Question: Let F be a bounded linear functional on L2 for which F|| = 1. (a) Define the hyperplane H = {fe L2: F(f) = 1}.

 Let F be a bounded linear functional on L2 for which

Let F be a bounded linear functional on L2 for which F|| = 1. (a) Define the "hyperplane" H = {fe L2: F(f) = 1}. Show that there is a unique point h E H that is nearest to the origin and that ||h|| > 1. (b) Prove that ||h|| = 1, where he H is defined in (a). Hint: According to Riesz' theorem there exists he LP such that F(f) = (h, f), for all f L. Use the fact that ||F|| = 1 to prove that ||h|| = 1. Hence, conclude that h E H and, therefore, that h = h. In particular ||h|| = |h|| = 1. Take note that we have shown here that h is the unique point on the unit sphere S = {fEL2: |f|| = 1} such that F(h) = 1. (c) Repeat (b) without using Riesz' theorem. (Hint: From (a) we know that h| > 1. We must prove that it is also true that h| (1 - Elg|| Notice that F(g) + 0 (why?) and by the linearity of F, we must have that Fg/(F9) = 1; i.e., g/(F9) EH. Now from (a) we know that Il g/(F(9)|| ||h||. Hence, complete the derivation.] Let F be a bounded linear functional on L2 for which F|| = 1. (a) Define the "hyperplane" H = {fe L2: F(f) = 1}. Show that there is a unique point h E H that is nearest to the origin and that ||h|| > 1. (b) Prove that ||h|| = 1, where he H is defined in (a). Hint: According to Riesz' theorem there exists he LP such that F(f) = (h, f), for all f L. Use the fact that ||F|| = 1 to prove that ||h|| = 1. Hence, conclude that h E H and, therefore, that h = h. In particular ||h|| = |h|| = 1. Take note that we have shown here that h is the unique point on the unit sphere S = {fEL2: |f|| = 1} such that F(h) = 1. (c) Repeat (b) without using Riesz' theorem. (Hint: From (a) we know that h| > 1. We must prove that it is also true that h| (1 - Elg|| Notice that F(g) + 0 (why?) and by the linearity of F, we must have that Fg/(F9) = 1; i.e., g/(F9) EH. Now from (a) we know that Il g/(F(9)|| ||h||. Hence, complete the derivation.]

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