2. Let X be a normed space. For each element (x, y) = XX, put || (x, y)||₁ == ||x|| + ||y||and || (x, y) || = max(||||, ||y||). Show that || ||1 and || are equivalent norms onXOX..2Show that if (a) is acoont seguonoo in Ithen it is also

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Answered: 2. Let X be a normed space. For each…

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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1. Let V be an inner product space. Suppose f,g E V such that f - g Lg (a) Prove that 1l12= ||f- g|2 + ||9||2.
Q5. Let (X, B(R), µ) be a Borel measure space. Suppose that {A, = ["+1, 4n=1]} be a family of elements of B(R). Show that A1 C A, C A3.. C A, C..,then evaluate u(U"1, 4n-1). =1l
Q9 Let V be an inner product space and T:V → V be a linear map. Suppose ||T (x) || = ||r||. Prove that T is one-to-one.
Use the inner product = f(x)g(x)dx in the vector space Cº [0, 1] to find , ||f||, ||g||, and the angle Of.g between f(x) and g(x) for f(x) = -5x² - 6 and g(x) = −9x + 5. (f,g) ||g|| Of.g radians.
3. Let x ER", A, B E Rnxn and || || be some vector norm (same notation is used for the corresponding operator norm). Show that || Ax|| ≤ ||A||||||, ||AB|| ≤ ||A|||| B|| and ||I|| = 1. Show by counterexample that ||- ||max and ||- ||F are not operator norms. ||AB|| 2 Hint: Find matrices A, B € R²×2 such that some of the properties in (1) do not hold. Note that ||I|| max = 1. In addition, the Frobenius norm satisfies the second inequality in (1): let AT = [a₁,..., an] and B = [b₁, ...,b], so that n n n n ·Σ (a{b;)² ≤ Σ ||a₁||²||6,;||³² = Σ || a;||²Σ||6; ij=1 i,j=1 i=1 j=1 n n = Σ |aki|² Σ |b₁j|² = || A||²|| B||²| i,k=1 j,l=1 (1) where the Cauchy-Schwarz inequality was used. 1/2
17. Show that the function d is a metric whered(x, y) = |x − y| / (1 + |x + y|)
Let X, Y be normed linear spaces. On X x Y define || (x,y) |l, = || x ||x + || y lly and I| (x,y) || . = max (||x||x<||y|ly). Prove that each defines a norm on X x Y and that they are equivalent.
(CS) is called the Cauchy-Schwarz inequality and (PI) is called the parallelogram identity.
Let ||. ||₁ and ||- |l2 be norms on vector space v if ||X→x||₁ → || x₂- Prove that ||. ||₁ and II. ||2 are equivalent x||2
3. Show the following result: Theorem. Let (X,d) be a metric space, A C X, and let u: A → R be an L-Lipschitz function. Then there is an L-Lipschitz extension.
9.
Prove that the map || · || : L(V, W) → R defined by {I > ||x|| | ||(x)7||} dns = is a norm on L(V, W).

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2. Let X be a normed space. For each element (x, y) = XX, put ||(x, y) || 1 max(x, y). Show that || ||1 and || || and ||(x, y) || : ΧΘ Χ. 2 =: Show that if (a) is convergo coquongo in 0 =: then it is also are equivalent norms on |||| + ||x || nuorgo OBUOLI903