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A functional analysis proof of titchmarsh's theorem on convolution

✍ Scribed by G.K. Kalisch


Publisher
Elsevier Science
Year
1962
Tongue
English
Weight
306 KB
Volume
5
Category
Article
ISSN
0022-247X

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✦ Synopsis


we mean as usual the space of complex-valued measurable functions defined on [0, 1] whose pth powers are integrable. By L~ [a, b] where 0 ~ a ~ b ~ 1 we shall mean here the closed subspace of L~[0, 1] consisting of functions vanishing a.e. on the complement of [a, b]. The support of a functionf defined on [0, 1] is defined as usual as the complement of the largest open subset of [0, 1] where f ----0 a.e. and we write Spt (f). The convolution off and g in LI[0 , 1] is defined as usual as

We recall that f,g E LI[0 , 1] and that convolution is associative. We write f*'~ for f, f, ..., f (n factors). We can now state the theorem mentioned in the title. TITCHMARSH'S THEOREM ON CONVOLUTION. If the functions f and g are in LI[0 , 1]; if f, g = 0 a.e. in [0, 1]; and if O ~ Spt (f) then g = 0 a.e. in [0, 1]. There are several proofs in the literature: Titchmarsh [1, 2], Crum [3], Dufresnoy [4, 5], Mikusifiski and Ryll-Nardzewski ((i): [6-9] and [I0, pp. 385-396]; (ii): [10, pp. 20-23]; (iii): [10, pp. 20-22] with [11]), Lions [12], Boas [20], Koosis [21], and Lax [22].

Our proof is entirely self-contained and is based on an interesting connection between this theorem and the linear transformation S defined on Lz[0, 1] by (Sf) (x) ----fof(y ) dy. If we write u = u(x) for the function identically 1 on [0, 1] then we can write Sf = u,f. Gelfand [13] raised the question of finding necessary and sufficient conditions in order that the linear combinations off, Sf, S% "." be dense in L~[0, 1]. Let us call a linear transformation A mapping a linear space L into itself cyclic if there exists an element f ~ L with the property that f, Af, AZf, "" and their linear combinations are dense


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