By Luc Tartar
After publishing an advent to the Navier–Stokes equation and oceanography (Vol. 1 of this series), Luc Tartar follows with one other set of lecture notes in response to a graduate path in elements, as indicated through the identify. A draft has been to be had on the net for many years. the writer has now revised and polished it right into a textual content obtainable to a bigger audience.
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Extra info for An Introduction to Sobolev Spaces and Interpolation Spaces
Up to a multiplication by a constant, the Lebesgue measure is the only nonzero Radon measure which is invariant by translation, so that it is uniquely deﬁned if we add the requirement that the volume of the unit cube be 1. For any locally compact3 commutative4 group there exists a nonzero Radon measure which is invariant by translation, unique up to multiplication by a constant, a Haar measure 5 of the group. For the additive group Z, a Haar measure is the counting measure; for the additive group RN , a Haar measure is the Lebesgue measure dx, and dt t is a Haar measure for the multiplicative group (0, ∞) (which is isomorphic to the additive group R by the logarithm, whose inverse is the exponential ).
5) and it is a Radon measure (and therefore a distribution). If a sequence an ∈ Ω converges to the boundary ∂Ω of Ω and cn is an arbitrary sequence, then µ = n cn δan is a Radon measure in Ω because in the formula µ, ϕ = c ϕ(a n n ), only a ﬁnite number of an belong to the compact support K n of ϕ. Physicists use the notation δ(x − a) instead of δa , and they deﬁne δ(x) as the “function” which is 0 for x = 0 and has integral 1; of course there is no such function and it is actually a measure, but after studying Radon measures and distributions one learns which formulas are right, and one can then decide quickly if a formula used by physicists can be proven easily, or if it is a questionable one, either by showing that it is false or by noticing that mathematicians do not know yet how to make sense out of the formal computations used by physicists in that particular case.
There exists C such that T, ϕ = C Ω ϕ(x) dx for all ϕ ∈ Cc∞ (Ω). Proof : By a connectedness argument it is enough to show the result with Ω replaced by any open cube Ω0 ⊂ Ω. One uses an induction on the dimension N , and one starts with the case N = 1, Ω0 being an interval (a, b). One b notices that if ϕ ∈ Cc∞ (a, b) satisﬁes a ϕ(x) dx = 0, then ϕ = dψ dx for a x ∞ function ψ ∈ Cc (a, b), and ψ is given explicitly by ψ(x) = a ϕ(t) dt. 7) with C = T, η , and that means T = C. 8) and one checks immediately that this indeed deﬁnes a distribution Tϕ on (a, b) because the bounds on derivatives of ϕ ⊗ ψ only involve a ﬁnite number of derivatives of ψ and the support of ϕ ⊗ ψ is the product of the supports of ϕ and of ψ, so that it stays in a ﬁxed compact set when the support of ψ stays in a ﬁxed compact set, ϕ being kept ﬁxed.