# Show that $\lim_{\varepsilon\to 0}I_\varepsilon = 0$

Consider $$I_\varepsilon :=\oint_{C_\varepsilon} z^αf(z)\,dz,$$ where $\alpha>−1$ is real, where $C_\varepsilon$ is a circle of radius $\varepsilon$ centered at the origin and $f(z)$ is analytic inside the circle.

Show that $\lim_{\varepsilon\to 0} I_\varepsilon =0$.

My attempt:

Since $f$ is analytic on the disk, $|f|$ is bounded---and indeed takes its maximum value on the circle (and not in the interior unless it is constant).

Note that using the parametrization $z = \varepsilon e^{i\theta}$, you get $$\oint_{C_\varepsilon} \frac{dz}{z} = 2\pi i.$$

Okay, so consider writing $z^\alpha = z^{1 + \alpha}/z$. Since $\alpha > -1$, $1 + \alpha > 0$. For fixed $\varepsilon > 0$, let $$M(\varepsilon) = \max\{|f(z)| : |z| = \varepsilon\}.$$

By the maximum modulus theorem, $M(\varepsilon) ≤ M(\delta)$ whenever $\varepsilon \leq \delta$ ---that is, $M$ is non-decreasing. Now, observe that (all integrals are over $C_\varepsilon$) $$\left|\oint z^\alpha f(z)\,dz\right|\leq \oint|z^{1+\alpha}f(z)|\,\frac{d|z|}{|z|}\leq \varepsilon^{(1+\alpha)}M(\varepsilon)2\pi.$$

Since $2\pi M(\varepsilon)$ is bounded and $1 + \alpha > 0$, you can take the limit as $\varepsilon\to 0$ and obtain the desired result.

My question Could it be proved without using the maximum modulus theorem?

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Have you done Cauchy integral formula? – user17762 May 10 '12 at 0:29
I suggest you learn TeX markup. It will make your questions much easier to read. – Alex Becker May 10 '12 at 0:34
Still do not see the Cauchy's integral formula – Daniela del Carmen May 10 '12 at 0:35

Maximum Modulus Theorem is not needed. It's just a straightforward estimate. All you need about $f$ is that it is bounded on some neighbourhood of $0$, say $|f(z)| \le B$ for $|z| \le r$. Then $|I_\epsilon| \le 2 \pi B \epsilon^{1+\alpha}$, so if $1 + \alpha > 0$, ...