Holomorphic bounded by log

Let f be a holomorphic function in \{z\Big| 0<|z|<1\}, continuous in \{z\Big| 0<|z|\le 1\} and |f(z)|\le \log\left(\frac{1}{|z|}\right). Prove that f\equiv 0.
Proof. Since f is continuous on the boundary, for z_0, s.t. |z_0|=1 we have |f(z_0)|=\lim_{z\to z_0}|f(z)|\le \lim_{z\to z_0}\log\left(\frac{1}{|z|}\right)=0 so f is 0 on the boundary. Consider now g(z)=zf(z). So g is also 0 on the boundary. Now observe that \lim_{z\to 0}|g(z)|=\lim_{z\to 0}|z||f(z)|\le \lim_{z\to 0}|z|\log\left(\frac{1}{|z|}\right)=0 so 0 is removable singularity for g. If we define g(0)=0, then g is holomorphic on \{z\Big||z|<1\}, continuous on the boundary. From the maximum modulus principle |g| attains its maximum at the boundary, and it is 0 there, so g\equiv 0 on \{z\Big||z|<1\}. Now take z_0\ne 0, and then f(z_0)=\frac{g(z_0)}{z_0}=0, so f\equiv 0.

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