What is the cardinality of $A=\left\{f\in C^1[0,1]:f(0)=0,\ f(1)=1,\ \left|f'(t)\right|\le 1\ forall\ t\ \in [0,1].\right\}$ What is the cardinality of the following set?
$A=\left\{f\in C^1[0,1]:f(0)=0,\ f(1)=1,\ \left|f'(t)\right|\le 1\ forall\ t\ \in [0,1].\right\}$
My try: Since $f\in C^1[0,1]$. So, $f$ satisfy the proposition of Lagrange's mean value theorem. So, there is $c\in (0,1)$: $f'(c)=1.$ That is slope of the tangent at $(c,f(c))$ is $1$.When I tried to sketch possible graphs with this much information. there exists a point in $(0,1)$ such that derivative has to be more than $1$. Only graph with this property is $y=x$.
 A: You are right that $f(x) = x$ is the only function in that set, but it is not sufficient to show that $f'(c)=1$ for some $c\in (0, 1)$.
Using the mean-value theorem you can argue that for $0 < x < 1$ there is a $c_1 \in (0, x)$ such that
$$
 \color{red}{f(x)} = f(0) + \underbrace{f'(c_1)}_{\le 1} \cdot \underbrace{(x-0)}_{> 0} \color{red}{\le} 0 + 1 \cdot (x-0) = \color{red}{x}
$$
and a $c_2 \in (x, 1)$ such that
$$
 \color{red}{f(x)} = f(1) + \underbrace{f'(c_2)}_{\le 1} \cdot \underbrace{(x-1)}_{< 0} \color{red}{\ge}  1 + 1 \cdot (x-1) = \color{red}{x}
$$
and therefore $f(x) = x$.
As you can see, the condition $|f'(t)| \le 1$ for all $t \in [0, 1]$ can be relaxed to  $f'(t) \le 1$ for all $t \in [0, 1]$.
A: Indeed, the conjecture is true. Using the mean value theorem in the interval $(x,y)\subset(0,1)$ we obtain that for all $x,y$
$$|f(x)-f(y)|\leq|x-y|$$
Setting $y=0$ and using the fact that $x>0$:
$$-x \leq f(x)\leq x$$
Similarly for $y=1$:
$$|f(x)-1|\leq 1-x\iff x\leq f(x)\leq2-x$$
and thus we find that $f(x)=x$.
