simple sup norm for a function on a unit circleHow to find the supremum of this?Geometric conditions equivalent to a set being the unit circle for some normWhy can we calculate the supremum of operator norm over unit circle?tangents at unit circle - parametrization leads to strange resultIs there a meaning to the notation “arg sup”?Show $sqrt2 = textsupA$ where $A = x in mathbb Q ;colon x^2 < 2$Supremum and infimum of a rational set bounded by irrational numbersMaximum and Supremum of IntervalsIntuition for $limsup$ and $liminf$For some $c geq 0$ $textsup c cdot f(x): textsome domain of $x$ $ = $c cdot textsup f(x): textsame domain of $x$ $

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simple sup norm for a function on a unit circle


How to find the supremum of this?Geometric conditions equivalent to a set being the unit circle for some normWhy can we calculate the supremum of operator norm over unit circle?tangents at unit circle - parametrization leads to strange resultIs there a meaning to the notation “arg sup”?Show $sqrt2 = textsupA$ where $A = x in mathbb Q ;colon x^2 < 2$Supremum and infimum of a rational set bounded by irrational numbersMaximum and Supremum of IntervalsIntuition for $limsup$ and $liminf$For some $c geq 0$ $textsup c cdot f(x): textsome domain of $x$ $ = $c cdot textsup f(x): textsame domain of $x$ $






.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;








2












$begingroup$


From my understanding , the supremum norm would just be the maximum value



So if I have a point z that lies on the unit circle $|z|=1$



Question:
what would the $sup_ |2z-1|$ be ....



Is it just the maximum value of $(2z-1)$ for all points on the unit circle?



I attempted it in to ways, not sure if either is correct



First method



$ sup_ |2z-1| = 1$ As you simply just take the $|z|=1 $ so $ |2(1)-1|=1$ I dont think this is correct



Second method:
You just let $z = x+iy$ where $|z|=1$



$|2(x+iy)-1| $ $= |(2x-1)-2iy| = $ $sqrt4x^2-4x-1+4y^2$ so if z lies on the unit circle $x^2 + y^2 = 1 $ or $ x = cosθ $, $y = sinθ$



So the sup would just be the maximum value for $sqrt3-4cosθ = sqrt7? $



Not sure what the maximum value for $sup_ |2z-1|$ would be here.










share|cite|improve this question











$endgroup$


















    2












    $begingroup$


    From my understanding , the supremum norm would just be the maximum value



    So if I have a point z that lies on the unit circle $|z|=1$



    Question:
    what would the $sup_ |2z-1|$ be ....



    Is it just the maximum value of $(2z-1)$ for all points on the unit circle?



    I attempted it in to ways, not sure if either is correct



    First method



    $ sup_ |2z-1| = 1$ As you simply just take the $|z|=1 $ so $ |2(1)-1|=1$ I dont think this is correct



    Second method:
    You just let $z = x+iy$ where $|z|=1$



    $|2(x+iy)-1| $ $= |(2x-1)-2iy| = $ $sqrt4x^2-4x-1+4y^2$ so if z lies on the unit circle $x^2 + y^2 = 1 $ or $ x = cosθ $, $y = sinθ$



    So the sup would just be the maximum value for $sqrt3-4cosθ = sqrt7? $



    Not sure what the maximum value for $sup_ |2z-1|$ would be here.










    share|cite|improve this question











    $endgroup$














      2












      2








      2





      $begingroup$


      From my understanding , the supremum norm would just be the maximum value



      So if I have a point z that lies on the unit circle $|z|=1$



      Question:
      what would the $sup_ |2z-1|$ be ....



      Is it just the maximum value of $(2z-1)$ for all points on the unit circle?



      I attempted it in to ways, not sure if either is correct



      First method



      $ sup_ |2z-1| = 1$ As you simply just take the $|z|=1 $ so $ |2(1)-1|=1$ I dont think this is correct



      Second method:
      You just let $z = x+iy$ where $|z|=1$



      $|2(x+iy)-1| $ $= |(2x-1)-2iy| = $ $sqrt4x^2-4x-1+4y^2$ so if z lies on the unit circle $x^2 + y^2 = 1 $ or $ x = cosθ $, $y = sinθ$



      So the sup would just be the maximum value for $sqrt3-4cosθ = sqrt7? $



      Not sure what the maximum value for $sup_ |2z-1|$ would be here.










      share|cite|improve this question











      $endgroup$




      From my understanding , the supremum norm would just be the maximum value



      So if I have a point z that lies on the unit circle $|z|=1$



      Question:
      what would the $sup_ |2z-1|$ be ....



