How to show directly $min frac{1}{2}|x-z|_2^2$ has a unique global minimizer over $Ax=b$?












3












$begingroup$


Consider the following problem:



$$
min_{Ax=b} frac{1}{2}|x-z|_2^2
$$
where $z in mathbb{R}^{n}$ is a given vector, $x in mathbb{R}^{n}$, $A in mathbb{R}^{m times n}$ and $b in mathbb{R}^{m}$.



Show that there exists a unique global minimizer?



My try:



It is clear that this problem is a convex optimization problem and the local minimizer is a global minimizer. Also, it is strogly convex so it has at most one global minimizer. However, can we show the uniqueness of the minimizer directly?
To that end, we need to need to assume $x^*, y^* in mathbb{R}^{n}$ be to global minimizer.



$$
frac{1}{2}|x^*-z|_2^2 leq frac{1}{2}|x-z|_2^2,,, forall x
$$



Also,
$$
frac{1}{2}|y^*-z|_2^2 leq frac{1}{2}|x-z|_2^2,,, forall x
$$



Using these two we need to come up with a contradiction. How?






optimization convex-analysis convex-optimization







share|cite|improve this question









$endgroup$












  • $begingroup$
    If you have strong convexity already for the objective function $a,$ then you know that $a((1 - t)x + ty) < (1 - t) a(x) + t a(y)$ for all $t in (0, 1),$ hence $x$ and $y$ cannot be global minimisers.
    $endgroup$
    – Will M.
    Dec 14 '18 at 4:26
















3












$begingroup$


Consider the following problem:



$$
min_{Ax=b} frac{1}{2}|x-z|_2^2
$$
where $z in mathbb{R}^{n}$ is a given vector, $x in mathbb{R}^{n}$, $A in mathbb{R}^{m times n}$ and $b in mathbb{R}^{m}$.



Show that there exists a unique global minimizer?



My try:



It is clear that this problem is a convex optimization problem and the local minimizer is a global minimizer. Also, it is strogly convex so it has at most one global minimizer. However, can we show the uniqueness of the minimizer directly?
To that end, we need to need to assume $x^*, y^* in mathbb{R}^{n}$ be to global minimizer.



$$
frac{1}{2}|x^*-z|_2^2 leq frac{1}{2}|x-z|_2^2,,, forall x
$$



Also,
$$
frac{1}{2}|y^*-z|_2^2 leq frac{1}{2}|x-z|_2^2,,, forall x
$$



Using these two we need to come up with a contradiction. How?






optimization convex-analysis convex-optimization







share|cite|improve this question









$endgroup$












  • $begingroup$
    If you have strong convexity already for the objective function $a,$ then you know that $a((1 - t)x + ty) < (1 - t) a(x) + t a(y)$ for all $t in (0, 1),$ hence $x$ and $y$ cannot be global minimisers.
    $endgroup$
    – Will M.
    Dec 14 '18 at 4:26














3












3








3





$begingroup$


Consider the following problem:



$$
min_{Ax=b} frac{1}{2}|x-z|_2^2
$$
where $z in mathbb{R}^{n}$ is a given vector, $x in mathbb{R}^{n}$, $A in mathbb{R}^{m times n}$ and $b in mathbb{R}^{m}$.



Show that there exists a unique global minimizer?



My try:



It is clear that this problem is a convex optimization problem and the local minimizer is a global minimizer. Also, it is strogly convex so it has at most one global minimizer. However, can we show the uniqueness of the minimizer directly?
To that end, we need to need to assume $x^*, y^* in mathbb{R}^{n}$ be to global minimizer.



$$
frac{1}{2}|x^*-z|_2^2 leq frac{1}{2}|x-z|_2^2,,, forall x
$$



Also,
$$
frac{1}{2}|y^*-z|_2^2 leq frac{1}{2}|x-z|_2^2,,, forall x
$$



Using these two we need to come up with a contradiction. How?






optimization convex-analysis convex-optimization







share|cite|improve this question









$endgroup$




Consider the following problem:



$$
min_{Ax=b} frac{1}{2}|x-z|_2^2
$$
where $z in mathbb{R}^{n}$ is a given vector, $x in mathbb{R}^{n}$, $A in mathbb{R}^{m times n}$ and $b in mathbb{R}^{m}$.



