Maximum principle on a closed strip for bounded holomorphic functions

Fix $\epsilon >0$ and define $g_{\epsilon }(z):=g(z)\exp (\epsilon (z^2-1))$. By [L2], $\mathrm{\mid }{g}_{\epsilon }(x+iy)\mathrm{\mid }=\mathrm{\mid }g(x+iy)\mathrm{\mid }\exp (\epsilon (x^2-y^2-1))$. On $x=0$ the exponential factor is at most $1$, and on $x=1$ it is $\exp (-\epsilon y^2)\le 1$, so both vertical boundary lines retain modulus at most $1$.

Choose $C\ge 0$ with $\mathrm{\mid }g\mathrm{\mid }\le C$. Since $x^2-1\le 0$ for $0\le x\le 1$, the horizontal sides at heights $y=\pm T$ satisfy $\mathrm{\mid }{g}_{\epsilon }(x\pm iT)\mathrm{\mid }\le C\exp (-\epsilon T^2)$. For all sufficiently large $T$, this is at most $1$.

The rectangle $R_T:=\{z:0<\mathrm{Re}z<1,\mathrm{\mid }\mathrm{Im}z\mathrm{\mid }<T\}$ is bounded, open and nonempty, and each coordinate of a segment between two of its points stays between that coordinate's endpoints, so the segment stays in $R_T$ and [L3] makes $R_T$ connected; it is therefore a bounded complex domain. With $T$ as in step 2.1, all four boundary sides of $R_T$ have $\mathrm{\mid }{g}_{\epsilon }\mathrm{\mid }\le 1$. The boundary maximum theorem [L1] therefore gives $\mathrm{\mid }{g}_{\epsilon }\mathrm{\mid }\le 1$ throughout $\overline{R_T}$.

Given $z\in S$, choose such a $T>\mathrm{\mid }\mathrm{Im}z\mathrm{\mid }$. Step 3.1 yields $\mathrm{\mid }g(z)\mathrm{\mid }\le \exp (-\epsilon \mathrm{Re}(z^2-1))$. Letting $\epsilon$ decrease to $0$ gives $\mathrm{\mid }g(z)\mathrm{\mid }\le 1$. This also covers both vertical boundary lines and the zero function.