By George Salmon

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**Extra info for A treatise on the analytic geometry of three dimensions**

**Sample text**

We see that 19K(z)I ~~ X, 3z > 0}. However, I gK(z) 15 M in that quadrant, so, by the first theorem of this §,19x(2) I < s throughout it. Let then ¶Rz > X and rjz > 0. We have If(z)I = z + iK I eB3z19K(z) I < z z+iKl eB z s. Suppose that 0 <, 3z < L. Then, if 1z > max (X, K) we have, by the previous relation, I f (z) I < 2eBL s. Here, s > 0 is arbitrary. Therefore f (x + iy) -,. 0 uniformly for 0 < y S L as x ->oo. We are done. D. The Paley-Wiener theorem Theorem. ~~

Assume that, for each e8-9, limsupl f(z)I < m. Then I f(z)I 5 m in -9. { ZE1J Remark. If -9 is a bounded domain, this is the ordinary maximum principle, and then the assumption that f (z) is bounded in -9 is superfluous. When -9 is unbounded, however, this assumption is really necessary, as the simplest examples show. Pick any n > 0 and fix it. According to the hypothesis, we can find a p > 0 such that I f (z) I < m + n for ze2? and I z I < p; we fix such a p and write Proof of theorem. 9 n{Izl>p}.

Let f(z) be regular in 3z > 0 and continuous up to the real axis. Suppose that log I f (z) I < O(I z I) for I z I large when 3z > 0, that limsup log If (iy) I = A, y and that -00 00 log, If W1 dx 1+x2 < oo. 3zlog+If(t)I dt. 3z/I z - tI2 = 91(i/(z - t)) is, for each tell, a positive and harmonic function in 3z > 0. For fixed z with positive real part we have, by calculus, f_' 1 7r 3z Z IZ dt =1, and, if z --+ xo a I8, z sup Ii-xol_a - 1 IZ - t1 2 t2 + 1 0 for each S > 0. Iz-t1 P(t)dt P(xo) for In our present situation P(t) = log+ If (t) I is continuous on R, so if we put UO z = I 00 zlog+If(t)I Iz-t1 2 dt for ,rjz > 0, U(z) is positive and harmonic in the upper half plane and Positive harmonic functions-representation as Poisson integrals 39 U(z) log' If (xo)I for z xoaR.