How is it valid to go from $\left|\frac{f(y)-f(x)}{y-x}\right|<C$ to $|f(y)-f(x)|\leq C|y-x|$?

Question

Let $A$ be the set of all functions $f$ such that, for some constant $C$, $|f(x)-f(y)|\leq C|x-y|$ for all $y$ in an interval around $x$ (in other words $f$ is Lipschitz of order 1 at $x$).

Question 11-34 (c) from Spivak's Calculus goes somewhat like this

Prove that if $f$ is differentiable at $x$ then $f\in A$.

Answer book says

If $f$ is differentiable at $x$ then $\forall\varepsilon>0\;\exists\delta$ such that $$|y-x|<\delta\Rightarrow\left|\frac{f(y)-f(x)}{y-x}-f'(x)\right|<\varepsilon$$ which means that, for $C=\varepsilon+|f'(x)|$, $$\frac{|f(y)-f(x)|}{|y-x|}<C$$ $$|f(y)-f(x)|\leq C|y-x|$$ $\square$.

This last step seems kinda shady. Since $x\neq y$ (because we have divided by it), how does one prove equality?

Answer 1

You have that $\dfrac{|f(x)-f(y)|}{|x-y|}<C$ for all $x\not=y$.

So $|f(x)-f(y)|<C|x-y|$ for all $x\not=y$.

For $x=y$ we have $|f(x)-f(y)|=|x-y|$ so $|f(x)-f(y)|\leq C|x-y|$ for all $x,y$.

Answer 2

$$\frac {|f(y)-f(x)|}{|y-x|}<C$$ which gives(since you know $y \ne x$) $$|f(y)-f(x)|<C|y-x|$$ "<" is a stronger condition than $"\le"$, so $$|f(y)-f(x)|\le C|y-x|$$

Answer 3

First, if $x\neq y$ then $$\left|\frac{f(y)-f(x)}{y-x}\right|< C \implies \frac{\left|f(y)-f(x)\right|}{\left|y-x\right|}< C \implies \left|f(y)-f(x)\right|<C\left|y-x\right|$$

Next, if $x=y$ then take the limit as $y$ approaches $x$

$$\lim_{y\to x}\left|f(y)-f(x)\right|= \lim_{y\to x}|y-x|=0$$

therefore $$\left|f(y)-f(x)\right|\le C\left|y-x\right|$$ for every $x,y$.