Convex set euclidean ball proof












1












$begingroup$


I was studying some notes on convex optimization and came across this formula



Euclidean ball with center $x_c$ and radius r:



$${{x_c + ru ∣ ∥u∥_2 leq 1}}$$



How do i even go about proving that this is a convex set. I am extremely new to this and i am not asking for answers. But i just need to even know how to start proving.



I am not even sure what does this mean



https://www.ics.uci.edu/~xhx/courses/CS206/convex_sets.pdf






optimization convex-analysis







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    I was studying some notes on convex optimization and came across this formula



    Euclidean ball with center $x_c$ and radius r:



    $${{x_c + ru ∣ ∥u∥_2 leq 1}}$$



    How do i even go about proving that this is a convex set. I am extremely new to this and i am not asking for answers. But i just need to even know how to start proving.



    I am not even sure what does this mean



    https://www.ics.uci.edu/~xhx/courses/CS206/convex_sets.pdf






    optimization convex-analysis







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      I was studying some notes on convex optimization and came across this formula



      Euclidean ball with center $x_c$ and radius r:



      $${{x_c + ru ∣ ∥u∥_2 leq 1}}$$



      How do i even go about proving that this is a convex set. I am extremely new to this and i am not asking for answers. But i just need to even know how to start proving.



      I am not even sure what does this mean



      https://www.ics.uci.edu/~xhx/courses/CS206/convex_sets.pdf






      optimization convex-analysis







      share|cite|improve this question











      $endgroup$




      I was studying some notes on convex optimization and came across this formula



      Euclidean ball with center $x_c$ and radius r:



      $${{x_c + ru ∣ ∥u∥_2 leq 1}}$$



      How do i even go about proving that this is a convex set. I am extremely new to this and i am not asking for answers. But i just need to even know how to start proving.



      I am not even sure what does this mean



      https://www.ics.uci.edu/~xhx/courses/CS206/convex_sets.pdf






      optimization convex-analysis




      optimization convex-analysis






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 28 at 15:30









      Robert Z

      101k1070143




      101k1070143










      asked Jan 28 at 15:27









      acemineraceminer

      230112




      230112






















          2 Answers
          2






          active

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          2












          $begingroup$

          Verify the definition (see page 1 of the linked pdf): let $x_1, x_2in B:={{x_c + ru ∣ |u|_2 leq 1}}$ that is,
          $$x_1=x_c+ru_1quad text{and}quad x_2=x_c+ru_2$$
          with $|u_1|_2 leq 1$ and $|u_2|_2 leq 1$,
          and let $thetain[0,1]$, then show that
          $$theta x_1+(1-theta)x_2=theta(x_c+ru_1)+(1-theta)(x_c+ru_2)\=x_c+r(theta u_1+(1-theta)u_2)$$
          belongs to $B$, that is $|theta u_1+(1-theta)u_2|_2leq 1$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            am i right to say that your last line is correct because of the nature of the original definition that the thetas sum up to 1?
            $endgroup$
            – aceminer
            Jan 28 at 15:53






          • 1




            $begingroup$
            @aceminer no, the last line is still a challenge. You asked 'how to start proving' and this answer shows you how to start.
            $endgroup$
            – LinAlg
            Jan 28 at 15:54






          • 1




            $begingroup$
            @aceminer No, in order to prove the last line you should use the definition of $|cdot|_2$.
            $endgroup$
            – Robert Z
            Jan 28 at 16:01








          • 1




            $begingroup$
            @RobertZ Thanks! it helped me solve my question
            $endgroup$
            – aceminer
            Jan 29 at 1:41



















          1












          $begingroup$

          Using the definition of convex, it is straightforward to show that if $C$ is convex then
          the translate ${ x+c | c in C}$ is convex.



          Using the definition of convex, it is straightforward to show that if $C$ is convex then
          the scaled set ${ rc | c in C}$ is convex.



