By A.K.Sharma

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**Extra info for Elementary Integral Calculus**

**Example text**

Suppose for example that X = R and let us put y = x − h. 3) |h| ≤ r =⇒ |f (x − h) − f (x)| ≤ r for every x ∈ R. 4) fh (x) = f (x − h) of f whose graphs are derived from the graph of f by horizontal translations. This said, the fact that d[fh (x), f (x)] ≤ r for every x means simply, in the notation of Chap. 5) dR (f, fh ) = f − fh R ≤ r. The existence, for every r > 0, of an r > 0 satisfying (3) thus means that as h tends to 0 the function fh (x) converges to f (x) uniformly on R. One would like to formulate uniform continuity on an arbitrary set X in a similar way, but in this case the function fh (x) is deﬁned only on the set = X formed from X by the horizontal translation of amplitude h, and convergence, uniform or not, no longer has a meaning.

E1 (x) + e−1 (x)] /2 − [e3 (x)/3 + e−3 (x)/3] /2 + . . 9) = −π/4 for 1/4 < |x| < 3/4, and by periodicity for the other values of x. 8), 1/4 ap = −1/4 = = = e−2πipx dx − 3/4 e−2πipx dx = 1/4 e−3πip/2 − e−πip/2 e−πip/2 − eπip/2 − = −2πip −2πip eπip/2 − e−πip/2 /2πip − e−πip eπip/2 − e−πip/2 /2πip = [1 − (−1)p ] sin(pπ/2)/πp, zero if p is even, and equal to 2(−1)(p−1)/2 /πp if p is odd; since we omitted a factor π/4, we ﬁnally have ap = 0 (p even) or (−1)(p−1)/2 /2p (p odd), which agrees with (8).

Let us choose an xk ∈ K in each of those of these balls Bk which actually intersect K. Since Bk is of radius r/2, so of diameter r, we have Bk ⊂ B(xk , r), so that the B(xk , r) cover K as desired. To prove the existence of F , it thus suﬃces to show that there exists a number r > 0 possessing the following property: (∗) for every x ∈ K the open ball B(x, r) is contained in one of the Ui . If this is so, then it is enough to choose a Ui containing B(xk , r) for each k to obtain the ﬁrst assertion of the theorem.