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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

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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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Why does it follow that g is purely imaginary inside the disk? (That was why I deleted my answer.) –  copper.hat Nov 3 '12 at 16:31
    
@copper.hat Nonconstant holomorphic maps are open maps. This means that no interior point of the disk can be mapped to a boundary point of the image. Hence the boundary of the image has to be contained in the image of the boundary, which in this case is $\{{\rm i}y : -1\leq y\leq 1\}$. –  Per Manne Nov 3 '12 at 17:55
    
@PerManne: Thanks, that is useful. I just found the Poisson integral representation which shows it explicitly, but I like the open map reasoning. –  copper.hat Nov 3 '12 at 18:25
    
@copper.hat Added the extra details, I answered couple days ago to the same question but with real instead of imaginary on the unit circle ;) –  N. S. Nov 3 '12 at 20:20
    
@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!) –  copper.hat Nov 3 '12 at 21:08
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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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Cauchy's integral theorem requires the path to be in the region of where $g$ is analytic. It is merely continuous on the boundary. –  copper.hat Nov 3 '12 at 16:43
    
Yes! I can't use cauchy integral formula , unless $f$ is holomorphic in a neighborhood of the disk –  Daniel Nov 3 '12 at 16:44
    
@copper.hat: continuity on the boundary and holomorphicity in the interior is sufficient. You may use an approximation argument from interior to boundary to prove that by yourself. Please refer to the related wikipedia page. –  23rd Nov 3 '12 at 16:45
    
Not that is matters here, but I get $-i \pi$ for the integral. –  copper.hat Nov 3 '12 at 17:03
    
@N.S. I have a question with your answer. I think that it's only a simple application of Cauchy integral formula , over the curve $ |z|=1$ but I can't see this, I have to compute the following $f(z)= \frac{1}{2\pi i} \int _{|z|=1} \frac{f(w)}{w-z}dw=\frac{1}{2\pi } \int _0 ^{2\pi} \frac{f(e^{it})}{e^{it}-z} e^{it} = \frac{1}{2\pi } \int _0 ^{2\pi} \frac{e^{it}+isin t}{e^{it}-z} e^{it}$ –  Daniel Nov 3 '12 at 17:03
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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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