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Let $f$ be an entire function that satisfies $|f(z)| \leq 1 + |z|$ for all $z\in \mathbb{C}$.
Show that $f(z) = az +b$ for fixed complex numbers $a,b$.

The hint tells us to try and use Cauchy's Integral Formula on an arbitrary circle.

This is my attempt:
Consider an arbitrary large circle with centre $0$ and let $\Upsilon$ be the contour around this circle.

Then $$\int_{\Upsilon} \frac{f(z)}{z^{n+1}} \mathrm{d}z = \frac{2\pi i}{n!}f^{(n)}(0).$$

Note that $$\int_{\Upsilon}\frac{f(z)}{z^{n+1}} \leq \int_{\Upsilon}\left|\frac{f(z)}{z^{n+1}} \right|\leq \int_{\Upsilon} \frac{1}{|z^{n+1}|} + \frac{1}{|z|^n}$$
and the right hand side is $0$ by Cauchy's Integral Formula.
I'm not sure if the first equality is valid, and not sure where to go from here.

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Here's the big missing idea: if a function's second derivative is identically zero, then the function is at most a linear polynomial. So you should show that the right hand side is zero when $n = 2$.

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  • $\begingroup$ Is the RHS always zero unless $n=0,1$? $\endgroup$ – Twenty-six colours Sep 6 '17 at 9:49
  • $\begingroup$ Yes consider the cauchy integral formula for the function identically 1. you obtain the n-th derivative of 1 to be $\frac{n!}{2\pi i}\int \frac{1}{z^{n+1}}$, hence the RHS is zero when $n>0$ and nonzero only when $n=0$. So in our case when $n=0,1$ (that number is called the winding number and says how many times the curves wraps around the point you are considering). $\endgroup$ – SC30 Sep 6 '17 at 10:07
  • $\begingroup$ I see; after showing then that the RHS is zero for $n=2$ (and any other integer after that), how do I conclude that the original LHS integral $$\int_{\Upsilon}\frac{f(z)}{z^{n=1}}$$ is identically zero for $n\geq 2$? I feel like I should show that the lower bound is also zero for this integral to conclude that it's zero. $\endgroup$ – Twenty-six colours Sep 6 '17 at 10:25
  • $\begingroup$ I missed that you were only providing an upper bound. You should bound the integral in absolute value (as is done in the other answer, which uses $R \to \infty$ to show the integral is $0$). $\endgroup$ – davidlowryduda Sep 6 '17 at 10:34
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By Cauchy integral formula around a circle centered at $0$ of radius $R$ we have $f^{(n)}(0)=\frac{n!}{2\pi i}\int \frac{f(z)}{z^{n+1}}dz$, taking modulus you have $|f^{(n)}(0)|\leq \frac{n!}{2\pi}\int |\frac{f(z)}{z^{n+1}}|d|z|=\frac{n!}{2\pi}\int \frac{|f(z)|}{R^{n+1}}d|z|\leq \frac{n!}{2\pi R^{n+1}}\int (1+R)d|z|=\frac{n!(1+R)2\pi R}{2\pi R^{n+1}}$ which tends to $0$ as $R\rightarrow \infty$ for $n\geq 2$. Since $f$ is entire we may consider the taylor series expansion centered in zero, the relations between the coefficient of the taylor series and the derivatives of $f$ let us conclude that $f(z)=a_0+a_1z$, which is what we wanted.

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  • $\begingroup$ Thank you. Why do we require that $R$ limits to $\infty$? As in, aren't we only showing that $f$ is a linear function when the circle becomes infinitely large? $\endgroup$ – Twenty-six colours Sep 6 '17 at 10:26
  • $\begingroup$ In the inequality we have that LHS doesn't depend on the radius of the circle, hence it holds even if we make the circle infinitely large, which means taking the limit. We use that to prove that the derivative in zero is zero, the fact that f is linear follows eventually. $\endgroup$ – SC30 Sep 6 '17 at 11:06
  • $\begingroup$ I see; thank you. Curious: why do you have the infinitesimals to be $\mathrm{d}|z|$ instead of $\mathrm{d}z$? $\endgroup$ – Twenty-six colours Sep 6 '17 at 11:21
  • $\begingroup$ Also, for the last equality, why does the integral $\int_{\Upsilon}dz$ evaluate to $2\pi R$? Since the integrand is entire ($g(z) = 1$), and the contour is a closed loop, shouldn't it equal to zero by Cauchy-Goursat? $\endgroup$ – Twenty-six colours Sep 6 '17 at 11:37
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Here is another approach, in case you're interested. Since $f$ is entire, we can write $f(z)=\sum_{n=0}^{\infty}a_nz^n,$ the power series converging for all $z.$ Now for any $r>0,$

$$2\pi (1+r)^2 \ge \int_0^{2\pi} |f(re^{it})|^2\,dt = \int_0^{2\pi} |\sum_{n=0}^{\infty}a_nr^ne^{int}|^2\,dt = 2\pi\sum_{n=0}^{\infty}|a_n|^2r^{2n}.$$

For this inequality to hold for all $r,$ we must have $a_n = 0$ for $n>1.$ Thus $f(z) = a_0 + a_1z$ as desired.

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