Basic Real Analysis by Anthony W. Knapp

By Anthony W. Knapp

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Sample text

Let bij be real numbers for i > 1 and j > 1. Suppose that (i) Li = limj bij exists in R uniformly in i , and (ii) L,; = limi bij exists in R for each j . , for each t > 0, there exist io and jo such that lbij - LI ir whenever i 1io and j jo, (e) Li = limi bij exists in R uniformly in j. REMARK. In applications we shall sometimes have additional information, typically the validity of (a) or (b). According to the statement of the proposition, however, the conclusions are valid without taking this extra information as an additional hypothesis.

As soon as k is large enough so that Ix,, - xo I < 6, we have Thus 1 + M,, is a uniform bound for the functions f,. 3. Uniform Convergence 23 The proof of uniform equicontinuity proceeds in the same spirit but takes longer to write out. Fix t > 0. The uniform continuity of f, for each n means that it makes sense to define I f ( x ) - f ( y ) I < twhcncvcr Ix-yl < S f and x and y are in the domain of f If Ix - y 1 < 6, ( t ) ,then If, ( x ) - f, ( y )1 < t . Put 6 ( t )= inf, 6, ( t ). Let us see that it is enough to prove that S ( t )> O: If x and y are in la, 61 with Ix - y I < S ( t ), then Ix - y 1 < S ( t ) 5 S,(t).

This proves (a). For (b), consider any subinterval [xipl, xi] of a partition, and let mi and Mi be the infimum and supremum of f on this subinterval. Also, let mi and Mi be the infimum and supremum of cf an this subinterval. If c > 0, then Mi/ = cMi and mi = cmi ,so that ( M : - m:)Axi = c(Mi - mi)Axi. If c 5 0 , then Mi = -cmi andmi = -cMi, so thatwe still have (M,'-mi)Axi = c(Mi -mi)Axi. 2%. For (c), we have f l d x 5 U ( P , f l ) 5 U ( P , f2) for all P. Taking the - - zb infimum on P in the inequality of the first and third members gives -b laf 2 d x .

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