Exercise 2. Let X = C[0, 1], where C[0, 1] is the set of continuous functions on [0, 1], and the metric p is the sup-norm, defined as P(f, g) = sup If (x) – g(x)|. Show that (C[0, 1], p) is a metric space.

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Please Answer Exercise 2. Advanced Thankyouu?❤
Definition. A metric space (X, e) is a set X with a distance function or metric p satisfying the following
nmnertiee
Property
(4) is called the triangle inequality.
Example.
1.
Let X = R and p(x, y) = |x – yl, the usual absolute value in R. It is easy to verify that (R. p) is a metric
space. The defined p satisfies all the properties.
(a) Since the absolute value only gives value of nonnegative real numbers, then it is guaranteed that
for all x, y eR, we have 0s |x –- yl. Thus, 0s p(x,y).
(b) Note that p(x,y) = |x – yl = 0 if and only if x - y = 0 and that is if and only if x = y.
(c) For all x, y ER,
P(x, y) = |x – yl = l(-1)(y – x)| = |(-1)lly – x| = ly – x| = p(y, x).
(d) Recall that the triangle inequality for real numbers states that if a, bER, then la + b| < Ja| + |bl.
For all real numbers x, y, z, we have,
Study Guide in_ Real Analysis
FM-AA-CIA-15 Rev. 0 03-June-2020
Math Elective 2 (Real Analysis)
Module 2: Abstract Integration
|x - yl = |(x – 2) + (z – y)| s lx – z| + |z –- yl.
Thus, p(x, y) < p(x, z) + p(z. y).
This is the Euclidean metric on R.
2. Consider X to be a non-empty set and define
p(x. y) = {6 fx =y
We check if the conditions for p to be a metric are satisfied.
(a) Since the possible values of p are only 0 and 1, thus, 0s p(x, y).
(b) The definition of p gives us that p(x, y) = 0 if and only if x = y.
(c) Note that if x * y then y + x and similarly, if x = y then y = x. Thus, p(x, y) = p(y,x).
(d) To prove triangle inequality for p, we have to consider
the cases of equality of any real
numbers x, y. z.
(i)
Let x = y = z. Then, p(x, y) = p(x, z) = p(z, y) = 0. Thus, the inequality
p(x, y) S p(x,z) + p(z, y)
Transcribed Image Text:Definition. A metric space (X, e) is a set X with a distance function or metric p satisfying the following nmnertiee Property (4) is called the triangle inequality. Example. 1. Let X = R and p(x, y) = |x – yl, the usual absolute value in R. It is easy to verify that (R. p) is a metric space. The defined p satisfies all the properties. (a) Since the absolute value only gives value of nonnegative real numbers, then it is guaranteed that for all x, y eR, we have 0s |x –- yl. Thus, 0s p(x,y). (b) Note that p(x,y) = |x – yl = 0 if and only if x - y = 0 and that is if and only if x = y. (c) For all x, y ER, P(x, y) = |x – yl = l(-1)(y – x)| = |(-1)lly – x| = ly – x| = p(y, x). (d) Recall that the triangle inequality for real numbers states that if a, bER, then la + b| < Ja| + |bl. For all real numbers x, y, z, we have, Study Guide in_ Real Analysis FM-AA-CIA-15 Rev. 0 03-June-2020 Math Elective 2 (Real Analysis) Module 2: Abstract Integration |x - yl = |(x – 2) + (z – y)| s lx – z| + |z –- yl. Thus, p(x, y) < p(x, z) + p(z. y). This is the Euclidean metric on R. 2. Consider X to be a non-empty set and define p(x. y) = {6 fx =y We check if the conditions for p to be a metric are satisfied. (a) Since the possible values of p are only 0 and 1, thus, 0s p(x, y). (b) The definition of p gives us that p(x, y) = 0 if and only if x = y. (c) Note that if x * y then y + x and similarly, if x = y then y = x. Thus, p(x, y) = p(y,x). (d) To prove triangle inequality for p, we have to consider the cases of equality of any real numbers x, y. z. (i) Let x = y = z. Then, p(x, y) = p(x, z) = p(z, y) = 0. Thus, the inequality p(x, y) S p(x,z) + p(z, y)
(i)
Let x = y = z. Then, p(x, y) = p(x, z) = p(z, y) = 0. Thus, the inequality
p(x, y) < p(x, z) + p(z. y)
holds.
(ii)
Let x * y + z. Then, p(x,y) = p(x, z) = p(z,y) = 1 and
p(x, z) + p(z. y) = 1 +1 = 2.
Thus, p(x,y) s p(x, z) + p(z, y).
(ii)
Let x + y and y = z. Then, p(x, y) = p(x, z) = 1, p(z, y) = 0, and
p(x, z) + p(z, y) = 1+ 0 = 1.
Thus, p(x, y) s p(x, z) + p(z, y).
(iv)
Let x = y and y # z. Then, p(x, y) = 0, p(x,z) = p(z, y) = 1 and
p(x, z) + p(z, y) = 1 +1 = 2.
PANGASINAN STATE UNIVERSITY
3
Study Guide inReal Analysis
FM-AA-CIA-15 Rev. 0 03-June-2020
Math Elective 2 (Real Analysis)
Module 2: Abstract Integration
Thus, p(x, y) s p(x, z) + p(z, y).
Hence, (X, p) is a metric space.
Exercise 2. Let X = C[0, 1], where C[0, 1] is the set of continuous functions on [0,1], and the metric p is the
sup-norm, defined as
P(f, g) = sup If (x) – g(x)|.
Show that (C[0, 1], p) is a metric space.
Transcribed Image Text:(i) Let x = y = z. Then, p(x, y) = p(x, z) = p(z, y) = 0. Thus, the inequality p(x, y) < p(x, z) + p(z. y) holds. (ii) Let x * y + z. Then, p(x,y) = p(x, z) = p(z,y) = 1 and p(x, z) + p(z. y) = 1 +1 = 2. Thus, p(x,y) s p(x, z) + p(z, y). (ii) Let x + y and y = z. Then, p(x, y) = p(x, z) = 1, p(z, y) = 0, and p(x, z) + p(z, y) = 1+ 0 = 1. Thus, p(x, y) s p(x, z) + p(z, y). (iv) Let x = y and y # z. Then, p(x, y) = 0, p(x,z) = p(z, y) = 1 and p(x, z) + p(z, y) = 1 +1 = 2. PANGASINAN STATE UNIVERSITY 3 Study Guide inReal Analysis FM-AA-CIA-15 Rev. 0 03-June-2020 Math Elective 2 (Real Analysis) Module 2: Abstract Integration Thus, p(x, y) s p(x, z) + p(z, y). Hence, (X, p) is a metric space. Exercise 2. Let X = C[0, 1], where C[0, 1] is the set of continuous functions on [0,1], and the metric p is the sup-norm, defined as P(f, g) = sup If (x) – g(x)|. Show that (C[0, 1], p) is a metric space.
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