7
$\begingroup$

I am working on a problem stating that

Find all entire functions $f$ that satisfy:

$|zf(z)-\sin z|\leq 1+|z|^{4/3}$ for all $z\in\mathbb{C}$.

I am stuck in this problem but I had some attempts:

(1):

Since we are dealing with an entire function, we want to apply Liouville’s theorem. However, the RHS is not a constant number.

So we consider a disc $D(0, R)$ for $R$ large enough. Then for $z\in D(0,R)$, we have $$|zf(z)-1|\leq |zf(z)-\sin z|\leq 1+|z|^{4/3}\leq 1+R^{4/3}.$$

Thus, by Liouville’s theorem, we know that the entire function $$g(z):=zf(z)-1$$ is constant.

Thus, we have $$zf(z)-1=C,\ \text{for some constant}\ C.$$

However, here comes the problem. If we continue, we would have $$f(z)=\dfrac{c+1}{z},$$

but then $$\lim_{z\rightarrow 0}f(z)\neq 0$$ and thus $f(z)$ has a singularity at $z=0$ which is not removable, and thus $f(z)$ cannot be entire.

So, such functions do not exist?

I don't think my argument here is correct since $g(z)$ is only constant on a large disc, but not the whole complex plane. However, this is the only way I can think of to have some entire function being bounded.

Other attempts could not yield me a constant on a side and an entire function on the other side.

For instance (2):

We can move $|z|^{4/3}$ to the LHS so that we have $$|zf(z)-\sin z|-|z|^{4/3}\leq 1,$$ but then I could not get a way to further shrink the LHS so that we have an entire function inside the complex norm.

We can also move everything to the RHS, so that we can indeed have something like $$-1\leq |z|^{4/3}-|zf(z)-\sin z|\leq \Big||z|^{4/3}-|zf(z)-\sin z|\Big|\leq |z^{4/3}-zf(z)+\sin z|,$$ but this inequality does not tell us anything since the RHS must be positive, so it is absolutely larger than $-1$.

Any hints, ideas would be greatly appreciated! Thank you.

Edit: The Whole Proof

This proof follows exactly from what Martin R suggested. I am just adding more details.

Define $$g(z):=zf(z)-\sin z.$$ As $f(z)$ is entire, $g(z)$ must also be entire. Thus, for any $R>0$ and $z_{0}\in\mathbb{C}$, $g$ is holomorphic in an open set containing the closure of the disc $D(z_{0}, R)$.

Thus, by Cauchy's Inequalities, we have $$|g^{(n)}(z_{0})|\leq\dfrac{n!\sup_{z\in \partial D}|g(z)|}{R^{n}}.$$ On the other hand, as $|g(z)|\leq 1+|z|^{4/3}$ for all $z\in\mathbb{C}$, we have $$\sup_{z\in \partial D}|g(z)|=1+R^{4/3},$$ so that $$|g^{(n)}(z_{0})|\leq\dfrac{n!(1+R^{4/3})}{R^{n}},\ \text{for all}\ z_{0}\in\mathbb{C}.$$

Taking $R\rightarrow 0$, we can conclude that as long as $n>4/3>1$, we have $|g^{n}(z_{0})|=0$ for $z_{0}\in\mathbb{C}$.

Thus, $g(z)$ is a polynomial of degree $1$, i.e. we can write $g(z)$ as $$g(z)=az+b\ \text{for all}\ z\in\mathbb{C}.$$

Then, we have $$g(0)=0=b,$$ so that $$g(z)=az\ \text{for all}\ z\in\mathbb{C}.$$

Thus, $$|az|\leq 1+|z|^{4/3}\ \text{for all}\ z\in\mathbb{C}.$$

If $z=0$, then $0\leq 1$ holds for all $a$. For $z\in\mathbb{C}\setminus\{0\}$, we can divide $|z|$ in both side so that $$|a|\leq |z|^{-1}+|z|^{1/3}.$$

For each $z\in\mathbb{C}\setminus\{0\}$, we can find a disc with center at $0$ and radius $|z|:=r$ so that $z$ lives on the boundary.