          Since the norm $| cdot |$ is a convex function, it is straightforward to show, from the
          definition, that the set ${x | |x| le 1}$ is convex.






          share|cite|improve this answer









          $endgroup$














            Your Answer





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            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            2












            $begingroup$

            Verify the definition (see page 1 of the linked pdf): let $x_1, x_2in B:={{x_c + ru ∣ |u|_2 leq 1}}$ that is,
            $$x_1=x_c+ru_1quad text{and}quad x_2=x_c+ru_2$$
            with $|u_1|_2 leq 1$ and $|u_2|_2 leq 1$,
            and let $thetain[0,1]$, then show that
            $$theta x_1+(1-theta)x_2=theta(x_c+ru_1)+(1-theta)(x_c+ru_2)\=x_c+r(theta u_1+(1-theta)u_2)$$
            belongs to $B$, that is $|theta u_1+(1-theta)u_2|_2leq 1$.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              am i right to say that your last line is correct because of the nature of the original definition that the thetas sum up to 1?
              $endgroup$
              – aceminer
              Jan 28 at 15:53






            • 1




              $begingroup$
              @aceminer no, the last line is still a challenge. You asked 'how to start proving' and this answer shows you how to start.
              $endgroup$
              – LinAlg
              Jan 28 at 15:54






            • 1




              $begingroup$
              @aceminer No, in order to prove the last line you should use the definition of $|cdot|_2$.
              $endgroup$
              – Robert Z
              Jan 28 at 16:01








            • 1




              $begingroup$
              @RobertZ Thanks! it helped me solve my question
              $endgroup$
              – aceminer
              Jan 29 at 1:41
















            2












            $begingroup$

            Verify the definition (see page 1 of the linked pdf): let $x_1, x_2in B:={{x_c + ru ∣ |u|_2 leq 1}}$ that is,
            $$x_1=x_c+ru_1quad text{and}quad x_2=x_c+ru_2$$
            with $|u_1|_2 leq 1$ and $|u_2|_2 leq 1$,
            and let $thetain[0,1]$, then show that
            $$theta x_1+(1-theta)x_2=theta(x_c+ru_1)+(1-theta)(x_c+ru_2)\=x_c+r(theta u_1+(1-theta)u_2)$$
            belongs to $B$, that is $|theta u_1+(1-theta)u_2|_2leq 1$.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              am i right to say that your last line is correct because of the nature of the original definition that the thetas sum up to 1?
              $endgroup$
              – aceminer
              Jan 28 at 15:53






            • 1




              $begingroup$
              @aceminer no, the last line is still a challenge. You asked 'how to start proving' and this answer shows you how to start.
              $endgroup$
              – LinAlg
              Jan 28 at 15:54






            • 1




              $begingroup$
              @aceminer No, in order to prove the last line you should use the definition of $|cdot|_2$.
              $endgroup$
              – Robert Z
              Jan 28 at 16:01








            • 1




              $begingroup$
              @RobertZ Thanks! it helped me solve my question
              $endgroup$
              – aceminer
              Jan 29 at 1:41














            2












            2








            2





            $begingroup$

            Verify the definition (see page 1 of the linked pdf): let $x_1, x_2in B:={{x_c + ru ∣ |u|_2 leq 1}}$ that is,
            $$x_1=x_c+ru_1quad text{and}quad x_2=x_c+ru_2$$
            with $|u_1|_2 leq 1$ and $|u_2|_2 leq 1$,
            and let $thetain[0,1]$, then show that
            $$theta x_1+(1-theta)x_2=theta(x_c+ru_1)+(1-theta)(x_c+ru_2)\=x_c+r(theta u_1+(1-theta)u_2)$$
            belongs to $B$, that is $|theta u_1+(1-theta)u_2|_2leq 1$.






            share|cite|improve this answer











            $endgroup$



            Verify the definition (see page 1 of the linked pdf): let $x_1, x_2in B:={{x_c + ru ∣ |u|_2 leq 1}}$ that is,
            $$x_1=x_c+ru_1quad text{and}quad x_2=x_c+ru_2$$
            with $|u_1|_2 leq 1$ and $|u_2|_2 leq 1$,
            and let $thetain[0,1]$, then show that
            $$theta x_1+(1-theta)x_2=theta(x_c+ru_1)+(1-theta)(x_c+ru_2)\=x_c+r(theta u_1+(1-theta)u_2)$$
            belongs to $B$, that is $|theta u_1+(1-theta)u_2|_2leq 1$.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Jan 28 at 15:40

























            answered Jan 28 at 15:33









            Robert ZRobert Z

            101k1070143




            101k1070143












            • $begingroup$
              am i right to say that your last line is correct because of the nature of the original definition that the thetas sum up to 1?
              $endgroup$
              – aceminer
              Jan 28 at 15:53






            • 1




              $begingroup$
              @aceminer no, the last line is still a challenge. You asked 'how to start proving' and this answer shows you how to start.
              $endgroup$
              – LinAlg
              Jan 28 at 15:54






