"continuously differentiable $\subseteq$ Lipshitz continuous" with $f(x) = x^2$

Question

In the Wiki, it says: continuously differentiable (i.e. class $C^1$) $\subseteq$ Lipshitz continuous.

Consider the simplest example ($x,y\in \mathbb{R}$): $$f(x) = x^2$$

1. It is not Lipshitz continuous. The reason is we cannot find $L$, such that $$\frac{f(y)-f(x)}{y-x}\leq L$$

2. $f(x)$ is differentiable with $f'(x) = 2x$, which is continuous. Therefore $f(x) = x^2$ is continuous differentiable and is in class $C^1$

However, this inference violates the one in the wiki. I guess the second inference might me wrong but why?

Answer 1

Over a compact connected set each $C^1$ function is Lipschitz continuous, because its derivative is bounded (Bolzano Weierstrass Theorem) by some constant. This is also the constant in the Lipschitz continuity - follows from the Mean value theorem.

Note: $f(x)=x^2$ is locally Lipschitz on $\mathbb R$.