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I am having trouble with the following problem and would appreciate some help. It is a piece of a proof of Cauchy's integral formula for a circle that appears in Dieudonne's Treatise on Analysis, Vol. I. I am interested in this proof because it proceeds directly from the assumption that $f$ is complex differentiable in an open neighborhood of the disk. All other proofs that I have seen take a more indirect route, often going through some form of Cauchy's theorem first.

Let $f$ be complex differentiable on an open neighbourhood of the closed unit disc. Assume that, for each given $\lambda \in (0,1)$ and $\epsilon > 0$, there are points $t_0 = 0 < t_1 < \cdots < t_m = 2\pi$, a number $\delta > 0$, and in each interval $[t_k,t_{k+1}]$ a point $\theta_k$ such that $$ | f(\zeta_k + x) - f(\zeta_k) - f^{\prime}(\zeta_k)x | \leq \epsilon |x| $$ whenever $|h| \leq \delta$ and $t_k \leq t \leq t_{k+1}$. Here $\zeta_k = z + \lambda(e^{i\theta_k} - z)$ and $\zeta_k + x = z + (\lambda+h)(e^{it} - z)$.

Define $$ g(\lambda) = \int_{0}^{2\pi} \frac{ f(z + \lambda(e^{it} - z)) - f(z)}{e^{it} - z} e^{it} dt. $$

By comparing each integral $$ A_k = \int_{t_k}^{t_{k+1}} \frac{ f( z + (\lambda + h)(e^{it} - z) ) - f( z + \lambda(e^{it} - z) ) }{e^{it} - z} e^{it} dt $$ to the corresponding expression $$ B_k = \frac{h}{i \lambda} ( f( z + \lambda(e^{i t_{k+1}} - z) ) - f( z + \lambda(e^{i t_k} - z) ) ) $$ show that, for each given $\lambda \in (0,1)$ and $\epsilon > 0$, there is a $\delta > 0$ such that $$ | g(\lambda + h) - g(\lambda) - 0 \cdot h | \leq \epsilon |h| $$ for $|h| \leq \delta$. That is, show that $g^{\prime}(\lambda) = 0$ for $\lambda \in (0,1)$.

Note $$ \sum_{k=0}^{m-1} (A_k - B_k) = g(\lambda + h) - g(\lambda) - 0 \cdot h. $$

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Are you sure that the $\epsilon$ at the end is the same $\epsilon$ as in the premise and not some multiple of it? (I looked at Problem 1 in Section IX.10 of the book; this seems to be the problem you're referring to, and to solve that problem it would suffice if the $\epsilon$ in your conclusion were some fixed multiple of the one in your premise, and that might be easier to prove.) –  joriki Mar 2 '11 at 3:30
    
That's true. The $\epsilon$ in the conclusion and the $\epsilon$ in the premise don't need to be the same. I was already aware of this, but I appreciate your suggestion. I have edited the question slightly to make it more clear that the $\epsilon$'s need not be the same. –  admchrch Mar 2 '11 at 5:10

1 Answer 1

The idea of the proof sketched in the question is to use conformality to relate the effect of normal perturbation of the circle of integration (which expands it from radius $\lambda$ to $\lambda+h$) to the tangential perturbation of said circle (which does nothing to the integral). Indeed, the complex differentiability gives $$\frac{ f( z + (\lambda + h)(e^{it} - z) ) - f( z + \lambda(e^{it} - z) ) }{e^{it} - z}e^{it} = he^{it}f'( z + \lambda(e^{it} - z) ) + o(h) \tag1$$ and $$\frac{ f( z + \lambda (e^{i(t+h)} - z) ) - f( z + \lambda(e^{it} - z) ) }{i\lambda } = he^{it}f'( z + \lambda(e^{it} - z) ) + o(h)\tag2$$ where the right-hand sides are the same. Now let $t_k=hk$, so that $t+h$ in (2) becomes $t_{k+1}$ and the sum over $k$ telescopes. The same summation in (1) gives a Riemann sum of $$ \int_0^{2\pi} \frac{ f( z + (\lambda + h)(e^{it} - z) ) - f( z + \lambda(e^{it} - z) ) }{e^{it} - z} e^{it} dt$$ which is thus shown to be $o(h)$.

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They invented the revival badge for you, +1. –  1015 Jun 23 '13 at 14:05

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