three line theorem when $M_0 = 0$ or $M_1 = 0$

Question

Let $F$ be a continuous function and bounded which is defined on $$S= \{z = x+iy\mid 0 \le x \le 1\}$$

which is holomorphic in the interior of $S$,if $$|F(iy)|\le M_0,|F(1+iy)|\le M_1$$

Then $$|F(x+iy)|\le M_0^{1-x}M_1^x$$

My question is when $M_0 = 0$,why the result holds,what comes into my mind is the identity priciple,but the function is not holomorphic on the boundary line?If we apply maximum modular pricinple,we need the region being bounded region?

Answer 1

If $M_0 = 0$ then $$ |F(x+iy)|\le \epsilon^{1-x}M_1^x $$ holds for all $\epsilon > 0$, which implies that $F$ is identically zero.

But in this case there is a simpler proof, using the Schwarz reflection principle and the identity theorem. Also the assumption that $F$ is bounded is not needed in this case.

If $f$ is holomophic in the interior of the strip $S$ and continuous on the imaginary axis with $F(iy) = 0$ for all $y \in \Bbb R$ then it can be extended to a holomorphic function on the strip $$ \{ x+iy \mid -1 < x < 1 \} \, , $$ The identity theorem shows that the extended function (and consequently $F$) is identical zero in its domain.