Show that there exists a unique global minimizer?



My try:



It is clear that this problem is a convex optimization problem and the local minimizer is a global minimizer. Also, it is strogly convex so it has at most one global minimizer. However, can we show the uniqueness of the minimizer directly?
To that end, we need to need to assume $x^*, y^* in mathbb{R}^{n}$ be to global minimizer.



$$
frac{1}{2}|x^*-z|_2^2 leq frac{1}{2}|x-z|_2^2,,, forall x
$$



Also,
$$
frac{1}{2}|y^*-z|_2^2 leq frac{1}{2}|x-z|_2^2,,, forall x
$$



Using these two we need to come up with a contradiction. How?






optimization convex-analysis convex-optimization




optimization convex-analysis convex-optimization






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 14 '18 at 2:28









SepideSepide

3038




3038












  • $begingroup$
    If you have strong convexity already for the objective function $a,$ then you know that $a((1 - t)x + ty) < (1 - t) a(x) + t a(y)$ for all $t in (0, 1),$ hence $x$ and $y$ cannot be global minimisers.
    $endgroup$
    – Will M.
    Dec 14 '18 at 4:26


















  • $begingroup$
    If you have strong convexity already for the objective function $a,$ then you know that $a((1 - t)x + ty) < (1 - t) a(x) + t a(y)$ for all $t in (0, 1),$ hence $x$ and $y$ cannot be global minimisers.
    $endgroup$
    – Will M.
    Dec 14 '18 at 4:26
















$begingroup$
If you have strong convexity already for the objective function $a,$ then you know that $a((1 - t)x + ty) < (1 - t) a(x) + t a(y)$ for all $t in (0, 1),$ hence $x$ and $y$ cannot be global minimisers.
$endgroup$
– Will M.
Dec 14 '18 at 4:26




$begingroup$
If you have strong convexity already for the objective function $a,$ then you know that $a((1 - t)x + ty) < (1 - t) a(x) + t a(y)$ for all $t in (0, 1),$ hence $x$ and $y$ cannot be global minimisers.
$endgroup$
– Will M.
Dec 14 '18 at 4:26










2 Answers
2






active

oldest

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1












$begingroup$

The set $C:={xinBbb R^n : Ax=b}$ is an affine subset, hence it is closed and convex. The projection of $z$ into $C$ has a unique solution i.e. $exists x_0in C$ such that
$$
||x_0 - z||_2 = min_{yin C} ||y-z||_2.
$$
It is easy to see that $x_0$ also solves your minimization problem. (You can prove the existence of $x_0$ by taking a minimizing sequence and use compactness argument to show that it converges. The statement is even true for a Hilbert space but the prove is a bit more involved.)



For the second part, $frac 12||cdot - z||^2_2$ is strictly convex. Assume for contradiction that there exist two distinct minimizers $x_1, x_2$ and show that $frac {x_1+x_2}2$ is an even better minimizer. (In fact, the uniqueness of the projection from the previous part already implies uniqueness of the minimizer.)






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    If we introduce $y=x-z$, the problem takes the form of a least squares problem
    $$
    min_{Ay=b-Az} frac12|y|_2^2,
    $$
    which has an unique global minimizer $y= A^{+}(b-Az)$. Thus, the global minimizer of the original problem is $x=A^{+}(b-Az)+z$.






    share|cite|improve this answer











    $endgroup$













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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      1












      $begingroup$

      The set $C:={xinBbb R^n : Ax=b}$ is an affine subset, hence it is closed and convex. The projection of $z$ into $C$ has a unique solution i.e. $exists x_0in C$ such that
      $$
      ||x_0 - z||_2 = min_{yin C} ||y-z||_2.
      $$
      It is easy to see that $x_0$ also solves your minimization problem. (You can prove the existence of $x_0$ by taking a minimizing sequence and use compactness argument to show that it converges. The statement is even true for a Hilbert space but the prove is a bit more involved.)



      For the second part, $frac 12||cdot - z||^2_2$ is strictly convex. Assume for contradiction that there exist two distinct minimizers $x_1, x_2$ and show that $frac {x_1+x_2}2$ is an even better minimizer. (In fact, the uniqueness of the projection from the previous part already implies uniqueness of the minimizer.)