      Is it just the maximum value of $(2z-1)$ for all points on the unit circle?



      I attempted it in to ways, not sure if either is correct



      First method



      $ sup_ |2z-1| = 1$ As you simply just take the $|z|=1 $ so $ |2(1)-1|=1$ I dont think this is correct



      Second method:
      You just let $z = x+iy$ where $|z|=1$



      $|2(x+iy)-1| $ $= |(2x-1)-2iy| = $ $sqrt4x^2-4x-1+4y^2$ so if z lies on the unit circle $x^2 + y^2 = 1 $ or $ x = cosθ $, $y = sinθ$



      So the sup would just be the maximum value for $sqrt3-4cosθ = sqrt7? $



      Not sure what the maximum value for $sup_ |2z-1|$ would be here.







      real-analysis maxima-minima supremum-and-infimum






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jun 30 at 8:42









      Bernard

      129k7 gold badges43 silver badges122 bronze badges




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      asked Jun 30 at 8:14









      Swag34214Swag34214

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          4 Answers
          4






          active

          oldest

          votes


















          6












          $begingroup$

          The first method doesn't work. The condition $|z| = 1$ doesn't imply $z = 1$, nor is $|2z - 1|$ the same as $2|z| - 1$.



          The second method is much better, however you seem to misplaced your $-$ signs. You should get
          $$|(2x - 1) colorred+ 2iy| = sqrt4x^2 - 4x colorred+ 1 + 4y^2 = sqrt5 - 4cos theta,$$
          which has a maximum of $sqrt9 = 3$.



          This could also be achieved through use of the triangle inequality. Consider
          $$|2z - 1| le |2z| + |-1| = 2|z| + 1 = 3,$$
          with equality if and only if $2z$ and $-1$ point in the same direction, i.e. when $z$ is a negative number. Substituting in $z = -1$ confirms that the value of $3$ is attained.






          share|cite|improve this answer











          $endgroup$












          • $begingroup$
            Ahh yes thank you for the clarification.
            $endgroup$
            – Swag34214
            Jun 30 at 8:38


















          2












          $begingroup$

          Geometrically, for $|z| = 1$, $|2z - 1|$ is the distance between a point on the circle of radius $2$ centered at the origin (in red), and the point $(1,0)$ (in blue).



          By looking at the attached image, we can see that the maximum such distance is $3$. This can more easily be seen by considering a plot of the circle of radius $3$ centered at $(1,0)$ (in green). The point $(-2,0)$ is a distance of $3$ away from $(1,0)$, and each other point on the circle of radius $2$ is closer to $(1,0)$.



          Thus, $$sup_ |2z - 1| = 3.$$






          share|cite|improve this answer











          $endgroup$




















            1












            $begingroup$

            The first method is definitely wrong as you suggested. This is because $|z_1 + z_2| neq |z_1| + |z_2|$. In other words: you cannot just replace all variables inside $||$ with their "length".



            The second method is correct since in every step you use a definition or some known fact. (Though you made a mistake when squaring $(2x-1)^2=4x^2-4x+1$. This gives you $sqrt5 - 4x)$. You do not need to represent $x$ using the angle $theta$. Just think about what $x$ would maximize this expression and can you find this value for $x$ on the unit circle. The conclusion should be that $x=-1$ which you can find on the unit circle (namely for $z=-1$).






            share|cite|improve this answer









            $endgroup$




















              1












              $begingroup$

              Note that $:|2z-1|=2Bigl|z-frac12Bigr|$, i.e. geometrically, twice the distance between the affix of $z$ and the point $;Abigl(frac12,0bigr)$ in the plane. It is attained at the intersection point of the diameter through $A$ (i.e. the $x$ axis) which has a negative abscissa: $(-1,0)$. Thus
              $$sup_|2z-1|=2cdotfrac 32=colorred3.$$






              share|cite|improve this answer









              $endgroup$















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                4 Answers
                4






                active

                oldest

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                4 Answers
                4






                active

                oldest

                votes









                active

                oldest

                votes






                active

                oldest

                votes









                6












                $begingroup$

                The first method doesn't work. The condition $|z| = 1$ doesn't imply $z = 1$, nor is $|2z - 1|$ the same as $2|z| - 1$.



