# 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.

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.

• These functions are differentiable ? Jul 10, 2022 at 6:36
• $g_i$'s are infinitely differentiable Jul 10, 2022 at 6:37
• @RichoddQsscraft what is $g_i$? Jul 10, 2022 at 6:39
• $g_i$s are obtained by "smoothing" $f_i$s. Since you're looking for a $C^1$ function, here's an explicit construction: math.stackexchange.com/a/410871/402166 Jul 10, 2022 at 6:42
• Ok thank you .... Jul 10, 2022 at 6:48

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.

• Thank you very much ..... Jul 10, 2022 at 6:48
• Derivative conditions is not satisfied... I think Jul 10, 2022 at 11:22
• You were right, the range needs to be modified accordingly. Jul 10, 2022 at 17:59

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

• yes it seems that your solution is correct. Jul 11, 2022 at 15:30
• Can you make it more simple by giving more simple examples Jul 11, 2022 at 15:30
• I believe that a simple example doesn't exist for this problem. Someone needs to find a smooth function by defining some parameters and calculating them. However, I think the solution I provided can be considered as a simple example. I just provided more details and that's why it became lengthy. Remember that any simple solutions of the kind $ax^b$ or $x^a+x^b$ doesn't work except f(x)=x (why?).
– Dan
Jul 12, 2022 at 0:04