maximizing concave function with parameter












0














We want to maximize the convex function:
$$vec{q} cdot vec{x} - lambda||vec{x} - vec{1}||_2^2$$ where $lambda$ is some parameter.



I'm looking at a solution that states that the maximum is a piecewise function depending on whether $||q|| leq 2 cdot lambda$ or otherwise.



The value of $x$ that maximizes the function above is claimed to be: $$begin{cases}
(1/2)vec{q}/lambda + vec{1} & ||vec{q}|| leq 2lambda \
vec{q}/||vec{q}|| + vec{1} & ||vec{q}||> 2lambda\
end{cases}$$



I didn't even consider how the norm of $vec{q}$ might relate to $lambda$ when trying to compute this myself. In fact, I arrived at only the solution for when $||vec{q}|| leq 2lambda$ by taking the derivative of $vec{q} cdot vec{x} - lambda||vec{x} - vec{1}||_2^2$ with respect to $vec{x}$ coordinate-wise, setting it equal to $0$, and then solving for $vec{x}$ again coordinate-wise.



Can someone please explain why we should care whether or not $||vec{q}|| > 2lambda$, and how to compute the $x$ that maximizes the above function when $||vec{q}|| > 2lambda$?



Edit: there is a constraint on $vec{x}$ that was mentioned elsewhere in the solution: $vec{x}$ is in the $2-$norm unit ball centered at $1$.






optimization convex-analysis convex-optimization maxima-minima







share|cite|improve this question





























    0














    We want to maximize the convex function:
    $$vec{q} cdot vec{x} - lambda||vec{x} - vec{1}||_2^2$$ where $lambda$ is some parameter.



    I'm looking at a solution that states that the maximum is a piecewise function depending on whether $||q|| leq 2 cdot lambda$ or otherwise.



    The value of $x$ that maximizes the function above is claimed to be: $$begin{cases}
    (1/2)vec{q}/lambda + vec{1} & ||vec{q}|| leq 2lambda \
    vec{q}/||vec{q}|| + vec{1} & ||vec{q}||> 2lambda\
    end{cases}$$



    I didn't even consider how the norm of $vec{q}$ might relate to $lambda$ when trying to compute this myself. In fact, I arrived at only the solution for when $||vec{q}|| leq 2lambda$ by taking the derivative of $vec{q} cdot vec{x} - lambda||vec{x} - vec{1}||_2^2$ with respect to $vec{x}$ coordinate-wise, setting it equal to $0$, and then solving for $vec{x}$ again coordinate-wise.



    Can someone please explain why we should care whether or not $||vec{q}|| > 2lambda$, and how to compute the $x$ that maximizes the above function when $||vec{q}|| > 2lambda$?



    Edit: there is a constraint on $vec{x}$ that was mentioned elsewhere in the solution: $vec{x}$ is in the $2-$norm unit ball centered at $1$.






    optimization convex-analysis convex-optimization maxima-minima







    share|cite|improve this question



























      0












      0








      0







      We want to maximize the convex function:
      $$vec{q} cdot vec{x} - lambda||vec{x} - vec{1}||_2^2$$ where $lambda$ is some parameter.



      I'm looking at a solution that states that the maximum is a piecewise function depending on whether $||q|| leq 2 cdot lambda$ or otherwise.



      The value of $x$ that maximizes the function above is claimed to be: $$begin{cases}
      (1/2)vec{q}/lambda + vec{1} & ||vec{q}|| leq 2lambda \
      vec{q}/||vec{q}|| + vec{1} & ||vec{q}||> 2lambda\
      end{cases}$$



      I didn't even consider how the norm of $vec{q}$ might relate to $lambda$ when trying to compute this myself. In fact, I arrived at only the solution for when $||vec{q}|| leq 2lambda$ by taking the derivative of $vec{q} cdot vec{x} - lambda||vec{x} - vec{1}||_2^2$ with respect to $vec{x}$ coordinate-wise, setting it equal to $0$, and then solving for $vec{x}$ again coordinate-wise.



