# Uniformly differentiable functions

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 $$x+h$$).
• @JamesBaxter of course $L_x(h)=f'(x)h$, how else? So, there is no room for choice in definition for your functions $L_x, r_x$. – Fedor Petrov Jan 22 at 16:01