Prove $|f(y)−f(x)−f′(x)(y−x)|≤ε|y−x|, ∀x,y∈[0,1], |x−y|<δ$.

Question

Suppose f is a differentiable function and f′ is continuous on [a,b],

prove that ∀ε > 0, there is δ > 0 such that $|f(y)−f(x)−f′(x)(y−x)|≤ε|y−x|, ∀x,y∈[0,1], |x−y|<δ$.

Since f is differentiable and f' is continuous, then by using the definitions of both, I can apply MVT to get to this conclusion, however I am having trouble writing it out so that it can be cohesive.

Answer 1

Let $x \in [a, b]$ then since $f(t)$ is differentiable at $x$ it follows that

$\lim_{y \to x} \frac{f(y)-f(x)}{y-x} = f^\prime(x)$

Translating this to $\epsilon, \delta$ we get that given $\epsilon > 0$ there exists a $\delta>0$ such that $|\frac{f(y)-f(x)}{y-x} - f^\prime(x)| < \epsilon$ if $|y - x|< \delta$. Which is equivalent to $|f(y)-f(x) - f^\prime(x)(y-x)| < \epsilon|y-x|$ when $|x-y| < \delta$.

I don't see why we need the mean value theorem here.

Answer 2

Hint: $\frac{|f(y) - f(x) -f'(x)(y-x)|}{|y-x|} = |\frac{f(y) - f(x)}{y-x} - f'(x)|$ To finish use MVT and the continuity of the derivative on a compact interval to attain a global $\delta$.

Answer 3

Apply the triangle inequality and use MVT. Then show that $|f'(c)-f'(x)|\le \epsilon$ for some $c\in(0,1),$ which is true since $f'$ is continuous.