Since, F is bounded both in LOR) and in lIR), it is Latnded in LLIR) forall fe[1,0) (Prdtem 31).
Since, F is bounded both in LOR) and in lIR), it is Latnded in LLIR) forall fe[1,0) (Prdtem 31).
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Transcribed Image Text:the extension oPerator of Theore m 8.6
Therefore Fis bounded in i(R)and also in
Since F is bounded both in LOR) and in LilIR), it is
Solve problem number 31
we Prove that H has a CompPact closure in.
Proof of (7)
Let
ct H be the unit ball in wTd: Let P be
he extension oPerator of Theore m 8.6
and set F=P(H),so that H=F;"
LE(I) (for atl 1<9 c00) bj apPlying Theorem 4-26-
1,1
Clearly, F is bounded in w(R), because
1,1
W CR)
CH, C, for some Ć
1,1
W (R)
(I) m.
La (I) by using (5).
Lounded in LLIR) forall ge[1,0) (Prdtem 31).
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