Cardinality of the set $S$

Question

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?

Answer 1

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 $\mathbb R$:

$$f(x)=\begin{cases}(1-2a)x&x<0\\ax^2+(1-2a)x&0\le x\le 2\\(1+2a)x-4a&x>2\end{cases}$$

which matches the original function $f$ on $[0,2]$ and linearly extends it on $(-\infty, 0)$ and $(2,\infty)$.

Answer 2

The cardinal of $S$, noted $\vert S \vert$ is equal to the continuum $\mathfrak c$, i.e. it has the cardinality of $\mathbb R$.

As mentioned in the other answers, there is an injection from a set having the cardinality of the continuum into $S$. Therefore the cardinal of $S$ is greater or equal to $\mathfrak c$.

Also, a continuous map is entirely defined when its values are given on the set of the rationals. Therefore the cardinal of the set of the continuous functions is equal to $\mathfrak c$. See this question for further details.

As $S$ is included into $\mathcal C(\mathbb R)$, $\vert S \vert$ is less or equal to $\mathfrak c$.

Finally, $\vert S \vert = \mathfrak c$, i.e. the cardinal of $\mathbb R$ (which is indeed uncountable) according to Cantor-Bernstein theorem.

Answer 3

You can think it by slope of tangent at each point of a curve.

Think, just take a differentiable function $g$ on $\mathbb{R}$ with $g(0)=0,g(2)=2$ such that $f(x)\lt g(x) $ for all $x\in (0,2) $ and $|g'(x)|\le \frac{3}{2} $ .

As there are uncountable points between $[f(x),g(x)]$ for each $x \in (0,2) $ .

We must find uncountably many differentiable functions $r(x)$ with $f(x) \lt r(x) \lt g(x) $ for $x\in (0,2) $ with $r(0)=0,r(2)=2$

Answer 4

Hint: If $V$ is a real vector space and $A\subseteq V$ is convex and has at least two points, then $\lvert A\rvert$ is at least the continuum (and at most $\lvert V\rvert$).

Answer 5

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) $0<g<3/2$ elsewhere.

Note that $\int_0^2 g > (4/3)\cdot(3/2) =2.$ 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 $|G'(x)| < 3/2$ everywhere. Note that $G'(x)=0$ for $x\notin [0,2].$

To finish, define

$$f_a(x) = G(x)+G(x-a),\, a>3.$$

Then $\{f_a\}$ is an uncountable collection of $C^\infty$ functions that satisfy the criteria.