# 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 $$y=x$$.

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 $$t \in [0, 1]$$.

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 $$f(x)=x$$.