By E. Combet

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**Example text**

Yn . e. , qth power summable, where p and q are conjugate to each other, then 1/p ∞ |xi + yi |p 1/p ∞ ≤ |xi |p i=1 1/p ∞ |yi |p + i=1 . i=1 (M3) If x(t) and y(t) belong to Lp (0, 1), then 1 0 1/p |x(t) + y(t)p dt 1 ≤ 0 1/p |x(t)p dt 1 + 0 1/p |y(t)|p dt . Proof: If p = 1 and p = ∞ the (M1) is easily seen to be true. Suppose 1 < p < ∞, then 1/p n |xi + yi | 1/p n ≤ p i=1 (|xi | + |yi |) p . 5) i=1 Moreover, (|xi | + |yi |)p = (xi | + |yi |)p−1 |xi | + (|xi | + |yi |)p−1 |yi |. Summing these identities for i = 1, 2, .

Thus for each > 0, there (n) (m) is an N such that for m, n > N , we have sup |ξi − ξi | < . It follows (n) i (n) that for each i, {ξi } is a Cauchy sequence. Let ξi = lim ξi n→∞ and let (n) ξi | x = {ξ1 , ξ2 , . }. Now for each i and n > N , it follows that |ξi − < . (n) (n) (n) Therefore, |ξi | ≤ |ξi | + |ξi − ξi | ≤ |ξi | + for n > N . , x ∈ l∞ and {ξi } converges to x in the l∞ norm. Hence, l∞ is complete under the metric deﬁned for l∞ . Problems 1. Show that in a metric space an ‘open ball’ is an open set and a ‘closed ball’ is a closed set.

For (ii) Let us suppose that g ∈ Γ. Let > 0 and let us partition [0, 1] into k equal intervals such that if x, x are in the same interval of the partitioning, |g(x) − g(x )| < /2 holds. Let us consider the ith subinterval i−1 ≤ x ≤ ki and consider the rectangle with sides g i−1 and g ki . For all k k − 2 ≤ y ≤ g k + 2. points within the rectangle the ordinates satisfy g i−1 k i−1 Thus ki , g ki is on the right-hand side of the rectangle and i−1 k ,g k is on the left-hand side of the rectangle. By joining these two points by a polygonal graph that remains within the rectangle and the line segments of which have slopes exceeding n in absolute value, we thus obtain a continuous function that is within of g and as because its slope exceeds n, it belongs to Γ ∼ Kn .