Cardinality of set $\{f\in C^1 (\mathbb R)\mid f(0)=0,f(2)=2, |f’(x)|\leq 3/2\}$ The Cardinality of set $\{f\in C^1(\mathbb R)\mid f(0)=0,f(2)=2, |f’(x)|\leq 3/2\}$ is
$1.$ empty set .
$2.$ non empty finite set.
$3.$ infinite set.
$4.$ uncountable set .
Function like $f(x)=x$ is in given set . But I am unable to find more functions like this . If derivative is less than $1$ then cardinality of given set is empty by uniqueness of fixed points, but here derivatives is less than or equal to $3/2$. Unable to find concept behind this . Please help . Thank you.
 A: The given set is uncountable. Consider the following family of piecewise linear functions $f_i:\mathbb{R}\rightarrow \mathbb{R}$ indexed by $i\in (1,\frac{3}{2})$
$$ f_i: x \mapsto \cases{ix \quad \text{if}\quad x\leq \frac{2}{i}\\
2 \quad\; \text{ if}\quad x>\frac{2}{i}}$$
Each function $f_i$ in the given family of piecewise linear functions can be smoothened at the boundary point to obtain a corresponding smooth function $g_i$. The family $(g_i)_{{i\in(1,\frac{3}{2})}}$ is uncountable and satisfies the given conditions.
A: By considering the range of all possible functions on a graph, this strategy comes to mind:
Take the family of test functions $f(x) = ax^b$, where $a>0,b>1$. This obviously satisfies the first condition. For the second one, we have $a2^b=2 \implies a = \frac{1}{2^{b-1}}$.
To satisfy the last condition, it is enough to satisfy $|abx^{b-1}| \leq \frac{3}{2}$. Its enough for $b<\frac{3}{2}$ as $|abx^{b-1}| \leq b$. Consider the set:
$$F = \{\frac{x^b}{2^{b-1}}: b \in (1,1.5)\}$$
Then this set of functions satisfy the given conditions. Hence 4 is correct.
A: Consider the set of functions
$f_{\alpha}(x)= \begin{cases} 
      x & x \leq e \\
      a\,\ln(x)^{\alpha} +b &x \geq e
   \end{cases}
$
for $\alpha \in (1,2)$, where $e\approx 2.718$ is the Euler number and $\ln(x)$ is the natural log with with $\ln(e)=1$
Obviousely $f_{\alpha}(0)=0$ and $f_{\alpha}(2)=2$. To make the functions differentiable at $x=e$, we must have:
$$ a + b = e$$
$$\frac{\alpha a}{e} = 1$$
Solving for $a$ and $b$ gives:
$$ a = \frac{e}{\alpha}$$
$$ b = e - \frac{e}{\alpha}$$
So the set of functions is
$f_{\alpha}(x) = \begin{cases} 
      x & x \leq e \\
      \frac{e}{\alpha}\,\ln(x)^{\alpha} + e - \frac{e}{\alpha} &x \geq e
   \end{cases}
$
for $\alpha \in (1,2)$.
Now we have,
$f^{'}_{\alpha}(x) = \begin{cases} 
      1 & x \leq e \\
      e \,\frac{\ln(x)^{\alpha-1}}{x} &x \geq e
   \end{cases}
$
which is continuous and hence $f_{\alpha}\in C^{1}$.
It is left is to show that $|f^{'}_{\alpha}(x)| \leq \frac{3}{2}$. For $x \leq e$ it is clear.
For $x \geq e$, the function
$g(x)=e \,\frac{\ln(x)^{\alpha-1}}{x}$ is decreasing, because for $x \geq e$ and $\alpha \in (1,2)$ we have,
$$ g^{'}(x) = e\,\frac{\ln(x)^{\alpha-2}(\alpha-1-\ln(x))}{x^2} < 0.$$
Therefore for $x \geq e$, $g(x) \leq g(e) = 1$. This means $f^{'}_{\alpha}(x) \leq 1$ for all $x$.
Since $\alpha \in (1,2)$, so the set is uncountable.
Not only I showed it for $|f^{'}_{\alpha}(x)| \leq \frac{3}{2}$ but also I showed there exists an uncountable set with condition $|f^{'}_{\alpha}(x)| \leq 1$ .
