Cardinality of the set $S$ Let $S=\{f \in C^1(\mathbb{R}): f(0)=0,f(2)=2 \; \mbox{and}\; |f'(x)|\leq\frac{3}{2} \forall x \in \mathbb{R}\}$,
find cardinality of $S$.

I think $S$ is an uncountable set. Am I right?
It should be possible to fit criteria for a parabola $f(x)=ax^2+bx+c$ for $x\in[0,2]$ in uncountably many ways:
$$0=f(0)=c\implies c=0$$
$$2=f(2)=4a+2b+c\implies 2a+b=1$$
so we will have $f(x)=ax^2+(1-2a)x$ and $f'(x)=2ax+1-2a$.
As $f'(x)$ is a linear function, it reaches maximum and minimum at the ends of the interval $[0,2]$, so we only need to make sure that $|f'(0)|\le 3/2$ and $|f'(2)|\le 3/2$. This means that the constraints are $|1-2a|\le 3/2, |1+2a|\le 3/2$. One can easy see that any $a$ in $[-1/4, 1/4]$ satisfies these inequalities, and there are uncountably many of those.
Now let's define the function $f$ on the whole $\mathbb R$:
$$f(x)=\begin{cases}(1-2a)x&amp;x&lt;0\\ax^2+(1-2a)x&amp;0\le x\le 2\\(1+2a)x-4a&amp;x&gt;2\end{cases}$$
which matches the original function $f$ on $[0,2]$ and linearly extends it on $(-\infty, 0)$ and $(2,\infty)$.
There is $g\in C^\infty(\mathbb R )$ such that
i) $g=0$ on $(-\infty,0]\cup [2,\infty)$
ii) $g=3/2$ on $[1/3,5/3]$
iii) $0&lt;g&lt;3/2$ elsewhere.
Note that $\int_0^2 g &gt; (4/3)\cdot(3/2) =2.$ We can thus choose $c\in (0,1)$ such that $\int_0^2 cg = 2.$
Now define $G(x) = \int_0^x cg(t)\,dt,$ $x\in \mathbb R.$ Then $G\in \mathbb C^\infty,$ $G(0)=0, G(2)=2.$ We also have $G'(x)=cg(x),$ which implies $|G'(x)|  &lt; 3/2$ everywhere. Note that $G'(x)=0$ for $x\notin [0,2].$
To finish, define
$$f_a(x) = G(x)+G(x-a),\, a&gt;3.$$
Then $\{f_a\}$ is an uncountable collection of $C^\infty$ functions that satisfy the criteria.