm looking for. I hope it would help.
            – mortal
            Nov 29 '18 at 22:39






          • 1




            No problem, but you should post that question as a separate question and giving the precise hypothesis.
            – Dante Grevino
            Nov 29 '18 at 22:43










          • thank you! I'll do it.
            – mortal
            Nov 29 '18 at 22:45
















          2














          It fails in $mathcal{H}=mathbb{R}^3$ with the usual inner product. Let's consider the following particular case: $y=(0,0,0)$, $x=(2,0,0)$, $z=(lambda,0,0)$ with $lambda^2>2$, $v=(0,1,0)$ and $t=1/2$. Then $langle z,v rangle =0 =langle tx+(1-t)y,vrangle $ but $||z||^2=lambda^2>2=t||x||^2$.






          share|cite|improve this answer





















          • thanks for that. but what if $leftlangle z , v rightrangle neq 0$? since I don't think that it would be my case..
            – mortal
            Nov 29 '18 at 22:29








          • 1




            I think there is no chance because, in general, your hypothesis is very weak. For example, modify the example above by $z=(lambda,1,0)$ with $lambda^2>3$, $x=(2,2,0)$.
            – Dante Grevino
            Nov 29 '18 at 22:35










          • if then could you please have a look on the case with $$ z = a - t cdot c , x = b - t cdot d , z = - t cdot d , v = a - b . $$ That is the case I'm looking for. I hope it would help.
            – mortal
            Nov 29 '18 at 22:39






          • 1




            No problem, but you should post that question as a separate question and giving the precise hypothesis.
            – Dante Grevino
            Nov 29 '18 at 22:43










          • thank you! I'll do it.
            – mortal
            Nov 29 '18 at 22:45














          2












          2








          2






          It fails in $mathcal{H}=mathbb{R}^3$ with the usual inner product. Let's consider the following particular case: $y=(0,0,0)$, $x=(2,0,0)$, $z=(lambda,0,0)$ with $lambda^2>2$, $v=(0,1,0)$ and $t=1/2$. Then $langle z,v rangle =0 =langle tx+(1-t)y,vrangle $ but $||z||^2=lambda^2>2=t||x||^2$.






          share|cite|improve this answer












          It fails in $mathcal{H}=mathbb{R}^3$ with the usual inner product. Let's consider the following particular case: $y=(0,0,0)$, $x=(2,0,0)$, $z=(lambda,0,0)$ with $lambda^2>2$, $v=(0,1,0)$ and $t=1/2$. Then $langle z,v rangle =0 =langle tx+(1-t)y,vrangle $ but $||z||^2=lambda^2>2=t||x||^2$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 29 '18 at 22:19









          Dante GrevinoDante Grevino

          94319




          94319












          • thanks for that. but what if $leftlangle z , v rightrangle neq 0$? since I don't think that it would be my case..
            – mortal
            Nov 29 '18 at 22:29








          • 1




            I think there is no chance because, in general, your hypothesis is very weak. For example, modify the example above by $z=(lambda,1,0)$ with $lambda^2>3$, $x=(2,2,0)$.
            – Dante Grevino
            Nov 29 '18 at 22:35










          • if then could you please have a look on the case with $$ z = a - t cdot c , x = b - t cdot d , z = - t cdot d , v = a - b . $$ That is the case I'm looking for. I hope it would help.
            – mortal
            Nov 29 '18 at 22:39






          • 1




            No problem, but you should post that question as a separate question and giving the precise hypothesis.
            – Dante Grevino
            Nov 29 '18 at 22:43










          • thank you! I'll do it.
            – mortal
            Nov 29 '18 at 22:45


















          • thanks for that. but what if $leftlangle z , v rightrangle neq 0$? since I don't think that it would be my case..
            – mortal
            Nov 29 '18 at 22:29








          • 1




            I think there is no chance because, in general, your hypothesis is very weak. For example, modify the example above by $z=(lambda,1,0)$ with $lambda^2>3$, $x=(2,2,0)$.
            – Dante Grevino
            Nov 29 '18 at 22:35










          • if then could you please have a look on the case with $$ z = a - t cdot c , x = b - t cdot d , z = - t cdot d , v = a - b . $$ That is the case I'm looking for. I hope it would help.
            – mortal
            Nov 29 '18 at 22:39






          • 1




            No problem, but you should post that question as a separate question and giving the precise hypothesis.
            – Dante Grevino
            Nov 29 '18 at 22:43










          • thank you! I'll do it.
            – mortal
            Nov 29 '18 at 22:45
















          thanks for that. but what if $leftlangle z , v rightrangle neq 0$? since I don't think that it would be my case..
          – mortal
          Nov 29 '18 at 22:29






          thanks for that. but what if $leftlangle z , v rightrangle neq 0$? since I don't think that it would be my case..
          – mortal
          Nov 29 '18 at 22:29






          1




          1




          I think there is no chance because, in general, your hypothesis is very weak. For example, modify the example above by $z=(lambda,1,0)$ with $lambda^2>3$, $x=(2,2,0)$.
          – Dante Grevino
          Nov 29 '18 at 22:35




          I think there is no chance because, in general, your hypothesis is very weak. For example, modify the example above by $z=(lambda,1,0)$ with $lambda^2>3$, $x=(2,2,0)$.
          – Dante Grevino
          Nov 29 '18 at 22:35












          if then could you please have a look on the case with $$ z = a - t cdot c , x = b - t cdot d , z = - t cdot d , v = a - b . $$ That is the case I'm looking for. I hope it would help.
          – mortal
          Nov 29 '18 at 22:39




          if then could you please have a look on the case with $$ z = a - t cdot c , x = b - t cdot d , z = - t cdot d , v = a - b . $$ That is the case I'm looking for. I hope it would help.
          – mortal
          Nov 29 '18 at 22:39




          1




          1




          No problem, but you should post that question as a separate question and giving the precise hypothesis.
          – Dante Grevino
          Nov 29 '18 at 22:43




          No problem, but you should post that question as a separate question and giving the precise hypothesis.
          – Dante Grevino
          Nov 29 '18 at 22:43












          thank you! I'll do it.
          – mortal
          Nov 29 '18 at 22:45




          thank you! I'll do it.
          – mortal
          Nov 29 '18 at 22:45


















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