      Can someone please explain why we should care whether or not $||vec{q}|| > 2lambda$, and how to compute the $x$ that maximizes the above function when $||vec{q}|| > 2lambda$?



      Edit: there is a constraint on $vec{x}$ that was mentioned elsewhere in the solution: $vec{x}$ is in the $2-$norm unit ball centered at $1$.






      optimization convex-analysis convex-optimization maxima-minima







      share|cite|improve this question















      We want to maximize the convex function:
      $$vec{q} cdot vec{x} - lambda||vec{x} - vec{1}||_2^2$$ where $lambda$ is some parameter.



      I'm looking at a solution that states that the maximum is a piecewise function depending on whether $||q|| leq 2 cdot lambda$ or otherwise.



      The value of $x$ that maximizes the function above is claimed to be: $$begin{cases}
      (1/2)vec{q}/lambda + vec{1} & ||vec{q}|| leq 2lambda \
      vec{q}/||vec{q}|| + vec{1} & ||vec{q}||> 2lambda\
      end{cases}$$



      I didn't even consider how the norm of $vec{q}$ might relate to $lambda$ when trying to compute this myself. In fact, I arrived at only the solution for when $||vec{q}|| leq 2lambda$ by taking the derivative of $vec{q} cdot vec{x} - lambda||vec{x} - vec{1}||_2^2$ with respect to $vec{x}$ coordinate-wise, setting it equal to $0$, and then solving for $vec{x}$ again coordinate-wise.



      Can someone please explain why we should care whether or not $||vec{q}|| > 2lambda$, and how to compute the $x$ that maximizes the above function when $||vec{q}|| > 2lambda$?



      Edit: there is a constraint on $vec{x}$ that was mentioned elsewhere in the solution: $vec{x}$ is in the $2-$norm unit ball centered at $1$.






      optimization convex-analysis convex-optimization maxima-minima




      optimization convex-analysis convex-optimization maxima-minima






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




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      edited Dec 11 '18 at 21:54

























      asked Dec 11 '18 at 5:19









      0k33

      12010




      12010






















          1 Answer
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          2














          The function (which is concave, not convex) is always maximized at $x= 1 + frac{q}{2 lambda}$ for $lambda > 0$. Perhaps there is some constraint on $x$. Also, the solution you suggest for the second case is a number, not a vector.






          share|cite|improve this answer



















          • 1




            you actually need $lambda>0$ as otherwise the function is unbounded (and not concave)
            – LinAlg
            Dec 11 '18 at 14:25






          • 1




            Yes, that’s right. I edit my answer.
            – mineiro
            Dec 11 '18 at 15:39










          • @mineiro you are so right. I've edited my question to add the constraint (and I copied down the second part of the piecewise totally wrong, I have no idea where I got it from). Even with the constraint, it's not immediately coming to me!
            – 0k33
            Dec 11 '18 at 21:56










          • I figured it out! never mind!
            – 0k33
            Dec 13 '18 at 2:09











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          1 Answer
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          1 Answer
          1






          active

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          active

          oldest

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          active

          oldest

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          2














          The function (which is concave, not convex) is always maximized at $x= 1 + frac{q}{2 lambda}$ for $lambda > 0$. Perhaps there is some constraint on $x$. Also, the solution you suggest for the second case is a number, not a vector.






          share|cite|improve this answer



















          • 1




            you actually need $lambda>0$ as otherwise the function is unbounded (and not concave)
            – LinAlg
            Dec 11 '18 at 14:25






          • 1




            Yes, that’s right. I edit my answer.
            – mineiro
            Dec 11 '18 at 15:39










          • @mineiro you are so right. I've edited my question to add the constraint (and I copied down the second part of the piecewise totally wrong, I have no idea where I got it from). Even with the constraint, it's not immediately coming to me!
            – 0k33
            Dec 11 '18 at 21:56










          • I figured it out! never mind!
            – 0k33
            Dec 13 '18 at 2:09
















