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There exist an holomorphic function $f$ on $|z|<1$ and continuous for $|z|\le 1$ such that $ f(e^{i\theta})= \cos \theta + 2i \sin \theta$?

I have no idea what I can do in this problem. :S

3 Answers 3

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There does not exist such an $f$. Suppose, on the contrary, such an $f$ exists. Using the boundary condition of $f$, a direct computation shows that $\int_{|z|=1}f(z)dz=-\pi i\ne 0$, which contradicts to Cauchy's integral theorem.

An alternative way to obtain a contradiction is to consider $g(z)=\frac{3z^2-1}{2}$, which share the same boundary condition with $zf(z)$ on the uint circle. Then if $f$ is holomorpic on the unit disk and continous on the closed unit disk, $g(z)$ and $zf(z)$ must coincide with each other, i.e. $f(z)=\frac{3z^2-1}{2z}$, which has a pole at $0$, a contradiction.

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    @Daniel: It is true. Remember $z \mapsto |z|f(z)$ is uniformly continuous on the closed unit disk. Hence for $r$ close to $1$, you have |rf(r e^{it})-f(e^{it})|< \epsilon, uniformly in $t$. Also, $\int_{|z|=r} f - \int_{|z|=1}f = i \int_0^{2 \pi} (r f(r e^{it})-f(e^{it})) e^{it} dt$. A little work shows that the integrals converge.2012-11-11
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If $f$ was holomorphic, then so would $g(z)=f(z)-z$. However, $g$ is purely imaginary (see below) and hence must be constant. However it is not constant, which is a contradiction. Hence such an $f$ does not exist.

Note: I glibly assumed that $g$ was imaginary on the closed unit disk, when in fact, all that I know apriori is that it is imaginary on the unit circle. @PerManne below has given a nice argument showing why it is imaginary on the unit disk. Here is another proof which gives an explicit representation of $g$.

I am using Theorem 5.25 in Rudin's Real & Complex Analysis. (The set $A$ in his statement is the collection of all functions analytic in the unit ball and continuous on the closed unit disk.) Then, if $g$ is analytic on the unit ball and continuous on the closed ball, then we can write $g(re^{i\theta}) = \frac{1}{2\pi} \int_{-\pi}^{\pi} P_r(\theta -t) g(e^{it}) dt$ where $P_r(\theta -t) = \frac{1-r^2}{1-2r\cos(\theta -t)+r^2}$ is the Poisson Kernel. Since the kernel is real valued, if follows that if $g$ takes purely imaginary values on the boundary then it takes purely imaginary values in the interior as well.

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Suppose by contradiction that such a function exists. Let $g(z)=f(z)-z$.

Then

$g(e^{i \theta})=i \sin(\theta)$

Then, $g$ is a holomorphic function on $|z| < 1$ whose range is imaginary.....

Added To get the contradiction, check my answer here.

Basically $h_1(z)=e^{g(z)}$ and $h_2(z)=e^{-g(z)}$ are constant on $|z|=1$, thus by applying the Maximum Modulus Principle to both, you get that $1 \leq |h_1(z) | \leq 1$. This proves that $h_1(z)$ is a constant function.

P.S. $g(z)=f(z)-2z$ satisfies the conditions of the above link, since for the proof $g$ doesn't need to be entire, just Analytic inside $|z| \leq 1$.

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    @N.S.: Very nice! And elementary, I had to resort to the Poisson kernel. (I initially misread the question as defining $g(re^{i\theta})$, hence my initial incorrect answer!)2012-11-03