$\lvert c_n \rvert = \frac {\max_{\lvert z \rvert = r} \vert f(z) \rvert} {r^n} (0 \lt r \lt R)$ then, $f(z)$ is of this form $f(z) = c_n*z^n$.

Question

$f(z) = \sum_{n = 0}^\infty c_nz^n$, $ \lvert z \rvert \lt R$. If at least one $c_n$ satisfies the following equation: $$\lvert c_n \rvert = \frac {\max_{\lvert z \rvert = r} \vert f(z) \rvert} {r^n} (0 \lt r \lt R)$$

then, $f(z)$ is of this form $f(z) = c_n*z^n$.

From Cauchy's integral formula we know $\lvert c_n \rvert \le \frac {\max_{\lvert z \rvert = r} \vert f(z) \rvert} {r^n}$, the only case in which the inequality becomes equality I think is when $\lvert f(z) \rvert$ is constant along $\lvert z \rvert = r$. Then I tried to deduce the form from here, but failed.

@Hongyan Did you read the answer ? Your linked question isn't correct, so don't try to reuse it here !!! — reuns, Oct 03 '17 at 08:29
Here the solution is to multiply by $e^{i \theta}$ to obtain $c_n =\Re(\frac{1}{2i\pi} \int_{|z| = r} \frac{f(z)}{z^{n+1}}dz)$ — reuns, Oct 03 '17 at 08:30
Of course I read the answer, so I said "failed". But I didn't get your solution, can you explain it a little more? — Hongyan, Oct 03 '17 at 08:43
You said $|c_n| = \frac{|f(z)|}{|z^{n}|}$ for $|z| = r$. I added another condition. — reuns, Oct 03 '17 at 09:14
@Hongyan Do you mean that for at least one $n$ the property $|c_n|=\frac{\mathrm{max}_{|z|=r}|f(z)|}{r^n}$ holds for all $0<r<R$ or just for one $r$ such that $0<r<R$. If you mean the first, I think I know a solution. For the later, I am not sure if the conclusion is true. — Christoph Mark, Oct 03 '17 at 09:45

score 1 · Accepted Answer · answered Oct 03 '17 at 10:10

OK, you use the Cauchy inequality for the coefficients $0\leq m<n$, and find that $$ |c_m|\leq\frac{\mathrm{max}_{|z|=r}|f(z)|}{r^m}=\frac{\mathrm{max}_{|z|=r}|f(z)|}{r^n}\cdot r^{n-m}=|c_n|r^{n-m} $$ for all $0<r<R$. Since $n-m>0$, if you let $r\to 0$, you find that $c_m=0$ for all $0\leq m<n$. In other words, the singularity of $g(z)=f(z)/z^n$ at $0$ is removable, and you have defined a holomorphic function $g$ on $\{|z|<r\}$. We cleary have $\sup_{|z|<r}|g(z)|=|c_n|$. On the other hand, we also have $$ g(z)=c_n+c_{n+1}z+\cdots\text{ and thus }g(0)=c_n\,. $$ This means that $g$ obtains its maximum. By the maximum modulus principle, $g$ must be constant equalt to $c_n$ everywhere on $\{|z|<r\}$. It follows that $f(z)=c_nz^n$ everywhere on $\{|z|<r\}$ and by the identity theorem everywhere on $\{|z|<R\}$.

$\lvert c_n \rvert = \frac {\max_{\lvert z \rvert = r} \vert f(z) \rvert} {r^n} (0 \lt r \lt R)$ then, $f(z)$ is of this form $f(z) = c_n*z^n$.

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