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Question is as follows :

Suppose that $f:\mathbb{C}\rightarrow \mathbb{C}$ is continuous such that $f^3,f^4$ are analytic in $\mathbb{C}$ then prove that $f$ is analytic in $\mathbb{C}$..

Choose $z_0\in \mathbb{C}$. Suppose that $f(z_0)\neq 0$. As $f$ is continuous, in a nbd around $z_0$ the value of $f$ is nonzero.

So, $f^4(z)/f^3(z)=f(z)$ in a small open ball centered at $z_0$.. So, $f$ is analytic at $z_0$.

Suppose $z_0$ is such that $f(z_0)=0$ then we can not say that $f^4(z)/f^3(z)=f(z)$ for all $z$ in a small open ball around $z_0$ as $f(z_0)=0$ we can not cancel.

But then we can certainly say that there is a ball where $f(z)$ is non zero except at $z_0$. Then in this ball we do have that $f^4(z)/f^3(z)=f(z)$. So, $f(z)$ is analytic in punctured disk around $z_0$.

I some how feel that function being analytic on punctured disk and function being continuous imply that function is analytic..

Could not figure out the proof for my supposedly correct statement.

As $f$ is analytic on punctured disk around $z_0$ we have laurent expansion for $f(z)$ as

$$f(z)=\cdots+\frac{a_{-2}}{(z-z_0)^2}+\frac{a_{-1}}{z-z_0}+a_0+a_1(z-z_0)+a_2(z-z_0)^2+\cdots$$

Suppose there is atleast one coefficient of $\dfrac{1}{z-z_0}$ that is non zero, then $|f(z)|\rightarrow \infty$ as $z\rightarrow z_0$..

But then, it is given that $f(z)$ is continuous.. continuous function on compact set is bounded so $f(z)$ has to be bounded on the disk where as we have $|f(z)|\rightarrow \infty$ if atleast one coefficient of negative powers of $(z-z_0)$ is non zero. So, this says all negative powers coefficients are zero..

So, we have $f(z)=a_0+a_1(z-z_0)+a_2(z-z_0)^2+\cdots$ in a nbd around $z_0$.

Could not deduce anything from here.

I think i am very close to final answer but could not see it immediately.

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  • $\begingroup$ @CameronWilliams : Thanks for showing interest in my question.. So, we define $g(z)=f(z)$ for $z\neq z_0$ and $g(z_0)=\lim_{z\rightarrow 0}f(z)$ then $g$ is analytic.. But then, $f$ is continuous so $\lim_{z\rightarrow z_0}f(z)=f(z_0)$ so, we have $g(z)=f(z)$ for $z\neq z_0$ and $g(z_0)=\lim_{z\rightarrow z_0}f(z)=f(z_0)$.. So, $f(z)$ is actually equal to $g(z)$.. So, $f(z)$ is analytic.. $\endgroup$ – user311526 Apr 20 '16 at 18:18
  • $\begingroup$ The result is actually true if we assume only $f$ is continuous and $f^3$ is analytic $\endgroup$ – zhw. Apr 20 '16 at 18:23
  • $\begingroup$ @zhw. : Is that so? How? $\endgroup$ – user311526 Apr 20 '16 at 18:25
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You could also use Riemann's removable singularity theorem: If $f$ is bounded and analytic in $\{0<|z-a|<r\},$  then  f  has a removable singularity at $a.$


Added in response to comment: Suppose $f$ is continuous on $\mathbb C$ and $f^3$ is analytic on $\mathbb C.$ Then $f$ is analytic on $\mathbb C.$

Proof: Suppose $f(a) \ne 0.$ We have

$$\frac{f(z)^3 -f(a)^3}{z-a} = \frac{f(z) -f(a)}{z-a} \cdot [f(z)^2 + f(z)f(a) + f(a)^2].$$

Now $f$ is continuous at $a,$ so for $z$ close to $a,$ the term in brackets is nonzero, and has limit $3f(a)^2.$ It follows that

$$\frac{(f^3)'(a)}{3f(a)^2}=f'(a).$$

Thus $f'(a)$ exists on the open set where $f\ne 0.$ You handle the case $f(a) = 0$ as already discussed.

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  • $\begingroup$ you have suggested that there is a more strong result... $f$ is continuous and $f^3$ is analytic then $f$ is analytic. Can you help me to see that... $\endgroup$ – user311526 Apr 20 '16 at 20:16
  • $\begingroup$ Oh.. This is great.. So, the other condition is unnecessary.. I remember using this kind of trick in some other question.. I will let you know if i recall... Thanks again.. $\endgroup$ – user311526 Apr 21 '16 at 3:53
  • $\begingroup$ math.stackexchange.com/questions/1734399/… is link of the question i was mentioning in last comment... So, why did they give that $f^4$ is analytic? Can we ignore the condition that $f$ is continuous? $\endgroup$ – user311526 Apr 21 '16 at 5:14
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"I somehow feel that function being analytic on punctured disk and function being continuous imply that function is analytic.."

Yes, that's true. Actually:

Suppose that $ \varnothing \neq \Omega \subset \Bbb C$ is an open convex set and let $z_0 \in \Omega$. If $f$ is continuous on $\Omega$ and $f$ is holomorphic on $\Omega \setminus \{z_0\}$, then $f$ is holomorphic on $\Omega$.

Proof:

Let $a \in \Omega$ and define:

$$F(z) = \int_{[a,z]} f(\xi) d\xi$$

Let $\eta \in \Omega$ and let $\epsilon > 0$. The continuity of $f$ gives a $\delta > 0$ such that $|z - \eta| < \delta \implies |f(z) - f(\eta)| < \epsilon$. Now let $z$ be such that $0 < |z - \eta| < \delta$. Then, where in the first step we will use Cauchy's theorem for triangles:

$$\left|\frac{F(z) - F(\eta)}{z - \eta} - f(\eta) \right|= \left| \frac{\int_{[\eta, z]} f(\xi) d\xi}{z - \eta} - f(\eta) \right| \\ = \left|\frac{\int_{[\eta, z]} (f(\xi) - f(\eta))d\xi}{z - \eta}\right| \\ \le \frac{\int_{[\eta, z]} |f(\xi) - f(\eta)|d\xi}{|z - \eta|} \le \max_{\xi \in [z, \eta]} \{|f(\xi) - f(\eta)| \}< \epsilon$$

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  • $\begingroup$ Can you give some reference for the statement aboyt $F'=f$.. $\endgroup$ – user311526 Apr 20 '16 at 19:05
  • $\begingroup$ @topgeomj I think that this result is called "Cauchy's theorem for a convex set" $\endgroup$ – user258700 Apr 20 '16 at 19:07
  • $\begingroup$ I will check that.. thanks :) $\endgroup$ – user311526 Apr 20 '16 at 19:09
  • $\begingroup$ @topgeomj you're welcome. I provided a proof anyway. $\endgroup$ – user258700 Apr 20 '16 at 19:25