Let $f:[0,1]\longrightarrow \mathbb{R}$ be continuous. Recall, the modulus of continuity of $f$ (of order $\alpha$) is defined as
$$\omega(f,\alpha):=\sup\{|f(x)-f(y)|:x,y\in[0,1], |x-y|\leq \alpha\},$$
for each $\alpha>0$. If $f$ is continuous, then $\omega(f,\alpha)\rightarrow 0$ when $\alpha \rightarrow 0$.
Now, let the set
$$C:=\{f:[0,1]\longrightarrow \mathbb{R}:f\textrm{ continuous, } 0=f(0)\leq f(x)\leq 1=f(1),\textrm{ for all } x\in [0,1]\}.$$
From the above considerations, we know that $\omega(f,\alpha)\rightarrow 0$ when $\alpha \rightarrow 0$ for every $f\in C$. Here, the question is: Can we give, in terms of $\alpha$, an upper bound for the supremum of these mudulii of continuity? That is to say, if we can give an upper bound for the number
$$\Omega(\alpha):=\sup\{\omega(f,\alpha): f\in C\}.$$
Of course, $\Omega(\alpha)\leq 1$ because $|f(x)-f(y)|\in[0,1]$ for each $x,y\in[0,1]$ and $f\in C$. But, It is true that
$$\Omega(\alpha)\rightarrow 0$$
when $\alpha \rightarrow 0$?
Thanks in advances for your comments.
Indeed $\Omega (\alpha) =1$ for all $\alpha >0$: it is clear for $\alpha \ge 1$. For each $\alpha<1$ fixed, let $f = f_\alpha \in C$ be given by
$$ f(x) = \begin{cases} 0 & \text{if }x\in [0,1-\alpha] \\ \text{linear} & \text{if }x\in [1-\alpha, 1].\end{cases}$$
(by linear we mean joining $(1-\alpha, 0)$ to $(1,1)$ by a straight line). Then $f\in C$ and $\omega (f, \alpha) = 1$ for this $f$.
In general if $C$ is any subset of continuous functions $f: [0,1]\to [-M, M]$ and $$\Omega_C (\alpha) := \sup \{ \omega(f, \alpha) : f\in C\}.$$
Then the subset $C$ is precompact (under the topology of uniform convergence) if and only if $\Omega_C (\alpha) \to 0$ as $\alpha \to 0$. This is really a restatement of the Arzela-Ascoli theorem.