Let \(F:[0,1]\to\mathbb C\) be continuous and define \[ f(z)=\int_0^1\frac{F(t)}{t-z}\,dt, \qquad z\in\mathbb C\setminus[0,1]. \]
- Prove that \(f\) is analytic on \(\mathbb C\setminus[0,1]\).
- Express the coefficients of the Laurent series of \(f\) at \(\infty\) in terms of \(F\), and use them to prove that \(F\) is uniquely determined by \(f\).
Solution
1. The function \(f\) is analytic on \(\CC\sm[0,1]\), with\[ f'(z) = \int_0^1\frac{F(t)}{(t-z)^2}\,dt. \]
Fix \(z_0\in\CC\sm[0,1]\) and let \[ d\coloneqq\operatorname{dist}(z_0,[0,1])>0. \] If \(\abs{z-z_0}<d/2\), then for every \(t\in[0,1]\), \[ \abs{t-z}\geq d/2, \qquad \abs{t-z_0}\geq d. \] For \(z\neq z_0\), \[ \frac{f(z)-f(z_0)}{z-z_0} = \int_0^1 \frac{F(t)}{(t-z)(t-z_0)} \,dt. \] As \(z\to z_0\), the integrand converges uniformly in \(t\in[0,1]\) to \[ \frac{F(t)}{(t-z_0)^2}. \] Indeed, continuity of \(F\) makes it bounded, while the two denominators are uniformly separated from zero on the indicated disk. Therefore the limit may be passed through the integral, giving \[ f'(z_0) = \int_0^1\frac{F(t)}{(t-z_0)^2}\,dt. \] Since \(z_0\) was arbitrary, \(f\) is analytic on \(\CC\sm[0,1]\).
2. For \(\abs z>1\),\[ f(z) = -\sum_{n=0}^{\infty} \left( \int_0^1 t^nF(t)\,dt \right) z^{-n-1}. \]
For \(t\in[0,1]\) and \(\abs z>1\), \[ \frac1{t-z} = -\frac1z\frac1{1-t/z} = -\sum_{n=0}^{\infty}\frac{t^n}{z^{n+1}}. \] Because \[ \abs{t/z}\leq\frac1{\abs z}<1, \] the geometric series converges uniformly for \(t\in[0,1]\). Multiplying by the bounded function \(F(t)\) and integrating term by term gives \[ \begin{aligned} f(z) &= -\sum_{n=0}^{\infty} \frac1{z^{n+1}} \int_0^1t^nF(t)\,dt. \end{aligned} \]
3. Thus the Laurent coefficient of \(z^{-n-1}\) at infinity is\[ \boxed{ a_{-n-1} = -\int_0^1t^nF(t)\,dt }, \qquad n=0,1,2,\ldots, \]and all coefficients of nonnegative powers of \(z\) vanish.
This is the coefficient comparison in the Laurent expansion from step 2.
4. If two continuous functions \(F,G:[0,1]\to\CC\) determine thesame transform \(f\), then for\[ H\coloneqq F-G \]one has\[ \int_0^1 t^nH(t)\,dt=0 \]for every \(n\geq0\).
The common transform has a unique Laurent expansion at infinity. Step 3 applied to \(F\) and \(G\) therefore gives equality of all their moments. Subtracting yields the displayed identities.
5. The identities in step 4 force \(H\equiv0\).
By linearity, step 4 implies \[ \int_0^1p(t)H(t)\,dt=0 \] for every complex polynomial \(p\).
By the Weierstrass approximation theorem, there are complex polynomials \(p_k\) converging uniformly on \([0,1]\) to the continuous function \(\overline{H}\). Hence \[ 0 = \lim_{k\to\infty} \int_0^1p_k(t)H(t)\,dt = \int_0^1\abs{H(t)}^2\,dt. \] The integrand is continuous and nonnegative, so this integral can vanish only if \(H(t)=0\) for every \(t\in[0,1]\).
6. Therefore \(F\) is uniquely determined by \(f\).
If \(F\) and \(G\) give the same \(f\), step 5 applied to \(H=F-G\) yields \(F=G\).
7. Q.E.D.
Step 1 proves analyticity, steps <1>2–<1>3 give the Laurent coefficients, and steps <1>4–<1>6 prove uniqueness.