Let $\gamma : [0,1] \longrightarrow \left \{z \in \Bbb C\ |\ \left \lvert z \right \rvert \leq 1 \right \}$ be a non-constant continuous mapping such that $\gamma (0) = 0.$ Let $f$ be an analytic function in the disk $\left \{z \in \Bbb C\ |\ \left \lvert z \right \rvert \lt 2 \right \},$ such that $f(0) = 0$ and $f(1) = 1.$ Then, there exists $\tau$ such that $0 \lt \tau \lt 1$ and such that for all $0 \lt t \lt \tau,$ we have that $f \left (\gamma (t) \right ) \neq 0.$
If the result is false then for each $0 \lt \tau \lt 1$ there exists $0 \lt t_{\tau} \lt 1$ such that $f \left (\gamma \left (t_{\tau} \right ) \right ) = 0.$ If we can somehow prove that the set $\left \{\gamma \left (t_{\tau} \right )\ |\ 0 \lt \tau \lt 1 \right \}$ is infinite then by Bolzano-Weierstrass theorem it has a limit point (since the set is bounded) and then by using identity theorem we have $f \equiv 0,$ a contradiction to the fact that $f(1) = 1.$ But how do I prove the infinitude of the set? Any help in this regard will be highly appreciated.
Thanks in advance.
You cannot prove that statement, since it is false. Suppose that$$\gamma(t)=\begin{cases}0&\text{ if }t<\frac12\\t-\frac12&\text{ otherwise.}\end{cases}$$Then $\gamma(0)=0$ and $\gamma$ is continuous and non-constant. Now, suppose that $f(z)=z$, in which case $f$ is analytic, $f(0)=0$, and $f(1)=1$. However, if $\tau=\frac12$, then $0<t<\tau\implies f\bigl(\gamma(t)\bigr)=0$.