          2














          The function (which is concave, not convex) is always maximized at $x= 1 + frac{q}{2 lambda}$ for $lambda > 0$. Perhaps there is some constraint on $x$. Also, the solution you suggest for the second case is a number, not a vector.






          share|cite|improve this answer



















          • 1




            you actually need $lambda>0$ as otherwise the function is unbounded (and not concave)
            – LinAlg
            Dec 11 '18 at 14:25






          • 1




            Yes, that’s right. I edit my answer.
            – mineiro
            Dec 11 '18 at 15:39










          • @mineiro you are so right. I've edited my question to add the constraint (and I copied down the second part of the piecewise totally wrong, I have no idea where I got it from). Even with the constraint, it's not immediately coming to me!
            – 0k33
            Dec 11 '18 at 21:56










          • I figured it out! never mind!
            – 0k33
            Dec 13 '18 at 2:09














          2












          2








          2






          The function (which is concave, not convex) is always maximized at $x= 1 + frac{q}{2 lambda}$ for $lambda > 0$. Perhaps there is some constraint on $x$. Also, the solution you suggest for the second case is a number, not a vector.






          share|cite|improve this answer














          The function (which is concave, not convex) is always maximized at $x= 1 + frac{q}{2 lambda}$ for $lambda > 0$. Perhaps there is some constraint on $x$. Also, the solution you suggest for the second case is a number, not a vector.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 11 '18 at 15:40

























          answered Dec 11 '18 at 6:34









          mineiro

          812




          812








          • 1




            you actually need $lambda>0$ as otherwise the function is unbounded (and not concave)
            – LinAlg
            Dec 11 '18 at 14:25






          • 1




            Yes, that’s right. I edit my answer.
            – mineiro
            Dec 11 '18 at 15:39










          • @mineiro you are so right. I've edited my question to add the constraint (and I copied down the second part of the piecewise totally wrong, I have no idea where I got it from). Even with the constraint, it's not immediately coming to me!
            – 0k33
            Dec 11 '18 at 21:56










          • I figured it out! never mind!
            – 0k33
            Dec 13 '18 at 2:09














          • 1




            you actually need $lambda>0$ as otherwise the function is unbounded (and not concave)
            – LinAlg
            Dec 11 '18 at 14:25






          • 1




            Yes, that’s right. I edit my answer.
            – mineiro
            Dec 11 '18 at 15:39










          • @mineiro you are so right. I've edited my question to add the constraint (and I copied down the second part of the piecewise totally wrong, I have no idea where I got it from). Even with the constraint, it's not immediately coming to me!
            – 0k33
            Dec 11 '18 at 21:56










          • I figured it out! never mind!
            – 0k33
            Dec 13 '18 at 2:09








          1




          1




          you actually need $lambda>0$ as otherwise the function is unbounded (and not concave)
          – LinAlg
          Dec 11 '18 at 14:25




          you actually need $lambda>0$ as otherwise the function is unbounded (and not concave)
          – LinAlg
          Dec 11 '18 at 14:25




          1




          1




          Yes, that’s right. I edit my answer.
          – mineiro
          Dec 11 '18 at 15:39




          Yes, that’s right. I edit my answer.
          – mineiro
          Dec 11 '18 at 15:39












          @mineiro you are so right. I've edited my question to add the constraint (and I copied down the second part of the piecewise totally wrong, I have no idea where I got it from). Even with the constraint, it's not immediately coming to me!
          – 0k33
          Dec 11 '18 at 21:56




          @mineiro you are so right. I've edited my question to add the constraint (and I copied down the second part of the piecewise totally wrong, I have no idea where I got it from). Even with the constraint, it's not immediately coming to me!
          – 0k33
          Dec 11 '18 at 21:56












          I figured it out! never mind!
          – 0k33
          Dec 13 '18 at 2:09




          I figured it out! never mind!
          – 0k33
          Dec 13 '18 at 2:09


















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