Note: Here all functions are $\mathbb R \to \mathbb R$. $Id$ denotes the identity function.

Let $g_i$ be a family of functions indexed by some (potentially uncountable) index set $I$. Given a function $f$, we say $g_i$ are uniformly $o(f)$ if for every $e > 0$ there exists some $d > 0$ such that $|g_i(x)| \le |ef(x)|$ for all $i$ in $I$ and $x$ in $B_d (0)$.

Call a function $f$ uniformly differentiable if for every $x$ in $\mathbb R$, $f(x+h) = f(x) + L_x (h) + r_x (h)$ for some linear functions $L_x$ and some uniformly $o(Id)$ functions $r_x$.

When is a differentiable function uniformly differentiable?


At first, we should have $L_x(h) =f'(x) h$ (this is the definition of the derivative, if we forget the uniformness.)

I claim that $f'$ must be uniformly continuous (seen from summing up the relations for $(x, h) $ and $(x+h, - h) $), and this is enough (seen from Lagrange intermediate value theorem $f(x+h) =f(x) +h f'(z) $, with some $z$ between $x$ and $x+h $).

  • $\begingroup$ Won’t the linear function hf’(z) vary with h? This isn’t allowed by the definition of the derivative.. $\endgroup$ – James Baxter Jan 22 at 15:59
  • $\begingroup$ @JamesBaxter of course $L_x(h)=f'(x)h$, how else? So, there is no room for choice in definition for your functions $L_x, r_x$. $\endgroup$ – Fedor Petrov Jan 22 at 16:01
  • $\begingroup$ Ohh sorry I get it now. Very nice.. $\endgroup$ – James Baxter Jan 22 at 16:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.