                The second method is much better, however you seem to misplaced your $-$ signs. You should get
                $$|(2x - 1) colorred+ 2iy| = sqrt4x^2 - 4x colorred+ 1 + 4y^2 = sqrt5 - 4cos theta,$$
                which has a maximum of $sqrt9 = 3$.



                This could also be achieved through use of the triangle inequality. Consider
                $$|2z - 1| le |2z| + |-1| = 2|z| + 1 = 3,$$
                with equality if and only if $2z$ and $-1$ point in the same direction, i.e. when $z$ is a negative number. Substituting in $z = -1$ confirms that the value of $3$ is attained.






                share|cite|improve this answer











                $endgroup$












                • $begingroup$
                  Ahh yes thank you for the clarification.
                  $endgroup$
                  – Swag34214
                  Jun 30 at 8:38















                6












                $begingroup$

                The first method doesn't work. The condition $|z| = 1$ doesn't imply $z = 1$, nor is $|2z - 1|$ the same as $2|z| - 1$.



                The second method is much better, however you seem to misplaced your $-$ signs. You should get
                $$|(2x - 1) colorred+ 2iy| = sqrt4x^2 - 4x colorred+ 1 + 4y^2 = sqrt5 - 4cos theta,$$
                which has a maximum of $sqrt9 = 3$.



                This could also be achieved through use of the triangle inequality. Consider
                $$|2z - 1| le |2z| + |-1| = 2|z| + 1 = 3,$$
                with equality if and only if $2z$ and $-1$ point in the same direction, i.e. when $z$ is a negative number. Substituting in $z = -1$ confirms that the value of $3$ is attained.






                share|cite|improve this answer











                $endgroup$












                • $begingroup$
                  Ahh yes thank you for the clarification.
                  $endgroup$
                  – Swag34214
                  Jun 30 at 8:38













                6












                6








                6





                $begingroup$

                The first method doesn't work. The condition $|z| = 1$ doesn't imply $z = 1$, nor is $|2z - 1|$ the same as $2|z| - 1$.



                The second method is much better, however you seem to misplaced your $-$ signs. You should get
                $$|(2x - 1) colorred+ 2iy| = sqrt4x^2 - 4x colorred+ 1 + 4y^2 = sqrt5 - 4cos theta,$$
                which has a maximum of $sqrt9 = 3$.



                This could also be achieved through use of the triangle inequality. Consider
                $$|2z - 1| le |2z| + |-1| = 2|z| + 1 = 3,$$
                with equality if and only if $2z$ and $-1$ point in the same direction, i.e. when $z$ is a negative number. Substituting in $z = -1$ confirms that the value of $3$ is attained.






                share|cite|improve this answer











                $endgroup$



                The first method doesn't work. The condition $|z| = 1$ doesn't imply $z = 1$, nor is $|2z - 1|$ the same as $2|z| - 1$.



                The second method is much better, however you seem to misplaced your $-$ signs. You should get
                $$|(2x - 1) colorred+ 2iy| = sqrt4x^2 - 4x colorred+ 1 + 4y^2 = sqrt5 - 4cos theta,$$
                which has a maximum of $sqrt9 = 3$.



                This could also be achieved through use of the triangle inequality. Consider
                $$|2z - 1| le |2z| + |-1| = 2|z| + 1 = 3,$$
                with equality if and only if $2z$ and $-1$ point in the same direction, i.e. when $z$ is a negative number. Substituting in $z = -1$ confirms that the value of $3$ is attained.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Jun 30 at 9:12

























                answered Jun 30 at 8:32









                Theo BenditTheo Bendit

                24.9k1 gold badge25 silver badges61 bronze badges




                24.9k1 gold badge25 silver badges61 bronze badges











                • $begingroup$
                  Ahh yes thank you for the clarification.
                  $endgroup$
                  – Swag34214
                  Jun 30 at 8:38
















                • $begingroup$
                  Ahh yes thank you for the clarification.
                  $endgroup$
                  – Swag34214
                  Jun 30 at 8:38















                $begingroup$
                Ahh yes thank you for the clarification.
                $endgroup$
                – Swag34214
                Jun 30 at 8:38




                $begingroup$
                Ahh yes thank you for the clarification.
                $endgroup$
                – Swag34214
                Jun 30 at 8:38













                2












                $begingroup$

                Geometrically, for $|z| = 1$, $|2z - 1|$ is the distance between a point on the circle of radius $2$ centered at the origin (in red), and the point $(1,0)$ (in blue).



