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If $f$ is entire on $\mathbb{C}$ and $|f(z)| \leq 100 \log_{e}|z|$ for each $z$ in $|z| \geq 2$ ,

If $f(i) = 2i$ then $f(1) = ?$,

I thought of applying ML inequality , Cauchys integral formula but i could not proceed with these tools in hand , i have read Contour integrations , ML inequality , Cauchy Goursat theorem .If you can cite any simple method for a begineer it would be good.

Actually i thought of using Liouville's theorem but due to the $\geq$ sign in modulus of $z$ i am unable to prove anything !

Thanks!

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    Are you taking the modulus of $z$ in the logarithm? – 2017-02-22
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    Oh..Isee .. will do an edit! – 2017-02-22
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    Did you think of this example or did you find it somewhere? I cannot think of a way to solve it, but I am far from perfect, just curious. – 2017-02-22
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    Actually it is ahomework question given a month ago , which i am trying to solve!@Noah – 2017-02-23
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    I am not sure but i think it will turn out that f is constant and hence we obtain result ? – 2017-02-23
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    I figured it out, but it uses the isolation of zeros theorem, have you seen that? – 2017-02-23
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    @Noah not yet, I thought of using Liouvilles theorem but then the problem was as i have $log(z)$ , so i need a function$(?)$ s.t $log(z) \leq (?)$, – 2017-02-23

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The proof will closely follow that of Louisville's theorem.

Fix some $z\in \Bbb C$, and let $R>|z|+2$. Thus we have that $B(0;2)\subset B(z; R)$. Furthermore, we have that for all $\zeta \in \partial B(z; R)$, since $|\zeta |>2$, $f(\zeta)\leq 100\log_e|\zeta|$. Since $\log_e $ is an increasing function, it then follows that $f(\zeta)\leq 100\log_e(R+|z|)$.

We thus have that \begin{eqnarray} |f'(z)| &&=\Big|\frac{1}{2\pi i}\int_{\partial B(z;R)} \frac{f(\zeta)}{(\zeta - z)^2}d\zeta\Big|\\ &&\leq \frac{1}{2\pi} \int_{\partial B(z;R)} \frac{|f(\zeta)|}{|(\zeta - z)^2|}|d\zeta|\\ &&\leq \frac{1}{2\pi} \int_{\partial B(z;R)} \frac{100\log_e(R+|z|)}{R^2}|d\zeta|\\ &&= 100\frac{\log_e(R+|z|)}{R} \end{eqnarray}

Since for a fixed $z$ this is true for all $R$ large enough, we may take the limit to see that $f'(z)=0$. Since $z$ was arbitrary, $\frac{d}{dz}f$ is identically zero, so $f$ is constant.

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    Why not $R \geq |z| + 2$ ? – 2017-02-24
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    Also how it follows that $f(\zeta)\leq 100\log_e(R+|z|)$ – 2017-02-24
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    It can be equal, it does not really matter though. – 2017-02-24
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    And since $|\zeta -z|=R$, $|\zeta|- |z|\leq R$ by the reverse triangle inequality, and $\log_e(|\zeta|)\leq \log_e(R+|z|)$. Since $f(\zeta)\leq 100\log_e(|\zeta|)$ the result follows. – 2017-02-24