If for every $z,$ either $|f(z)| \le 1$ or $|f'(z)| \le 1,$ then $f$ is a linear polynomial I am working on the following exercise:
Let $f$ be entire and assume that for every $z,$ either  $|f(z)| \le 1$ or $|f'(z)| \le 1$ (or both). Then $f$ is a linear polynomial.
I have a few questions about this. First, I believe I solved it and would like someone to verify that my proof is correct:
Proof. We can use the generalised Liouville theorem. It states that if $f$ is bounded by $A + B|z|^n$ then $f$ is a polynomial of degree at most $n-1$.
Applying this theorem to $|f'|\le 1$ we get that $f'$ is a polynomial of degree at most $0$ at points where $|f'|\le 1$. Hence at these points $f$ is a polynomial of degree at most $1$.
At the other points $f$ has degree at most $0$.
Hence $f$ has degree at most $1$ everywhere.

Is this proof correct?

My other question about this exercise is this:
The hint I have is to write $f$ as
$$ f(z) = f(z_0) + \int_{z_0}^z f'(w) dw$$
where $z_0 = t_0 z$ where $t_0 = \sup \{t_1 \mid 0 \le t \le t_1, |f(tz)|\le 1 \}$.
So one expresses $f(z)$ as an integral along a line from $z_0$ to $z$. Taking the absolute value on both sides and using the triangle inequality:
$$ |f(z)| \le |f(z_0)| + \int_{z_0}^z |f'(w)| dw$$
I was tempted to continue by adding
$$ \le |f(z_0)| + \int_{z_0}^z 1 dw$$
but there is no reason why $f'$ should be bounded by $1$ between $z_0$ and $z$.
So my second question is:

How do I use this hint? Is there a mistake in the hint? Should it be
$f'$ in the definition of $t_0$?


I think there’s a mistake in the hint.
Inspired by the hint,I think we can prove as follow:
$\forall x\in \mathbb{C}\backslash\{0\}\ $,suppose $\ \vert f(z)\vert>1\ $,then$\ \vert f’(z)\vert\leq 1.\ $
$Let\ t_0=inf\lbrace t_1\in [0,1]\vert\ \vert f’(tz)\vert\leq 1, \forall t\in[t_1,1]\rbrace.$
Since$\ f(z)\ $is entire,we know that both$\ f(z)\ $and$\ f’(z)\ $are continuous,so$\ \vert f(t_0z)\vert=\vert f’(t_0z)\vert =1\ $
By Rectangle Integral Formula,we have$\ f(z)=f(z_0)+\int_{z_0}^{z}f’(\omega)\,d\omega,\ $ where $\ z_0=t_0z,\ $the integral line is the line segment connecting$\ z_0\ $and$\ z\ $
Let A=max$\{\vert f(0)\vert ,\ 1\},\ $then$\ \vert f(z)\vert \leq A+\vert\int_{z_0}^{z}f’(\omega)\,d\omega\vert\leq A+\vert z\vert$
