- Analysis Functional 1 (35.32 KiB) Viewed 33 times
Analysis Functional
-
answerhappygod
- Site Admin
- Posts: 899603
- Joined: Mon Aug 02, 2021 8:13 am
Analysis Functional
Analysis Functional
Exercise 5. We put E = C[0, 1] ||f||00 = = {f: [0] →R: f is continuous} sup f(x). if fe E T€[0,1 = ||f||1 1. Prove that (E, ||-||) is a normed vector space. 2. Let (Pk)k the sequence of functions defined by f(x)\da, if ƒ € E. k P(x) = Σα", x € (0, 3). n=0 (i) Prove that (Pk)k converge to a function P in the normed vector space (E. ||-||). 3. Let (fn)n the sequence of functions given by fn(t) = (3t)" for all t = [0, 1]. (i) Evaluate ||fn||1 and ||fn||- (ii) Prove that the norms |||| and ||-||1 are not equivalent on E.
Exercise 5. We put E = C[0, 1] ||f||00 = = {f: [0] →R: f is continuous} sup f(x). if fe E T€[0,1 = ||f||1 1. Prove that (E, ||-||) is a normed vector space. 2. Let (Pk)k the sequence of functions defined by f(x)\da, if ƒ € E. k P(x) = Σα", x € (0, 3). n=0 (i) Prove that (Pk)k converge to a function P in the normed vector space (E. ||-||). 3. Let (fn)n the sequence of functions given by fn(t) = (3t)" for all t = [0, 1]. (i) Evaluate ||fn||1 and ||fn||- (ii) Prove that the norms |||| and ||-||1 are not equivalent on E.