TITLE: Cardinality of $\{ f\in C'[0,1] : f(0)=0, f(1)=1, |f'(t)|\leq 1 \forall t\in[0,1]\}$... NBHM $2007$
QUESTION [6 upvotes]: Question is to find :
What is the cardinality of the following set :
$$A=\{ f\in C'[0,1] : f(0)=0, f(1)=1, |f'(t)|\leq 1 \forall t\in[0,1]\}$$
I would like to see for some time that all $f\in C'[0,1]$ which i am looking for are just polynomials as any continuous function can be approximated uniformly by polynomials.
with $f(0)=0$ condition I would like to say $f$ do not have constant term.
with the condition that $f'(t)$ being bounded by $1$ I would like to say that $f$ do not have higher powers
ie., suppose $f=x^3+x^2+1$ then, $f'=3x^2+2$ whihch is not bounded by $1$ 
I do not really mean this should how the polynomials of higher degrees look like but if $f$ has degree $n>1$ then, $f(x)=nx^{n-1}+g(x)$ and this would be not bounded by $1$.
I have not yet used $|f'(t)|\leq 1 \forall t\in[0,1]$ but i belive this would also give same result.
Just for simplicity Suppose $f(x)=x^3$ then, $|f'(t)|=\frac{|f(x)-f(y)|}{|x-y|}=\frac{|x^3-y^3|}{|x-y|}=|x^2+xy+y^2|$
But i can choose $x,y$ accordingly suhc that this is not bounded by $1$
So, $f(x)$ do not have higher powers and because $f(1)=1$, coefficient of $x$ has to be $1$
i.e., $$A=\{ f\in C'[0,1] : f(0)=0, f(1)=1, |f'(t)|\leq 1 \forall t\in[0,1]\}=\{x\}$$
I would like some one to verify if this is sufficient or do i have to rewrite the arguments (I am sure i have to, but i am not sure how to do so).
Please help me to clear this ambiguity.. Thank you

REPLY [15 votes]: Suppose that $f(c)\ne c$ for some $c\in(0,1)$. If $f(c)>c$, the mean value theorem ensures that there is an $x\in(0,c)$ such that $f\,'(x)>1$, and if $f(c)<c$, it ensures that there is an $x\in(c,1)$ such that $f\,'(x)>1$. Thus, the identity function is the only member of $A$.
{"set_name": "stack_exchange", "score": 6, "question_id": 577068}