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I want to show the following:

If $f(z) $ is a continuous function on a connected open subset of the complex plane and $f(z)^2$ is an analytic function, then $f(z)$ is analytic.

Clearly if $f(z) \neq 0$ then $$\frac{f(z+h)-f(z)}{h}=\frac{f(z+h)^2-f(z)^2}{h}.\frac{1}{f(z+h)+f(z)}$$

which shows $f^\prime(z) $ exists if $f(z)\neq0$.

What do I do when $f(z) =0$?

Michael Hardy
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jim
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2 Answers2

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a) Since $f^2$ is holomorphic, the Cauchy-Riemann criterion (expressed in Wirtinger's wonderfully concise notation) $\partial f^2/\partial \bar z=2f\partial f/\partial \bar z=0$, shows that $f$ is holomorphic at all points $z$ where $f(z)\neq 0$, since there $\partial f/\partial \bar z=0$. [This is the part solved by Don Antonio in more classical notation]

b) Since $f^2$ is holomorphic its zeros are isolated and so are those of $f$ (they are the same!)
But $f$ is locally bounded at those potential singularities, because $f^2$ is, and so Riemann's theorem on removable singularities permits you to conclude that the singularity is bogus and that actually $f$ is holomorphic also at the zeros of $f$ .

Conclusion: $f$ is holomorphic everywhere on its domain.

Georges Elencwajg
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  • so in your b> part : zeros of $f^2$ are same as the zeros of $f$ and zeros of $f$ are isolated (as zeros of $f^2$ are isolated ). so $f$ is analytic possibly at these zeros and then use riemanns thrm – jim Feb 18 '13 at 14:09
  • Dear jim: yes, precisely. – Georges Elencwajg Feb 18 '13 at 14:12
  • sorry for not accepting your answer i upvoted it thanks for the explanation – jim Feb 18 '13 at 14:13
  • Dear jim: it doesn't matter in the least and I'm happy that the green check went to our very active and competent friend Donantonio. – Georges Elencwajg Feb 18 '13 at 14:16
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    +1 for using the derivative wrt $,z,$ notation, which I didn't want to use because I thought it'd make things harder to understand to the OP, but no doubt renders a neater proof. – DonAntonio Feb 18 '13 at 14:26
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    Dear DonAntonio, I am sure you knew Wirtinger's notation. The decision to use more advanced notation/concepts is always a very problematic one and your choice is very reasonable. I find it optimal that readers now may select themselves the version they are more comfortable with. – Georges Elencwajg Feb 18 '13 at 14:50
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Suppose

$$f(x,y)=u(x,y)+iv(x,y)=:u+iv\Longrightarrow f^2=u^2-v^2+2uvi$$

Since $\,f^2\,$ is analytic then the Cauchy-Riemann equations apply here:

$$(u^2-v^2)'_x=(2uv)'_y\;\;\;,\;\;\;(u^2-v^2)'_y=-(2uv)'_x$$

But

$$(u^2-v^2)'_x=2uu_x-2vv_x\;\;\,\;\;(2uv)'_y=2u_yv+2uv_y\\(u^2-v^2)'_y=2uu_y-2vv_y\;\;,\;\;-(2uv)'_x=-2u_xv-2uv_x$$

So equalling:

$$I\;\;\;\;uu_x-vv_x=\;\;\;\;uv_y+vu_y\Longleftrightarrow\;\, u(u_x-v_y)-v(u_y+v_x)=0\\II\;\;\;\;uu_y-vv_y\;=-uv_x-vu_x\Longleftrightarrow u(u_y+v_x)+v(u_x-v_y)=0$$

Can you take it now from here (i.e., check the CR equations for $\,f=u+iv\,$) ?

DonAntonio
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  • if $u$ and $v$ are both not $0$ then $Ax=0$ has nonzero solution which implies $\det(A) =0\implies (u_x-v_y)^2+(u_y+v_x)^2=0$ from this cr equation follows. here $A= \begin{bmatrix} u_x-v_y & u_y+v_x \ u_y+v_x & u_x-v_y \

    \end{bmatrix}$

     – jim Feb 18 '13 at 13:45
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    The only delicate point is precisely to prove analyticity at the points where $u=v=0$, i.e. at the points where $f$ vanishes. [And that is the reason why I decided to answer this question: in order that users have a possible solution also for that case on record. I also wanted to advertise the power of Wirtinger's notation] Anyway, +1 for DonAntonio. – Georges Elencwajg Feb 18 '13 at 13:56
  • Hi, I also have the same question so don't want to ask it again. Wondering how you "take it from here now". I dont get any cancellation from adding or subtracting the two equations so how can I deduce the terms in the parentheses are zero? – user24907 Apr 18 '18 at 20:37
  • I think.. this is wrong answer because we don't know $u, v$ are partially differentiable fucntion. we just know that $f^2$ is analytic – hew Jul 19 '19 at 07:55
  • @hew You seem to be missing the point: never in the answer I did assume $;f;$ is analytic: I only used the given data that $;f^2;$ is analytic and thus the CR equations apply to it...and from this I deduce equations from which we can have the CR equations for $;f;$ . – DonAntonio Jul 19 '19 at 08:30
  • @DonAntonio Could you explain how can we show that there exist $u_x, u_y, v_x, v_y$?? – hew Jul 19 '19 at 08:33
  • @Hew That's trivial from the fact that the partial derivatives (of $;f^2$) exist because of the condition on $;f^2;$ ...! – DonAntonio Jul 19 '19 at 08:37
  • @DonAntonio Why that's trivial? For example, square of Dirichlet function F is constantly 1 but F is not conti and not differentiable.(Of course , in this problem we assume f is conti. – hew Jul 19 '19 at 08:41
  • I think that in general, although $h^2$ is differentiable, $h$ need not differentiable. – hew Jul 19 '19 at 08:43
  • is there possibility that $u_x$ doesn't exist but $f^2$ is analytic? In fact, i can't get counterexample yet.. but we need more argument. – hew Jul 19 '19 at 08:46
  • @hew Ok, now I see your point...and I'm far from this stuff now. I think I'll have to think it over a little, though I think the continuity of $;f;$ is paramount here. Thanks. – DonAntonio Jul 19 '19 at 12:00
  • One can show that $u, v \in C^\infty(\mathbb R^2)$ using the fact that $C^\infty(\mathbb R^2)$ is an algebra over $\mathbb R$. Note that the assumption that $f(z_0) \ne 0$ is necessary to deal with square root. – Luke May 13 '22 at 20:46