By L. P. Lebedev, I. I. Vorovich, G. M. L. Gladwell (auth.)

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E. 2) to integrals: (In If(x) + g(x)IPdQ) lip::; (In If(x)IPdQ) lip + (In Ig(x)IPdQ) liP. 3) The space lJ'(Q) is the completion of GJ(Q). 1), this means that elements in LP(Q) are equivalence classes of Cauchy sequences of continuous functions. nlfn(x)-fm(X)IPdQ)l/P -+Oasm,n-+oo, and that two sequences {fn(x)} and {gn(x)} are equivalent if Let us examine elements in LP(Q). First, we say loosely that if f(x) E GJ(Q), then f(x) E LP(Q), but this is not strictly accurate. The elements of lJ'(Q) are not functions, but equivalence classes of Cauchy sequences of functions.

Suppose this is false then, given any c> 0 we i=l can find a combination N N Lib;! = with Y = LbiXi i=l 1 i=l such that lIyll < c. This means that we can find a sequence {y(m)} of the form N y(m) N = L bim)xi with i=l L Ib~m)1 = 1 i=l such that lim lIy(m) II = O. m-+oo N Since L Ibi m)1= 1 we have Ib~m)1 S 1 for i = 1,2,···, N. We consider the se- i=l quence bP), b~2), b~3), ... ; this is an infinite sequence in the closed and bounded set Ib~m)1 S 1. Since this set is compact, the sequence {b~m)} contains a subsequence converging to a number dl such that Id11 S 1.

But A (B) can be zero only if all the ar. ) are zero. 1) are zero. 1 Show that equality holds in Holder's inequality iff there is a constant M such that k = 1,2, ... ,n. 10 Some inequalities > 1, then Proof. The case p = 1 is trivial. 2 Show that if 1 ::; p < 00, satisfies the conditions for a norm in IRN. 3 Show that if 1::; p < 00, then provides a suitable norm in the space lP of all sequences {xd with 00 L: IXilP < 00. 4 Verify that IRN and IP are complete with the respective norms. 7), namely 00 Xi = L: j=l aijXj + Ci 00.