# Where can I find a quick proof that a function is analytic if and only if it admits a serie expansion?

I am currently studying functional analysis following the Aliprantis book. There, they define functions $f: O \longrightarrow X$ where $O$ is an open subset of $\mathbb{C}$ and $X$ is a Banach space. They also define the following concepts:

$f$ is analytic at a point $\lambda_0$ when $f$ is differentiable in an open subset containing $\lambda_0$

and

Every analytic function can be written as a power series

$$\sum_{n=0}^\infty (\lambda-\lambda_0)^nx_n$$

with $x_n\in X$.

My problem is that they don't prove the second statment because it's identical to the regular case (when $f$ is defined as $f:O\longrightarrow \mathbb{C}$).

Which book do you recommend to see this proof? I am not very comfortable with complex analysis and I don't know where to search.

Thank you.

• Rudin: Real and Complex Analysis. Mar 23, 2017 at 12:50

First prove that once differentiable implies infinitely differentiable, and then look at the series

$$f(x)=\sum_{n=0}^\infty\frac{f^{(n)}(a)}{n!}(x-a)^n$$

Once you know that this expression is well defined, the proof follows the proof of Taylor's Theorem. The fact that once differentiable implies infinitely can be found in any complex analysis textbook, as it is a fundamental and foundational result.

• Proving that once differentiable implies infinitely differentiable is the hard part. Mar 23, 2017 at 15:13

Le be $D(z_0,r)\subset O$. Take $z\in D(z_0,r)$ and $\rho\in(0,r)$ s.t. $z\in D(z_0,\rho)$. As $f$ is complex-differentiable, by the Cauchy integral formula:

$$2\pi i f(z) = \int_{|w-z_0|=\rho}\frac{f(w)}{w - z}\,dw.$$ But for $|z - z_0| < |w - z_0| = \rho$, $1/(w-z)$ is the sum of a convergente series: $$\frac1{w - z} = \frac1{w - z_0 - (z - z_0)} = \frac1{w - z_0}\,\frac1{1 - (z - z_0)/(w - z_0)} = \frac1{w - z_0}\,\sum_{k=0}^\infty\left(\frac{z - z_0}{w - z_0}\right)^k,$$ Now, using the uniform convergence of the series: \eqalign{2\pi i f(z) &= \int_{|w-z_0|=\rho}\left(\sum_{k=0}^\infty f(w){(z - z_0)^k\over(w - z_0)^{k+1}}\right)dw\cr &= \sum_{k=0}^\infty\left(\int_{|w-z_0|=\rho}\frac{f(w)}{(w - z_0)^{k+1}}\,dw\right)(z - z_0)^k.\cr } I.e., $f =$ a power series converging in the disk $D(z_0,\rho)$, so $C^\infty$, and by the unicity of coefficients: $${f^{(k)}(z_0)\over k!} = \frac1{2\pi i}\int_{|w-z_0|=\rho}\frac{f(w)}{(w - z_0)^{k+1}}\,dw.$$ As the series is convergent for all $\rho\in(0,r)$, the radius of convergence is at least $r$.

Suppose $f$ is holomorphic in $|z-z_0| < r$. Let $x^*$ be a continuous linear functional on $X$. Then $x^*\circ f$ is a holomorphic scalar function in $|z-z_0| < r$, and you can argue by justifying the following steps: $$x^*(f(z)) = \sum_{n=0}^{\infty}(z-z_0)^n\frac{1}{2\pi i}\oint_{|w-z_0|=s < r}\frac{1}{(w-z_0)^{n+1}}x^*(f(w))dw \\ = \sum_{n=0}^{\infty}(z-z_0)^nx^*\left(\frac{1}{2\pi i}\oint_{|w-z_0|=s < r}\frac{1}{(w-z_0)^{n+1}}f(w)dw\right) \\ = x^*\left(\sum_{n=0}^{\infty}(z-z_0)^{n}\frac{1}{2\pi i}\oint_{|w-z_0|=s < r}\frac{1}{(w-z_0)^{n+1}}f(w)dw\right)$$ The first step is the ordinary McClaurin series expansions for a scalar function. The second step is argued by approximating the integral by a Riemann sum. The third step is justified by showing that the vector sum converges because (a) $f$ is uniformly bounded on the contour, and (b) $|\frac{z-z_0}{w-z_0}| < 1$ for $s$ closed enough to $r$. Of course, you'll need the fact that $X$ is a Banach space to obtain a limit. Because this holds for all $x^*$, the conclusion is that $$f(z) = \sum_{n=0}^{\infty}(z-z_0)^{n}\frac{1}{2\pi i}\oint_{|w-z_0|=s < r}\frac{1}{(w-z_0)^{n+1}}f(w)dw, \;\;\; |z-z_0| < r.$$ This is the desired vector Laurent series.