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    A holomorphic function on $\DD\setminus\{0\}$ with $|f(z)|\le\log(1/|z|)$ is identically zero

    P-FHQAB

    Area
    Complex Analysis
    Topics
    Maximum Modulus Principle, Removable Singularities, Singularities
    Status
    draft

    Show that if \(f\) is holomorphic on \(\DD\smz\) and \(\abs{f(z)} \leq \log\qty{1\over \abs{z}}\), then \(f\equiv 0\).

    Solution

    • Claim: \(f\) has a removable singularity at \(z_0=0\).

    • Thus \(f\) extends to some holomorphic \(F\) defined on \(\DD\).

    • By continuity, \(\abs{F}\) satisfies the same inequality as \(f\).

    • Now \(\lim_{\abs z \to 1}\abs{F(z)}\leq \lim_{\abs z\to 1} \log\qty{1\over \abs{z}} = 0\), so by the MMP on \(\DD\), \(F\equiv 0\) on \(\DD\smz\).

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