Chapter 1

Odds and Ends

1.1. Banach Spaces, Operators, and Linear Functionals

Proble m 1.1.1. / / ||^+2/|| = ||#|| + IMI for two vectors x and y in a normed

space, then show that \\ax + (3y\\ = a\\x\\ + (3\\y\\ for all scalars a, (3 0.

Solution: If a (3, then

a\\x\\ + 0\\y\\ \\ax + Py\\ = \\a(x + y) + (0 - a)y\\

a\\x + y\\-(a-0)\\y\\=a\\x\\+0\\y\\.

Hence, \\ax + (3y\\ = a\\x\\ + ;9||y|| holds for all a, /? 0. I

Proble m 1.1.2. LetT: V — W be a surjective one-to-one operator between

two vector spaces. Show that T~1 (the inverse ofT) is a linear operator.

Solution: Let S\ W — V denote the inverse of T. That is, we have TS = Iw

and ST = Iy. Let i, w G W and let A be a scalar.

For the additivity of 5 note that

T[{Sv + Sw)} = TSv + TSw = v + w = T[S{v + w)].

Since T is one-to-one, it follows that S(v-\-w) = S(v)-\-S(w). That is, S is additive.

Similarly, for the homogeneity of S observe that

T[S{Xw)} =Xw = XTSw = T[XSw],

and so S(Xw) — XS(w). Therefore, S is a linear operator. I

Proble m 1.1.3. IfT: X — y is a bounded operator between normed spaces,

then show that

\\T\\ =mm{M0: \\Tx\\ M\\x\\ for allxeX}.

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http://dx.doi.org/10.1090/gsm/051/01