By William F. Donoghue

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**Example text**

The elegant proof is due to L. Ggrding. + Theorem: Let be an arbitrary sequence of real numbers; then there exists a function F(x), C" on the whole axis, such that F'k'(0) = a, for all k. PROOF: equal to Let $(x) be a C"-function which vanishes for 1x1 and let 6, = k '&olakl. The function + 1 for 1x1 < + =- 1 and is 51 10. SMOOTH FUNCTIONS has the required properties. Only finitely many terms of the series are nonzero on any closed interval [c, d ] not containing the origin, since +(bkX) vanishes for 1x1 > l / b k , a quantity which converges to 0.

This is called Harnack's inequality and it enables us to establish the following remarkable theorem. 9. HARMONIC FUNCTIONS AND THE POISSON INTEGRAL 47 Theorem: Let uk(x) be a sequence of positive harmonic functions in a region G and xoa point of G such that the values U k ( x 0 ) are bounded: then there exists a subsequence uk,(x) converging uniformly on compact subsets of G to a positive, harmonic limit in G. PROOF: It is enough to show that the sequence is uniformly bounded on any compact subset of G which is connected and which contains xo.

From the Taylor formula, then, from which the Leibnitz formula follows immediately. In particular we have 11. TAYLOR'S FORMULA 55 The Taylor expansion exists in general for smooth functionsf(x) defined in a region of R". Suppose, for simplicity, that f ( x ) is C"; we'may write where gk(X) is the function j i ( 3 j / d x k ) ( x t )df. It is legitimate to differentiate under the integral sign; we find and this may be differentiated again. It is easy to see that gk(0) = (a@xk)(0), and hence or more generally where the functions g&x) are also C".