A related question to the Weierstrass approximation theorem

Question

Suppose $f : [a,b] \to \mathbb{R}$ is a continuous function. Then, Weierstrass Approximation Theorem asserts that there is a polynomial on $[a,b]$ which is as close to $f$ as we want (in a uniformly approximation sense).

It is my question :

Let $S$ be a set $\big\{ \big| \frac{f(y) - f(x)}{y-x} \big| : a \leq x < y \leq b \big\}$, which is a set of absolute values of slopes between every two distinct points of $f$.

Then, is the set $S$ bounded above?

Answer 1

No. Let $f(x) = \sqrt{x}$ on, for example, $[0,1]$. The average rate of change of $f$ is unbounded for pairs of points near $0$.

Consider also $$ f(x) = \begin{cases} 1 - \sqrt{1-x^2} ,& 0 \leq x < 1 \\ 1 - \sqrt{1 - (x-2)^2} ,& 1 \leq x \leq 2 \end{cases} $$ Its graph is two quarter circles meeting at a vertical tangent. The average rates of change are unbounded for sample points near (and on the same side of) $1$.

Answer 2

The set is unbounded for a generic continuous function $f$. Take for example $f(x)=\sqrt{x}$ on $[0,1]$. If your set is bounded, your function is called Lipschitz continuous.

Answer 3

No.

$f(x)=\sqrt{x}$, let $[a,b]=[0,1]$. $f$ is continuous but $\forall n \in \mathbb{N}$ : $\exists x \in [0,1]$ such that $\frac{f(x)-f(0)}{x} \geqslant n$