Integral Equations

∫1Since f(x) is an integrable function, the integralF (x) =∫ x−1−1|f(x)|p −11 (x)dx converges, andf(t)dtp 1 (t)is absolutely continuous (as a function in the form of an integral depending on the upper limitof integration), F ′ (x) = f(x)p −11 (x), and2π∫1−1The latter equalities yieldU n (x)f(x) dxp 1 (x) = 2 π∫1−1U n (x)F ′ (x)dx = − 2 π∫1−1U ′ n(x)F (x)dx.∫1−1∫1U n(x)[ϕ(x) ′ − F (x)]dx = −n−1dxT n (x)[ϕ(x) − F (x)]p 1 (x)= 0, n ≥ 1.Hence, ϕ(x) − F (x) = C = const. The explicit form of F ′ (x) yields ϕ ′ (x) = f(x)p −11 (x);therefore, ϕ ′ (x)p 1 (x) ∈ L (1)2 and ϕ ′ (x) ∈ L (2)2 .Now, let us prove the inverse statement. Let ϕ ∈ ˜W 21 and be represented by the Fourier–Chebyshev series (366). Form the Fourier serieswhereη n = 2 π∫1−1∞∑p 1 (x)ϕ ′ (x) = η n U n (x), |x| ≤ 1,n=1U n (x)(p 1 (x)ϕ ′ (x)) dxp 1 (x) = − 2 π= 2n π∫1−1∫1−1T n (x) ϕ(x)dxp 1 (x) .U ′ n(x)ϕ(x)dx =Now, in order to prove that ξ ∈ h 2 , it is sufficient to apply the Bessel inequality to η n = nξ n ,whereξ n = 2 ∫1ϕ(x)T n (x) dxπp 1 (x) .✷−120.1 Solution to integral equations with a logarithmic singularity ofthe kernelConsider the integral operator with a logarithmic singularity of the kernelLϕ = − 1 π∫1−1ln |x − s|ϕ(s) dsp 1 (s) , ϕ ∈ ˜W 1 2 ,76