                By looking at the attached image, we can see that the maximum such distance is $3$. This can more easily be seen by considering a plot of the circle of radius $3$ centered at $(1,0)$ (in green). The point $(-2,0)$ is a distance of $3$ away from $(1,0)$, and each other point on the circle of radius $2$ is closer to $(1,0)$.



                Thus, $$sup_ |2z - 1| = 3.$$






                share|cite|improve this answer











                $endgroup$

















                  2












                  $begingroup$

                  Geometrically, for $|z| = 1$, $|2z - 1|$ is the distance between a point on the circle of radius $2$ centered at the origin (in red), and the point $(1,0)$ (in blue).



                  By looking at the attached image, we can see that the maximum such distance is $3$. This can more easily be seen by considering a plot of the circle of radius $3$ centered at $(1,0)$ (in green). The point $(-2,0)$ is a distance of $3$ away from $(1,0)$, and each other point on the circle of radius $2$ is closer to $(1,0)$.



                  Thus, $$sup_ |2z - 1| = 3.$$






                  share|cite|improve this answer











                  $endgroup$















                    2












                    2








                    2





                    $begingroup$

                    Geometrically, for $|z| = 1$, $|2z - 1|$ is the distance between a point on the circle of radius $2$ centered at the origin (in red), and the point $(1,0)$ (in blue).



                    By looking at the attached image, we can see that the maximum such distance is $3$. This can more easily be seen by considering a plot of the circle of radius $3$ centered at $(1,0)$ (in green). The point $(-2,0)$ is a distance of $3$ away from $(1,0)$, and each other point on the circle of radius $2$ is closer to $(1,0)$.



                    Thus, $$sup_ |2z - 1| = 3.$$






                    share|cite|improve this answer











                    $endgroup$



                    Geometrically, for $|z| = 1$, $|2z - 1|$ is the distance between a point on the circle of radius $2$ centered at the origin (in red), and the point $(1,0)$ (in blue).



                    By looking at the attached image, we can see that the maximum such distance is $3$. This can more easily be seen by considering a plot of the circle of radius $3$ centered at $(1,0)$ (in green). The point $(-2,0)$ is a distance of $3$ away from $(1,0)$, and each other point on the circle of radius $2$ is closer to $(1,0)$.



                    Thus, $$sup_ |2z - 1| = 3.$$







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited Jun 30 at 9:44

























                    answered Jun 30 at 8:31









                    Eh WhaEh Wha

                    1415 bronze badges




                    1415 bronze badges





















                        1












                        $begingroup$

                        The first method is definitely wrong as you suggested. This is because $|z_1 + z_2| neq |z_1| + |z_2|$. In other words: you cannot just replace all variables inside $||$ with their "length".



                        The second method is correct since in every step you use a definition or some known fact. (Though you made a mistake when squaring $(2x-1)^2=4x^2-4x+1$. This gives you $sqrt5 - 4x)$. You do not need to represent $x$ using the angle $theta$. Just think about what $x$ would maximize this expression and can you find this value for $x$ on the unit circle. The conclusion should be that $x=-1$ which you can find on the unit circle (namely for $z=-1$).






                        share|cite|improve this answer









                        $endgroup$

















                          1












                          $begingroup$

                          The first method is definitely wrong as you suggested. This is because $|z_1 + z_2| neq |z_1| + |z_2|$. In other words: you cannot just replace all variables inside $||$ with their "length".



                          The second method is correct since in every step you use a definition or some known fact. (Though you made a mistake when squaring $(2x-1)^2=4x^2-4x+1$. This gives you $sqrt5 - 4x)$. You do not need to represent $x$ using the angle $theta$. Just think about what $x$ would maximize this expression and can you find this value for $x$ on the unit circle. The conclusion should be that $x=-1$ which you can find on the unit circle (namely for $z=-1$).