Thus, the above inequality can be rewritten into $$|a|\leq \dfrac{1}{r}+r^{1/3}\ \text{for all}\ r>0.$$ Since this inequality holds for all $r>0$, we have $$|a|\leq\min\Big\{r>0:\dfrac{1}{r}+r^{1/3}\Big\}.$$

To find the minimum, define $$h(r):=\dfrac{1}{r}+r^{1/3},$$ so that $$h'(r)=-r^{-2}+\dfrac{1}{3}r^{-2/3}=r^{-2}\Big(-1+\dfrac{1}{3}r^{4/3}\Big),$$ and we have the critical point $r_{\min}=3^{3/4}$. Also, for $r>r_{\min}$, $h'(r)>0$, and $r<r_{\min}$, $h'(r)<0$.

Thus, $h(r)$ achieves local minimum for $r>0$ at $r_{\min}$ with the local minimum value $$h(r_{\min})=\dfrac{4}{3^{3/4}}.$$

Therefore, the entire function $f(z)$ also the form: $$f(z)=az,\ \text{where}\ |a|\leq \dfrac{4}{3^{3/4}}.$$

I'd like to express my appreciation to Martin R who always patiently answers my dumb questions. Please upvote his post, I own him too much. ^ ^

$\endgroup$
  • $\begingroup$ Liouvill'es Theorme is not applicable to $B(0,R)$. Unless you have a bounded anlaytic function on the entire complex plane you cannot apply this theorem. The correct answer for this question is $f(z)=a+\frac {\sin \, z} z$. $\endgroup$ – Kabo Murphy Jul 23 at 12:35
  • $\begingroup$ @KaviRamaMurthy yes you are right. I followed a wrong direction. I actually once thought about using Cauchy.... $\endgroup$ – JacobsonRadical Jul 23 at 12:51
11
$\begingroup$

There are several problems with your approach. For example, the estimate $$ |zf(z)-1|\leq |zf(z)-\sin z| $$ is wrong at every zero of $\sin(z)$. Also Liouville's theorem applies only to bounded functions in $\Bbb C$, not to bounded functions in a disk (no matter how large).

On the other hand, the inequality $$ |zf(z)-\sin z|\leq 1+|z|^{4/3} $$ implies that $zf(z)-\sin z$ is a polynomial of degree at most one. This can be shown with Liouville's theorem and induction (see for example Show that an entire function bounded by $|z|^{10/3}$ is cubic), or using Cauchy's inequalities for the derivatives. Therefore $$ zf(z)-\sin z = a + bz $$ with constants $a, b \in \Bbb C$. Finally show that $a=0$ and $$ |b| \le \min \{ 1/r + r^{1/3} \mid r > 0 \} = \frac{4}{3^{3/4}} \approx 1.755 \, . $$.

$\endgroup$
  • $\begingroup$ you are definitely right. I never thought about Cauchy's inequality. Thank you so much! $\endgroup$ – JacobsonRadical Jul 23 at 12:44
  • $\begingroup$ one more question about the intuition behind this question. how did you come up with an idea that $zf(z)-\sin z$ is a polynomial? is such an inequality sort of a sign? $\endgroup$ – JacobsonRadical Jul 23 at 12:49
  • $\begingroup$ @JacobsonRadical: Yes, the polynomial growth restriction immediately suggests such an approach. $\endgroup$ – Martin R Jul 23 at 12:52
  • $\begingroup$ how do you conclude that you find all entire functions? for instance, perhaps $f(z)=z^{1/3}$ also works? $\endgroup$ – JacobsonRadical Jul 23 at 12:54
  • 1
    $\begingroup$ @JacobsonRadical: $z^{1/3}$ itself cannot be defined as a continuous function around $z=0$. $\endgroup$ – Martin R Jul 23 at 13:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.