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What is the cardinality of the following set:
$$A = \{f \in C^1[0,1] : f (0) = 0, f (1) = 1, |f'(t)| \leq 1\text{ for all }t \in [0, 1]\}$$

I am not getting any clue proceed further to solve this problem and totally stuck.

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$$\int_0^1 f'(t) dt = f(1) - f(0) = 1 - 0 =1$$ $$\left \vert \int_0^1 f'(t) dt \right \vert \leq \int_0^1 \left \vert f'(t) \right \vert dt \leq 1$$ Hence, $f'(t) = 1$ almost everywhere i.e. $f(t) = t$ almost everywhere. With continuity, you can try to prove that this is probably the only function. –  user17762 Jan 2 '13 at 2:45
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Let $f(t)$ be such a function, and let $g(t)=t-f(t)$. Note that $g(0)=0$ and $g(1)=0$. We show that $g(t)=0$ for all $t$ in our interval.

Suppose that $g(t)\lt 0$ at $t=a$. By the Mean Value Theorem, $$\frac{g(a)-g(0)}{a-0}=g'(c)$$ for some $c$ between $0$ and $a$. It follows that $g'(c)=1-f'(c)\lt 0$. Thus $f'(c)\gt 1$.

Suppose that $g(t)\gt 0$ at $t=a$. By the Mean Value Theorem, $$\frac{g(1)-g(a)}{1-a}=g'(c)$$ for some $c$ between $a$ and $1$. The left-hand side is negative. It follows that $g'(c)=1-f'(c)\lt 0$, which again implies that $f'(c)\gt 1$.

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Hint: If there exists a point $f'(t) < 1$, then there exists a neighborhood $N=N_r(t)$ such that $\forall n \in N, f'(n) < 1$, by the continuity of $f'$. Use this to show that $f(1) < 1$.

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Using the mean-value theorem you can deduce that $f(1)-f(t)\le 1-t$ for all $0\le t\le 1$, and that $f(t)-f(0) \le t$.

Together, these mean $f(t)=t$ for all $t \in [0,1]$, and there's only one function that satisfies that.

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