            • 1




              $begingroup$
              @aceminer No, in order to prove the last line you should use the definition of $|cdot|_2$.
              $endgroup$
              – Robert Z
              Jan 28 at 16:01








            • 1




              $begingroup$
              @RobertZ Thanks! it helped me solve my question
              $endgroup$
              – aceminer
              Jan 29 at 1:41


















            • $begingroup$
              am i right to say that your last line is correct because of the nature of the original definition that the thetas sum up to 1?
              $endgroup$
              – aceminer
              Jan 28 at 15:53






            • 1




              $begingroup$
              @aceminer no, the last line is still a challenge. You asked 'how to start proving' and this answer shows you how to start.
              $endgroup$
              – LinAlg
              Jan 28 at 15:54






            • 1




              $begingroup$
              @aceminer No, in order to prove the last line you should use the definition of $|cdot|_2$.
              $endgroup$
              – Robert Z
              Jan 28 at 16:01








            • 1




              $begingroup$
              @RobertZ Thanks! it helped me solve my question
              $endgroup$
              – aceminer
              Jan 29 at 1:41
















            $begingroup$
            am i right to say that your last line is correct because of the nature of the original definition that the thetas sum up to 1?
            $endgroup$
            – aceminer
            Jan 28 at 15:53




            $begingroup$
            am i right to say that your last line is correct because of the nature of the original definition that the thetas sum up to 1?
            $endgroup$
            – aceminer
            Jan 28 at 15:53




            1




            1




            $begingroup$
            @aceminer no, the last line is still a challenge. You asked 'how to start proving' and this answer shows you how to start.
            $endgroup$
            – LinAlg
            Jan 28 at 15:54




            $begingroup$
            @aceminer no, the last line is still a challenge. You asked 'how to start proving' and this answer shows you how to start.
            $endgroup$
            – LinAlg
            Jan 28 at 15:54




            1




            1




            $begingroup$
            @aceminer No, in order to prove the last line you should use the definition of $|cdot|_2$.
            $endgroup$
            – Robert Z
            Jan 28 at 16:01






            $begingroup$
            @aceminer No, in order to prove the last line you should use the definition of $|cdot|_2$.
            $endgroup$
            – Robert Z
            Jan 28 at 16:01






            1




            1




            $begingroup$
            @RobertZ Thanks! it helped me solve my question
            $endgroup$
            – aceminer
            Jan 29 at 1:41




            $begingroup$
            @RobertZ Thanks! it helped me solve my question
            $endgroup$
            – aceminer
            Jan 29 at 1:41











            1












            $begingroup$

            Using the definition of convex, it is straightforward to show that if $C$ is convex then
            the translate ${ x+c | c in C}$ is convex.



            Using the definition of convex, it is straightforward to show that if $C$ is convex then
            the scaled set ${ rc | c in C}$ is convex.



            Since the norm $| cdot |$ is a convex function, it is straightforward to show, from the
            definition, that the set ${x | |x| le 1}$ is convex.






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              Using the definition of convex, it is straightforward to show that if $C$ is convex then
              the translate ${ x+c | c in C}$ is convex.



              Using the definition of convex, it is straightforward to show that if $C$ is convex then
              the scaled set ${ rc | c in C}$ is convex.



              Since the norm $| cdot |$ is a convex function, it is straightforward to show, from the
              definition, that the set ${x | |x| le 1}$ is convex.






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                Using the definition of convex, it is straightforward to show that if $C$ is convex then
                the translate ${ x+c | c in C}$ is convex.



                Using the definition of convex, it is straightforward to show that if $C$ is convex then
                the scaled set ${ rc | c in C}$ is convex.



                Since the norm $| cdot |$ is a convex function, it is straightforward to show, from the
                definition, that the set ${x | |x| le 1}$ is convex.






                share|cite|improve this answer









                $endgroup$



                Using the definition of convex, it is straightforward to show that if $C$ is convex then
                the translate ${ x+c | c in C}$ is convex.



                Using the definition of convex, it is straightforward to show that if $C$ is convex then
                the scaled set ${ rc | c in C}$ is convex.



                Since the norm $| cdot |$ is a convex function, it is straightforward to show, from the
                definition, that the set ${x | |x| le 1}$ is convex.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 28 at 16:20









                copper.hatcopper.hat

                128k559161




                128k559161






























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