By Herbert K. Weiss

This document includes en research of wartime files of assaults through German opponents on US Bombers, and of wrestle among US and German opponents.

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Where η = η(r) is the unique solution of cos x = 2r/(1 + r2 ) on the interval (0, π/2). v. 2π −π g(x − t) dt = f (x), tan t/2 29 a. e. Then he remarked that n Sn (f, x) = aj cos jx + bj sin jx j=1 1 = 2π π g(x + t) −π 1 cos(n + 1/2)t − dt. tan t/2 sin t/2 Therefore lim Sn (f, x) = f (x) a. v. n→∞ g(x + t) −π cos nt dt = 0, t a. e. v. g(x + t) dt t exists a. e. 8) He noticed that this is not due to the smallness of the integrand. In fact, he knew that there exists a continuous function f and a set of positive measure A so that f (x + t) − f (x − t) dt = +∞ t for x ∈ A.

Take α = 1/ε and f − ϕ p < ε2 . We obtain m{Ωf (x) > ε} ≤ C(1/2ε). It follows easily that m{Ωf (x) > 0} = 0. 5 Summability As we have said, Du Bois Reymond constructed a continuous function whose Fourier series diverges at some point. Lipot Fej´er proved, when he was 19 years old, that in spite of this we can recover a continuous function from its Fourier series. Recall that if a sequence converges, there converges also, and to the same limit, the series formed by the arithmetic means of his terms.

Then the uniform boundedness of the norm of Sn follows if R can be extended to a continuous operator on Lp [−π, π]. 6 The conjugate function 25 The operator R is related to the conjugate harmonic function. Consider a power series (an + ibn )z n . n>0 Its real and imaginary part for z = eit are (an cos nt − bn sin nt); u= v= n>0 (an sin nt + bn cos nt). n>0 We say that v = u is the conjugate series to u. The operator H that sends u to v must satisfy H(cos nt) = sin nt; H(sin nt) = − cos nt. It is the same to say H(eint ) = −i sgn(n)eint .