Uniform continuity and differentiation.

Question

For (a) I have said

By MVT:

$|f(x)-f(y)|\le K|x-y|$

Choose $\delta=\epsilon/K$

$|f(x)-f(y)|=|(f(x)-f(y))/(x-y)||x-y|=|f'(c)||x-y|\le K|x-y| \le K\delta=\epsilon$

For (b) I have said

$f'(x)=-4x^3/(1+x^4)^2$ so $|f'(x)| \le B$ where be is some B in the real numbers, eg I believe 2 would be a suitable choice for B.

Any help would be appreciated, thank you.

Answer 1

(a) Be careful, MVT does not say $$|f(x) - f(y) | \leq K|x-y|.$$ It says there exists $c \in \mathbb{R}$ (in this case) such that $$|f(x) - f(y) = |f'(c)||x-y|.$$

(b) The maximum of $f'(x)$ occurs at $x = -\sqrt[4]{3}$, so $| f'(x)| \leq 3^{3/4} $. Use the standard tools in calculus to find this maximum.

Answer 2

In (a), using MVT we know that given $[a, b] \subset \mathbb{R}$, for any $x, y \in [a, b]$, $\exists c \in [a, b]$ such that $|f(x)-f(y)| = |f'(c)| |x-y|$. Since $|f'(c)|<K$, we conclude that $|f(x)-f(y)| \leq K |x-y|$ for any $x, y \in \mathbb{R}$.

Therefore, $f$ is Lipschitz continuous in $\mathbb{R}$, and therefore uniformly continuous.