Question

Topology

C O, 1 and be the supremum norm (a) Prove that (X || |) is a Banach space. You can assume that (X, | |) is a normed vector space (over R) |f|0supE0.1 \5(x)|.| 4. Let X C (b) Show that || |o0 that the parallelogram identity fails.] on X is not induced by any inner product. Hint: Check for all E[0, 1]. Show that {gn}n>1 (0, 1] BI= {gE X |9||<1} is a compact (c) For every 2 1, define gn(x) converges pointwise on [0,1], but does not converge uniformly оn (d) Hence, determine whether or not subset of (X, 0). Justify your answer (е) Define T : X —Xbу Т(и) (t) u(s) ds for all t e [0, 1] and u E X. (i) Let T2u T(Tu). Using that T2(u)(t) - s) u(s) ds, deduce that |T2u T2||0 < (1/2)||u Fix f€ X. Define V : X —> X by W(u) — f +T (и) for u € X. (A) Using (e) (i), show that V2 is a contraction - v|| for all u, v E X. 11 (B) Hence, show that (u) has unique solution u E X. 21 =

Answer 1

gup 1fiæ)) 2E to, 11 X Co, 13 , (a) 5pce e ongy su (x,-e)p mortmmed vecto Compsele Patove hak it % (x _We bea caeehy seauence in LeL Given o Aa Matwra&IuM bet k So arc gup 37 | (x)uf - (*) F v e to 13..--) Ond Caeehy Seavence IR ince R | Sn (x) - $(x)|-> o n Ca)- (x) in iR taKing ixed and twe et , ) n (x)<E V m m> k gmd | In x)- Sx)| <e ncE to,1 3 -UFII n conveacz to f This ho ug thar uni. ince n Continusyg , S0 eaeh Henee Bana eh Space A bmpRe-te noam QinerSpace g caed Banach Space).

F(x) - o 1-, 5ae ci So Eo, 1, W = Sup (fCx)1 - eto,11 Nouo 1. , (S-9J( Sup1+e) (x) xeto, So 1 Sup iB-F doe motrai5ypave Hcnce Raw. veateeogoeam indicced doeg nob nnetpcodcct 2n-ac 1 In- 2 uniforr måy Nousi -Uhen the point oe Ramit herre disConti n uovs at J pomt does nol on verge te unifarmim on Eo,1