Does there exists such real analytic function? (NBHM 2016)

Question

a)Let $f(z)=e^x+iv$ then Cauchy Riemann equation will give us contradiction thus this cannot be true as $e^x=v_y \text{and} 0=v_x$, now $v_x=0 \implies v=g(y)$ and first equation then gives $g'(y)=e^x$ which is not true.

b) is true take the zero function.

c) This is not true since $f$ is entire and bounded thus constant and $f(0)=1 \implies f(z)=1 \quad \forall z\in \Bbb{C} $ but that contradicts $|f(z)|\le e^{-|z|}\quad \forall z$.

Am I correct?

Why does $f(0)=1 \rightarrow f(z)=1$ $ \forall z \in\mathbb{C}$? — siegehalver, Jan 23 '16 at 16:41
But $f(z)=e^x+iv(x,y)$ might satisfy Cauchy Riemann, so your answer to (a) needs amending. — Thomas Andrews, Jan 23 '16 at 16:41
@ChristopherHalverson It does if you know that $f$ is constant, as previously concluded. — Thomas Andrews, Jan 23 '16 at 16:42
@ThomasAndrews I somehow missed that when reading it. Thanks. — siegehalver, Jan 23 '16 at 16:44
I suspect (b) meant to ask for whether non-constant analytic functions of this sort exist. :) There are such functions - indeed, if $f$ is entire and bounded on the real axis, then $f(z^2)$ is entire and bounded on the real and imaginary axes. — Thomas Andrews, Jan 23 '16 at 16:47
In c) how can you say that $f$ is bounded? That inequality is not enough to state that. — K.Power, Jan 23 '16 at 16:51
It's not a hard conclusion to make, but the OP should give more details about how he concludes that $f$ is bounded. @K.Power — Thomas Andrews, Jan 23 '16 at 16:52
Oh yes, I didn't notice the "$|z| \geq 1$" disclaimer. Apologies — K.Power, Jan 23 '16 at 16:57
@lomberlego Hi, when you click on ask question , see there you will get an option of uploading image — Savannah, Aug 24 '17 at 16:48
i have tried and upload but t image image is no shown in the screen@maths man — jasmine, Aug 25 '17 at 12:27

score 1 · Answer 1 · answered Feb 24 '16 at 16:19

1

For 1) note that real part of analytic function. must be harmonic but here $e^x$ is not harmonic. Other you have correctly explained

answered Feb 24 '16 at 16:19

PAMG

4,440

how is the boundedness explained in part 3)? – vidyarthi Dec 27 '16 at 08:43
@vidyarthi:see Mathmore's answer... – Styles Oct 29 '17 at 19:13

score 0 · Answer 2 · answered Jan 19 '17 at 16:58

0

Explanations for (a) and (b) are correct.

For (c), Observe that $\forall |z| \ge 1$, $|f(z)| \le e^{-|z|} \lt 1$.

Now we will use the result that the continuous image of compact set is compact.

Consider the set $|z| \le 1$ then this is closed and bounded in $\Bbb C$ and hence compact. As $f$ is entire, the image of this set is also compact and hence closed and bounded.

So $f$ is bounded and entire on $\Bbb C$ $\Rightarrow f$ is constant. Given that $f(0)=1$. Hence $f(z) = 1 \; \forall z \in \Bbb C$. But then $1=|f(2)| \not\le e^{-2}.$ Hence such a function satisfying above properties does not exist.

answered Jan 19 '17 at 16:58

Error 404

6,006

Does not the function $f(z)=e^{-|z|}$ satisfy the condition of $(c)$? – Babai Jul 26 '17 at 22:27
@Babai is $f(z)=e^{-|z|}$ an analytic function? – Error 404 Jul 27 '17 at 02:18
@Mathmore:How do made the conclusion that $f$ is entire? – Styles Oct 29 '17 at 18:45
@PKStyles Any analytic function on $\Bbb C$ is entire(Definition of entire functions). Here $f: \Bbb C \to \Bbb C$ is given analytic. Thus it is entire. – Error 404 Oct 30 '17 at 11:22

Does there exists such real analytic function? (NBHM 2016)

2 Answers2