We are given Zorn’s Lemma, which is taken as an axiom of set theory: Lemma If X is a nonempty partially ordered set with the property that every totally ordered subset of X has an upper bound in X, then X has a maximal element.

Once we make wave functions by quantizing some classical system, we have states that live in Hilbert space. At this point it essentially doesn’t matter where we came from; now we’re in Hilbert space and we’ve left our classical starting point behind.

It extends the methods of vector algebra and calculus from the two-dimensional Euclidean plane and three-dimensional space to spaces with any finite or infinite number of dimensions. A Hilbert space is an abstract vector space possessing the structure of an inner product that allows length and angle to be measured.

In summary, a Hilbert space is a normed, linear space with an inner product which is complete in the norm induced by the inner product. Consider a vector space over R (or C). This can be made into an inner product space by specifying an inner product ⟨∗,∗⟩, which takes two vectors and returns an element of R (or C).

An infinite-dimensional space can have many different norms. Hilbert spaces with their norm given by the inner product are examples of Banach spaces. While a Hilbert space is always a Banach space, the converse need not hold. Therefore, it is possible for a Banach space not to have a norm given by an inner product.

A sequence {xn} in a Banach space X is weakly convergent to a vector x ∈ X if f (xn) converges to f (x) for every continuous linear functional f in the dual X ′. The Banach space X is weakly sequentially complete if every weakly Cauchy sequence is weakly convergent in X.

So every cauchy sequence converges. Hence C1([a,b]) is a Banach Space.

A normed linear space is a metric space with respect to the metric d derived from its norm, where d(x, y) = x − y. Definition 5.1 A Banach space is a normed linear space that is a complete metric space with respect to the metric derived from its norm.

for every y ∈ Y, there exists x ∈ X with Tx = y, and x ≤ Cy. Then Y is a Banach space. Corollary 2.1. Let X be a Banach space, and let Y be a closed linear subspace of X.

Every (real or complex) vector space admits a norm. Indeed, every vector space has a basis you can consider the corresponding «ℓ1» norm.

Theorem. Every real Cauchy sequence is convergent.

Cauchy sequences. = 2 M2 < ϵ. ) is a Cauchy sequence.

A sequence {an}is called a Cauchy sequence if for any given ϵ > 0, there exists N ∈ N such that n, m ≥ N =⇒ |an − am| < ϵ. |an − L| < ϵ 2 ∀ n ≥ N. Thus if n, m ≥ N, we have |an − am|≤|an − L| + |am − L| < ϵ 2 + ϵ 2 = ϵ.

If a sequence (an) is Cauchy, then it is bounded. If a subsequence of a Cauchy sequence converges to x, then the sequence itself converges to x.

Theorem 2.4: Every convergent sequence is a bounded sequence, that is the set {xn : n ∈ N} is bounded. Remark : The condition given in the previous result is necessary but not sufficient. For example, the sequence ((−1)n) is a bounded sequence but it does not converge.

A sequence is called a Cauchy sequence if the terms of the sequence eventually all become arbitrarily close to one another. That is, given ε > 0 there exists N such that if m, n > N then |am- an| < ε. Note that this definition does not mention a limit and so can be checked from knowledge about the sequence.

normed space (Rn, ·) is complete since every Cauchy sequence is bounded and every bounded sequence has a convergent subsequence with limit in Rn (the Bolzano-Weierstrass theorem). The spaces (Rn, ·1) and (Rn, ·∞) are also Banach spaces since these norms are equivalent.

So we define a sequence as a sequence an is said to converge to a number α provided that for every positive number ϵ there is a natural number N such that |an – α| < ϵ for all integers n ≥ N. For example, 1n converges to 0.

Cauchy sequences have amazing properties that can be used to understand the behavior of a system as time progresses. They are heavily used in fields like satellite design, manufacturing, construction, treatment plants, and so on.

2 Answers. Short answer (with some extra text to fill it out): Yes. On any real or complex vector space X for which a norm ‖⋅‖ is defined, part of the definition is that ‖x‖ is a real number for each x∈X.

If a normed linear space X has a complete linear subspace Y of finite codimension n in X, then X is complete, and X is naturally isomorphic (as an LCS) with Y ⊕ ℂ n . The proof of this is quite easy, and proceeds by induction in n.

The space Rn with the lp-norm is a normed vector space. Proof. The space Rn is an n-dimensional vector space, so we just need to verify the properties of the norm. x + y1 = |x1 + y1| + |x2 + y2| + ··· + |xn + yn| ≤ |x1| + |y1| + |x2| + |y2| + ··· + |xn| + |yn| ≤ x1 + y1.

In many applications, however, the metric space is a linear space with a metric derived from a norm that gives the “length” of a vector. Such spaces are called normed linear spaces. For example, n-dimensional Euclidean space is a normed linear space (after the choice of an arbitrary point as the origin).

This is an example of a metric space that is not a normed vector space: there is no way to define vector addition or scalar multiplication for a finite set.

A norm can be turned into a metric, via d (x,y) = w (x − y). This is called the induced metric.