Given the family C[0, 1] of continuous functionsƒ [0, 1] → R, and the norm||f|| :=sup [ƒ(x)].x= [0,1]Let us define a distance functionp(ƒ, g) = ||ƒ — g||, \ƒ,g € C[0, 1]Our aim is to prove that < C[0, 1], p > is a metric space.By the Extreme Value Theorem, ||ƒ|| isfinite for every continuous functionfe C[0, 1]. The distance function,p(f, g) = ||ƒ - g|| ≥ 0 for allf, g € C[0, 1]. For f = g on domain[0, 1], then p(f, g) = 0. This provesthat p is positive definite, a propertythat every metric function satisfies.O TrueO FalseRationale:

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Answered: Given the family C[0, 1] of continuous…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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2. Let [0, 1] := {ƒ : [0,1] → R : f is a bounded function on [0, 1]}. Let ||f|| sup f(x)| x= [0,1] for fel [0, 1]. Show that ( [0, 1], || - ||) is a Banach space.
Let (X, || · ||) be a Banach space, J CR be bounded and F C C1 (J, X ). If sup ||f(x)|| < , f€F, rɛJ then F is uniformly equicontinuous in BC(J, X).|
(1.2) Are the following functions metrics on X? Substantiate all your answers. (a) d(r, y) = |a2 – y2| , X = R (b) d(x, y) = |a3 – y³|, X = R (c) d(x, y) = e"-v, X = R.
17. Show that the function d is a metric whered(x, y) = |x − y| / (1 + |x + y|)
Let f be a mapping from [1,+0[ to [1, +"[ defined by f(x) = x+1/x. Then f has a unique fixed point O f-f(y)|s2|x-yl and f has no fixed point O f(x)-f(y)Is2|x-y| and f has a fixed point, None of the choices,
Use logarithmic differentiation to find the derivative of y.. 14) у %3D1 х - 3 1 A) B) (Inx In(x - 3)) X - 3 x X- 3 1 3 D) 2 x - 3 X - 3
How do I show (b)?
2. Show that, on the space of continuous functions C[0, 1] the distances defined by the norms = max| |ƒ(x)\, ||f||₁ = √| | \ƒ(1)|dt, ||f||2 = √√√²\ƒ(1)|²³dt, ||f|| = max are mutually non-equivalent.
Show that (1) => (2) and (2) => (1). (1) X is connected. (2) There is no continuous function f : X → {-1,1}, where {-1,1} is equipped with discrete topology. Note: The function need not be surjective.
Show that d(f₁g) = sup { | f(s)-g(s)/: se}} defines metric on the space B(S) a of Bounded real valued functions on a sut S.
Let (X, dx), and (Y, dy) be metric spaces, and suppose that f : X → Y is continuous on X, and X is compact. (a) Show that f(X) is compact. (b) Show that if f is bijective then f-¹ is continuous as well.
Could you teach me how to show this in detail?

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Given the family C[0, 1] of continuous functions ƒ [0, 1] → R, and the norm ||f|| = sup |f(x)]. x= [0,1] Let us define a distance function p(ƒ, g) = ||ƒ — g||, f, g € C[0, 1] - Our aim is to prove that < C[0, 1], p > is a metric space.