Question

Please solve this question.

Problem 1. We denote by the set of all sequences (UK)x=1,2,... = (U1, U2, ...) (ux E C) u= satisfying luxl <00. Moreover, we define k=1 (u, v) = xox(u, v E f). k=1 (1) Prove that is a vector space. (2) Prove that is a inner product space with respect to (5.). (3) Construct the norm induced from the inner product (-). Also, prove that this norm satisfies the conditions of norm. (4) Prove that is a Hilbert space. (5) Fix n E N. We set L = {(U1, U2, ..., Un, 0,0,...): UEC (k = 1,2,...,n)}. Then, prove that L is a closed subspace of (, and write Lt explicitly.

Sorry please neglect the bottom picture

which says "moreover ...".

I am happy to upbote if you solve (1)-(5).

Answer 1

• 2 = { us (uus. buce: lex lo conoy llighet u = (h, us, ...), 82 (0, 2, ...) E d², ;, ;6C Then I tugel (& and I lovalca ha le Mant Mow. kal Z luxe tvalg 2 ( 2 1 x) 42 Man 1² * To val") <(EM/M) (MM) [Minkowski inequality] - Utv VER2 & uuel2 adi utva (mus, ...)+(...) = (1+V, Ust... --), by defn of e² =(2, tly, az tuz, ....) ui, vite Mit via v. tui 2 utu in) & let u, v, w E l with us (miles, ...) N2(V, V2, ...), W, (w,,W2,...), 4;,Q;, Wu te - u +(utw) = (mius, ...)+((, , , ...) + (W, 12,...) (uu,...) + (2, tW,,V2 + W, , - - -) (n+w, tw, uz + UtW , . .-) - (+,,+0,-) +w,,W2---)

+ altv) tw tu, u, we? i et u (ay, uz, I E d2 ; uitd. Then (unuz, ...)+(0,0,...) = (1+0, uz to, - - -) (m, uz, .--) ituito. U to and g (0, 0, -s-) E l ? .. otot... 20 (2. I 0(0,0,...) G l 2 sit. uto=u&u El² 1 cet us (m. ,.--) El 2, U: EC. thes (mics, --) + (-4,-U2, ...) -lo, o,..--) now, uel 2 ay luelea. K.21 и и 7 I turl(& Jual atual - u =(-4,-4, .-) El 2 1. for each u El 27 - UE l 2 s.it u +(-u) = 0 wil eet cec and a 2 (we, us, . -) e R2. Il cul ca at rund ca. Ra ka

al y Elcuxľ ( DC K21 -) 66e2 & CAC, ut l2. & eod.tc, ure2 ay (cm, curi: sy o cu el? vii) cet c, d ec, ufe². c(du) (da, , duz, ...) sledu, oduz, - - 2 cd (m, uz, - ~) se. z(ed)u viü et cec, u, ve d² with u2(m, ez, ~-) v2(0,, U2,- -) ui, vifc. (d)e(u + v) ze (mus,...) + (re,, 22,..)) 2e (entu, ilgtu, 10.-) acem uteulicly tarzi v 12 (cu o cuz, - ~)+(cv, CV 2,-) ceu, un,....) te (vi,w2,-2) acutev v CAC, & utd2 ix) ut c id el, a 2 (ui, , r ) El², Del², uitl & C fd)u a (e+d) (minus in ) 2(e+d) u (eta) us, - -) {cu, tdu , cuz tary 2

(cur, cur, .-) + ( day, deez, .-) clu, uz, - -) + dlu, uz, .-) 2 cut du v cid el, url2 *) tEc is identity Then for all unlu,us. .--) E lè , ui el ius 3. (mus, .--) I is identity in C, 1. ui sui o (m, un 2 (1.4, 1.4, . ~-) -) vi su . & a fl 2. e2 is a vector space. K21 e2 2) & f is defined by (u, v) - 2 by (u, v) = { un VK UUE 8 let u, ut e? ces (una 2 (ra) (iv) = Tunin u 21 EC. СИ (w, ul. K 211 ) Evicta I the 4. 22 uuuu kn C 거 Tutu =(u, v). Mill EC, UiT

net (and be any Cauchy sequence in e?, where am (s ....). Then for every Exo 7 N&t. & in, n7N d(am, an) = 11am-xull 12 = Lam-an, diman) J거 1426,697) It follows that for every j = 1, 2, have & sila? _ lukt & minen ارا۔ we choose a a fixed j, from @ we oo see that the in then is a Cauchy sequence (m) ris It converges since & is complete say . as mto. Using these termits, we define &=(2, 4, 17 and show that a E l and an x. from we have tm,nyN El & {") 1 ² < E , k21,2,.. ko 121 rething now k , we obtain for myn < 62 , k21,2, [km)

we get 21 s (m) ; 1² c. er J에 Y (m) hij El 2 completeness of e2 let kya then for m YN an-x - die since ant l2 , it follows by Minkowski inequality ant (2 km) el Furthermore the series in o represents so that implies that ex. Sin since (am) was an arbitrary Cauchy sequence in e e2, this proves inequality that that ar anam llam-xl 3 Mm ca 1² is complete inner product space inl., e2 in a Hilbert space