Infimum over a class of holomorphic functions on the unit disk I am hoping the following is true but don't yet know.  Thoughts and ideas are appreciated.


  Let $k \ge 2$ be an integer. Let $0&lt;\epsilon &lt;1$. Then there is a holomorphic function $g: \mathbb{D} \to \mathbb{C}$ s.t. $$ \inf_{z \in \mathbb{D}} |1+z+z^kg(z)| &gt; 1-\epsilon$$


Note: When $k=2$ the conclusion can be seen to follow using the Riemann Mapping Theorem.
How about this?

By the maximum/minimum principle, we know that $$\inf_{z \in \mathbb{D}} \left| 1+ z + z^k g(z) \right| = \min_{\left| z \right| =1} \left| 1 + z + z^k g(z) \right|.$$ Continuing this, it follows that \begin{eqnarray*}
\min_{\left| z \right| =1} \left| 1 + z + z^k g(z) \right| &amp; &gt; &amp; \min_{\left| z \right| =1} \left| z + z^k g(z) \right| \\
&amp;=&amp; \min_{\left| z \right| =1} \left| z \right| \cdot \left| 1 + z^{k-1}g(z) \right| \\
&amp;=&amp; \min_{\left| z \right| =1} \left| 1 + z^{k-1}g(z) \right| \\
&amp; &gt; &amp; \min_{\left| z \right| =1} \left( \left| 1 \right| - \left| z^{k-1}g(z) \right| \right) \\
&amp;=&amp; 1 - \min_{\left| z \right| =1} \left| z^{k-1}g(z) \right| \\
&amp;=&amp; 1 - \min_{\left| z \right| =1} \left| g(z) \right|, 
\end{eqnarray*} choosing $g$ such that $\min_{\left| z \right| =1} \left| g(z) \right| = \varepsilon$ yields the result.
How about this?

By the maximum/minimum principle, we know that $$\inf_{z \in \mathbb{D}} \left| 1+ z + z^k g(z) \right| = \min_{\left| z \right| =1} \left| 1 + z + z^k g(z) \right|.$$ Continuing this, it follows that \begin{eqnarray*}
\min_{\left| z \right| =1} \left| 1 + z + z^k g(z) \right| &amp; &gt; &amp; \min_{\left| z \right| =1} \left| z + z^k g(z) \right| \\
&amp;=&amp; \min_{\left| z \right| =1} \left| z \right| \cdot \left| 1 + z^{k-1}g(z) \right| \\
&amp;=&amp; \min_{\left| z \right| =1} \left| 1 + z^{k-1}g(z) \right| \\
&amp; &gt; &amp; \min_{\left| z \right| =1} \left( \left| 1 \right| - \left| z^{k-1}g(z) \right| \right) \\
&amp;=&amp; 1 - \min_{\left| z \right| =1} \left| z^{k-1}g(z) \right| \\
&amp;=&amp; 1 - \min_{\left| z \right| =1} \left| g(z) \right|, 
\end{eqnarray*} choosing $g$ such that $\min_{\left| z \right| =1} \left| g(z) \right| = \varepsilon$ yields the result.