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By Tao T.

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10. Since not every function in Lp is a simple function with finite measure support, we thus see that the space of simple functions with finite measure support with the Lp norm is an example of a normed vector space which is not complete. 8. Show that the space of simple functions (not necessarily with finite measure support) is a dense subspace of L∞ . Is the same true if one reinstates the finite measure support restriction? 9. e. countably generated). e. has a countable dense subset) for all 1 ≤ p < ∞.

We often want to distinguish “large” functions in V from “small” ones, especially in analysis, in which “small” terms in an expression are routinely discarded or deemed to be acceptable errors. One way to do this is to assign a magnitude or norm f V to each function that measures its size. Unlike the situation with scalars, where there is basically a single notion of magnitude, functions have a wide variety of useful notions of size, each measuring a different aspect (or combination of aspects) of the function, such as height, width, oscillation, regularity, decay, and so forth.

If p ≥ 1, then we can take C = 1 (this fact is also known as Minkowski’s inequality). Proof. The claims (i), (ii) are obvious. 16) and is left as an exercise. 16). By the non-degeneracy property we may take f Lp and g Lp to be non-zero. Using the homogeneity, we can normalise f Lp + g Lp to equal 1, thus (by homogeneity again) we can write f = (1 − θ)F and g = θG for some 0 < θ < 1 and F, G ∈ Lp with F Lp = G Lp = 1. Our task is now to show that |(1 − θ)F (x) + θG(x)|p dµ ≤ 1. 20) |(1 − θ)F (x) + θG(x)|p ≤ (1 − θ)|F (x)|p + θ|G(x)|p .

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An epsilon of room: pages from year three of a mathematical blog by Tao T.

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