Where does this inequality come from: $\frac{|f(x) - f(x-h)|}{h} \le ||f'||_{L^\infty(x, x-1)}$?

Question

I came across this inequality today: $$\frac{|f(x) - f(x-h)|}{h} \le ||f'||_{L^\infty(x, x-1)}$$

I realise if we let $h \to 0$ we obtain the derivative on the left hand side so I can see it has something to do with that. But what's the significance of the right hand side and why is the $L^\infty$ norm used?

Answer 1

Assuming continuity and differentiability of $f$:

This comes from the Mean Value Theorem: If $f$ is continuous on $[a,b] \in \mathbb{R}$:

$$\exists c \in [a,b]: f'(c)=\frac{f(b)-f(a)}{b-a}$$

Now:

$||f'||_{L^{\infty}}$ is the infinity norm, which is the supremum value of $|f'|$ on some interval.

So, your formula is saying that the absolute value of the secant's slope about $x$ is less than or equal to the maximum absolute value of the derivative over the same interval $x\pm h$.