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# A Course in Functional Analysis by John B Conway By John B Conway

This publication is an introductory textual content in useful research. in contrast to many sleek remedies, it starts with the actual and works its option to the extra basic. From the reports: "This ebook is a wonderful textual content for a primary graduate direction in useful analysis....Many fascinating and critical purposes are included....It comprises an abundance of workouts, and is written within the enticing and lucid variety which we now have come to anticipate from the author." --MATHEMATICAL studies

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Example text

Eap (T). • 4. 14. Proposition. If T is a compact operator on Yf, A. ) - 1 is a bounded operator on Yf. PRooF. Since A. )h I � c I h I for all h in Yf. )hn -+ f. )hn - (T - A)h m II and so { hn } is a Cauchy sequence. Hence hn -+ h for some h in Yf. )h = f. ) is closed and, by (2. )*] l. = Yf, by hypothesis. )h = f. )Af = f for all f in Yf. )Af I = I f 11 . So I Af I � c - 1 I f I and A is bounded. )h - h]. Since A. )h = h. )- 1 . 15. Corollary. It will be proved in a later chapter that if A. ¢ap (T) and A.

Proposition. ), f (Kf)(x) = k(x, y)f(y)dJ-L (y) is a compact opera tor and II K II � II k II 2 . The following lemma is useful for proving this proposition. The proof is left to the reader. 8. Lemma. ). If k and K are as in the preceding proposition, then ( k, c/> ii ) = ( K ei' ei ) . 7. First we show that K defines a bounded operator . (Y) ) · (J i f(y) l 2 dJJ. ( Y) ) dJJ. 8. Thus I ( Kei' ei ) 1 2 . ), there are at most a countable number of i and j such that ( k, cf> ii ) # 0; denote these by { t/Jk m: 1 � k, m < oo }.

Yt is a complete metric space, this is equivalent to showing that T(ball Je) has compact closure. Let e > 0 and choose n such that I T - Tn II < e/3. Since Tn is compact, there are vectors h 1 , , hm in ball Je such that Tn(ball Je) c U j 1 B(Tn hi ; e/3). So if II h II � 1, there is an hi with I Tn hi - Tn h II < ef3. Thus· II Thj - Th II � II Thj - Tn hj II + II Tn hj - Tn h II + II Tnh - Th II < 2 11 T - Tn II + e/3 < e. Hence T(ball Je) c U j 1 B(Thi ; e). • The proof of (c) is left to the reader.