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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?

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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 $).

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  • $\begingroup$ 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

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