                          share|cite|improve this answer









                          $endgroup$















                            1












                            1








                            1





                            $begingroup$

                            The first method is definitely wrong as you suggested. This is because $|z_1 + z_2| neq |z_1| + |z_2|$. In other words: you cannot just replace all variables inside $||$ with their "length".



                            The second method is correct since in every step you use a definition or some known fact. (Though you made a mistake when squaring $(2x-1)^2=4x^2-4x+1$. This gives you $sqrt5 - 4x)$. You do not need to represent $x$ using the angle $theta$. Just think about what $x$ would maximize this expression and can you find this value for $x$ on the unit circle. The conclusion should be that $x=-1$ which you can find on the unit circle (namely for $z=-1$).






                            share|cite|improve this answer









                            $endgroup$



                            The first method is definitely wrong as you suggested. This is because $|z_1 + z_2| neq |z_1| + |z_2|$. In other words: you cannot just replace all variables inside $||$ with their "length".



                            The second method is correct since in every step you use a definition or some known fact. (Though you made a mistake when squaring $(2x-1)^2=4x^2-4x+1$. This gives you $sqrt5 - 4x)$. You do not need to represent $x$ using the angle $theta$. Just think about what $x$ would maximize this expression and can you find this value for $x$ on the unit circle. The conclusion should be that $x=-1$ which you can find on the unit circle (namely for $z=-1$).







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Jun 30 at 8:34









                            araomisaraomis

                            8041 silver badge10 bronze badges




                            8041 silver badge10 bronze badges





















                                1












                                $begingroup$

                                Note that $:|2z-1|=2Bigl|z-frac12Bigr|$, i.e. geometrically, twice the distance between the affix of $z$ and the point $;Abigl(frac12,0bigr)$ in the plane. It is attained at the intersection point of the diameter through $A$ (i.e. the $x$ axis) which has a negative abscissa: $(-1,0)$. Thus
                                $$sup_|2z-1|=2cdotfrac 32=colorred3.$$






                                share|cite|improve this answer









                                $endgroup$

















                                  1












                                  $begingroup$

                                  Note that $:|2z-1|=2Bigl|z-frac12Bigr|$, i.e. geometrically, twice the distance between the affix of $z$ and the point $;Abigl(frac12,0bigr)$ in the plane. It is attained at the intersection point of the diameter through $A$ (i.e. the $x$ axis) which has a negative abscissa: $(-1,0)$. Thus
                                  $$sup_|2z-1|=2cdotfrac 32=colorred3.$$






                                  share|cite|improve this answer









                                  $endgroup$















                                    1












                                    1








                                    1





                                    $begingroup$

                                    Note that $:|2z-1|=2Bigl|z-frac12Bigr|$, i.e. geometrically, twice the distance between the affix of $z$ and the point $;Abigl(frac12,0bigr)$ in the plane. It is attained at the intersection point of the diameter through $A$ (i.e. the $x$ axis) which has a negative abscissa: $(-1,0)$. Thus
                                    $$sup_|2z-1|=2cdotfrac 32=colorred3.$$






                                    share|cite|improve this answer









                                    $endgroup$



                                    Note that $:|2z-1|=2Bigl|z-frac12Bigr|$, i.e. geometrically, twice the distance between the affix of $z$ and the point $;Abigl(frac12,0bigr)$ in the plane. It is attained at the intersection point of the diameter through $A$ (i.e. the $x$ axis) which has a negative abscissa: $(-1,0)$. Thus
                                    $$sup_|2z-1|=2cdotfrac 32=colorred3.$$







                                    share|cite|improve this answer












                                    share|cite|improve this answer



                                    share|cite|improve this answer










                                    answered Jun 30 at 8:53









                                    BernardBernard

                                    129k7 gold badges43 silver badges122 bronze badges




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                                        Magento 2 “No Payment Methods” in Admin New OrderHow to integrate Paypal Express Checkout with the Magento APIMagento 1.5 - Sales > Order > edit order and shipping methods disappearAuto Invoice Check/Money Order Payment methodAdd more simple payment methods?Shipping methods not showingWhat should I do to change payment methods if changing the configuration has no effects?1.9 - No Payment Methods showing upMy Payment Methods not Showing for downloadable/virtual product when checkout?Magento2 API to access internal payment methodHow to call an existing payment methods in the registration form?