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        The set $C:={xinBbb R^n : Ax=b}$ is an affine subset, hence it is closed and convex. The projection of $z$ into $C$ has a unique solution i.e. $exists x_0in C$ such that
        $$
        ||x_0 - z||_2 = min_{yin C} ||y-z||_2.
        $$
        It is easy to see that $x_0$ also solves your minimization problem. (You can prove the existence of $x_0$ by taking a minimizing sequence and use compactness argument to show that it converges. The statement is even true for a Hilbert space but the prove is a bit more involved.)



        For the second part, $frac 12||cdot - z||^2_2$ is strictly convex. Assume for contradiction that there exist two distinct minimizers $x_1, x_2$ and show that $frac {x_1+x_2}2$ is an even better minimizer. (In fact, the uniqueness of the projection from the previous part already implies uniqueness of the minimizer.)






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          The set $C:={xinBbb R^n : Ax=b}$ is an affine subset, hence it is closed and convex. The projection of $z$ into $C$ has a unique solution i.e. $exists x_0in C$ such that
          $$
          ||x_0 - z||_2 = min_{yin C} ||y-z||_2.
          $$
          It is easy to see that $x_0$ also solves your minimization problem. (You can prove the existence of $x_0$ by taking a minimizing sequence and use compactness argument to show that it converges. The statement is even true for a Hilbert space but the prove is a bit more involved.)



          For the second part, $frac 12||cdot - z||^2_2$ is strictly convex. Assume for contradiction that there exist two distinct minimizers $x_1, x_2$ and show that $frac {x_1+x_2}2$ is an even better minimizer. (In fact, the uniqueness of the projection from the previous part already implies uniqueness of the minimizer.)






          share|cite|improve this answer









          $endgroup$



          The set $C:={xinBbb R^n : Ax=b}$ is an affine subset, hence it is closed and convex. The projection of $z$ into $C$ has a unique solution i.e. $exists x_0in C$ such that
          $$
          ||x_0 - z||_2 = min_{yin C} ||y-z||_2.
          $$
          It is easy to see that $x_0$ also solves your minimization problem. (You can prove the existence of $x_0$ by taking a minimizing sequence and use compactness argument to show that it converges. The statement is even true for a Hilbert space but the prove is a bit more involved.)



          For the second part, $frac 12||cdot - z||^2_2$ is strictly convex. Assume for contradiction that there exist two distinct minimizers $x_1, x_2$ and show that $frac {x_1+x_2}2$ is an even better minimizer. (In fact, the uniqueness of the projection from the previous part already implies uniqueness of the minimizer.)







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 14 '18 at 4:07









          BigbearZzzBigbearZzz

          8,67621652




          8,67621652























              1












              $begingroup$

              If we introduce $y=x-z$, the problem takes the form of a least squares problem
              $$
              min_{Ay=b-Az} frac12|y|_2^2,
              $$
              which has an unique global minimizer $y= A^{+}(b-Az)$. Thus, the global minimizer of the original problem is $x=A^{+}(b-Az)+z$.






              share|cite|improve this answer











              $endgroup$


















                1












                $begingroup$

                If we introduce $y=x-z$, the problem takes the form of a least squares problem
                $$
                min_{Ay=b-Az} frac12|y|_2^2,
                $$
                which has an unique global minimizer $y= A^{+}(b-Az)$. Thus, the global minimizer of the original problem is $x=A^{+}(b-Az)+z$.






                share|cite|improve this answer











                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  If we introduce $y=x-z$, the problem takes the form of a least squares problem
                  $$
                  min_{Ay=b-Az} frac12|y|_2^2,
                  $$
                  which has an unique global minimizer $y= A^{+}(b-Az)$. Thus, the global minimizer of the original problem is $x=A^{+}(b-Az)+z$.






                  share|cite|improve this answer











                  $endgroup$



                  If we introduce $y=x-z$, the problem takes the form of a least squares problem
                  $$
                  min_{Ay=b-Az} frac12|y|_2^2,
                  $$
                  which has an unique global minimizer $y= A^{+}(b-Az)$. Thus, the global minimizer of the original problem is $x=A^{+}(b-Az)+z$.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Dec 14 '18 at 4:50

























                  answered Dec 14 '18 at 4:18









                  AVKAVK

                  2,0961517




                  2,0961517






























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