Cardinality of set $\{f\in C^1 (\mathbb R)\mid f(0)=0,f(2)=2, |f’(x)|\leq 3/2\}$

Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

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 $|f^{'}_{\alpha}(x)| \leq 1$ .