Normed Spaces, Completeness and Banach Spaces
A vector space with a length function, plus the question of whether every Cauchy sequence in it converges. Completeness is what lets an existence proof finish by taking a limit and knowing the limit is still in the space.
What a learner can do afterwards
- Verify the norm axioms for a proposed norm
- Give a normed space that is not complete and name the limit it is missing
- Explain why all norms agree in finite dimensions and why that argument fails in infinite dimensions
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A norm gives vectors length and obeys definiteness (only zero has norm 0), homogeneity (||cv|| = |c| ||v||) and the triangle inequality (||u+v|| <= ||u|| + ||v||). On Rn, three norms are the sum of the absolute entries, the Euclidean norm (square root of the sum of squares), and the max norm (largest absolute entry).
A normed space is complete when every Cauchy sequence converges in it. On polynomials over [0, 1], use the sup norm, the largest absolute value. The Taylor sums p_n(x) = sum_{j=0}^n x^j/j! approach e^x: their error is at most sum_{j>n} 1/j!, which tends to 0 independently of x. This uniform convergence makes p_n Cauchy in sup norm. The limit e^x is not a polynomial, so the space is incomplete.
Equivalent norms bound each other by positive constants. On R3, M = max(|x|,|y|,|z|), E = sqrt(x^2+y^2+z^2): M^2 <= E^2 <= 3M^2 gives M <= E <= sqrt(3) M. In a basis, N(v) <= sum_i |v_i|N(e_i), so N is continuous. Its positive minimum and finite maximum on the compact Euclidean unit sphere compare norms by scaling. In infinite dimensions the closed unit ball is not compact: polynomials x^n have sup norm 1 but L1 norm 1/(n+1).
A complete normed space is a Banach space. Completeness lets an existence proof finish: build a Cauchy sequence of approximations, take its limit, and the limit stays in the space. The Lp spaces from the last stop are complete for every p at least 1, so they are Banach spaces. A closed subspace of a Banach space is again a Banach space.
Decimal truncations of sqrt(3) are rational and Cauchy but have an irrational limit, so the rationals are incomplete with absolute-value distance. Normed spaces use distance ||u-v||.
Let N1 and N2 be norms with N2(v) <= C N1(v) for a fixed C > 0. For any tolerance epsilon > 0, N1(u-v) < epsilon/C gives N2(u-v) < epsilon. Thus every N1-Cauchy sequence is N2-Cauchy: choose late terms using the tolerance epsilon/C.
A norm supplies lengths, completeness keeps every limit inside the space, and Banach spaces are where approximation arguments land safely.
Original lesson, questions and flashcards authored for this dataset.
Original notes, questions and flashcards authored for this dataset.
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Where it sits
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