By S. Ya. Khavinson

This publication offers with difficulties of approximation of constant or bounded services of a number of variables by way of linear superposition of services which are from an identical type and feature fewer variables. the most subject is the gap of linear superpositions $D$ regarded as a subspace of the distance of constant capabilities $C(X)$ on a compact area $X$. Such houses as density of $D$ in $C(X)$, its closedness, proximality, and so forth. are studied in nice element. The method of those and different difficulties in response to duality and the Hahn-Banach theorem is emphasised. additionally, huge awareness is given to the dialogue of the Diliberto-Straus set of rules for locating the easiest approximation of a given functionality via linear superpositions.

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9i, ... ) E b if and only if 9i E b (Xi), i = 1,.... The space b is a Banach space. 35) g~ L hi(x)gi o cpi(x), 1 By the assumptions, A(b) c C(X). Let b = A- 1 (C(X)). Then, bis a closed subspace of b, and so it is a Banach space. Consider the restriction A of A onto b: b--+ C(X). 17) also hold for bounded Borel functions gi, not merely for continuous functions. - Let c be a subspace in b generated by the sequences g = (gi, ... , gi, ... ), where 9i E C (Xi)· Clearly, b :::> c. 36) remains the same on C(Xi) as on b(Xi) and equals the total variation llvill· Thus, for v =A*µ= vi II · ) , .

And hence when it is substituted into Fm it yields Fm(x) '¢ x. Thus, the function v(x) ~f vn(x) simultaneously "services" all words in Gn \Gn-i· Note that in the construction of the functions vn(x) the segment [a, b] C [O, +oo) was arbitrary. Take a sequence of segments {[an, bn]}, where 0 < an+i < bn+i < an < bn :::; 1 {bi = 1). 26) l b,. dµ = 1. a,. 27) completes the proof of the theorem. 3. There exist compact sets X, Xi, ... , Xn,... and surjective mappings 'Pi : x --+ xi, i = 1, ... 'n, ...

A COUNTEREXAMPLE 45 v (x, F2) is chosen so that all the values that appear when we apply the construction process above lie to the right of those in Mi. This can be achieved as follows. Let and let Po be the largest entry in the matrix Mi. 4) the function vo(x) = ebx has the fastest growth. 25) Yo> max ( vbnd(p), uivhnd(p), ... , vbnpl(p), uivhnd ... Upvbnpl(p)), then, starting the process for v (x, F2) with yo, we shall only deal with points that lie to the right of p. Therefore, we can only deal with functions inn (Mi), and the process will produce the function v (x, F2) that belongs ton {M2), where M2 :::>Mi.