Spring 2010 - Problem 6

Cauchy's integral formula, Fubini's theorem

Let μ\mu be a finite, positive, regular Borel measure supported on a compact subset of the complex plane C\C and define the Newtonian potential of μ\mu to be

Uμ(z)=C1zwdμ(w).U_\mu\p{z} = \int_\C \frac{1}{\abs{z - w}} \,\diff\mu\p{w}.
  1. Prove that UμU_\mu exists at Lebesgue-almost all zCz \in \C and that

    KUμ(z)dxdy<\iint_K U_\mu\p{z} \,\diff{x} \,\diff{y} < \infty

    for every compact KCK \subseteq \C. Hint: Fubini.

  2. Prove that for almost every horizontal or vertical line LCL \subseteq \C, μ(L)=0\mu\p{L} = 0 and KUμ(z)ds<\int_K U_\mu\p{z} \,\diff{s} < \infty for every compact subset KLK \subseteq L where ds\diff{s} denotes Lebesgue linear measure on LL. Hint: Fubini and (1). (Here a.e. vertical line means the vertical lines through (x,0)\p{x, 0} for a.e. xRx \in \R. Likewise for horizontal lines.)

  3. Define the Cauchy potential of μ\mu to be

    Sμ(z)=C1zwdμ(w),S_\mu\p{z} = \int_\C \frac{1}{z - w} \,\diff\mu\p{w},

    which trivially exists whenever Uμ(z)<U_\mu\p{z} < \infty. Let RR be a rectangle in C\C whose four sides are contained in lines LL having the conclusion of (2). Prove that

    12πiRSμ(z)dz=μ(R).\frac{1}{2\pi i} \int_{\partial R} S_\mu\p{z} \,\diff{z} = \mu\p{R}.

    Hint: Fubini and Cauchy.

Solution.

See Spring 2010 - Problem 6.