- #1
jostpuur
- 2,116
- 19
Let [tex]f:]a,b[\to\mathbb{R}[/tex] be a differentiable function. For each fixed [tex]x\in ]a,b[[/tex], we can define a function
[tex]
\epsilon_x: D_x\to\mathbb{R},\quad\quad \epsilon_x(u) = \frac{f(x+u) - f(x)}{u} \;-\; f'(x)
[/tex]
where
[tex]
D_x = \{u\in\mathbb{R}\backslash\{0\}\;|\; a < x+u < b\}.
[/tex]
Now we have [tex]\epsilon_x(u)\to 0[/tex] when [tex]u\to 0[/tex] for all [tex]x[/tex], but let us then define a following collection of functions for all [tex]|u|<b-a[/tex].
[tex]
\epsilon_u:E_u\to\mathbb{R},\quad\quad \epsilon_u(x) = \epsilon_x(u)
[/tex]
where
[tex]
E_u = \{x\in ]a,b[\;|\; a < x + u < b\}.
[/tex]
For all [tex]\delta > 0 [/tex] there exists [tex]U>0[/tex] so that [tex]]a+\delta, b-\delta[\subset E_u[/tex] when [tex]|u| < U[/tex]. So now it makes sense to ask, that under which conditions does the collection [tex]\epsilon_u|_{]a+\delta, b-\delta[}[/tex] approach zero uniformly when [tex]u\to 0[/tex], for all [tex]\delta > 0 [/tex]?
For example, could f being continuously differentiable be enough?
[tex]
\epsilon_x: D_x\to\mathbb{R},\quad\quad \epsilon_x(u) = \frac{f(x+u) - f(x)}{u} \;-\; f'(x)
[/tex]
where
[tex]
D_x = \{u\in\mathbb{R}\backslash\{0\}\;|\; a < x+u < b\}.
[/tex]
Now we have [tex]\epsilon_x(u)\to 0[/tex] when [tex]u\to 0[/tex] for all [tex]x[/tex], but let us then define a following collection of functions for all [tex]|u|<b-a[/tex].
[tex]
\epsilon_u:E_u\to\mathbb{R},\quad\quad \epsilon_u(x) = \epsilon_x(u)
[/tex]
where
[tex]
E_u = \{x\in ]a,b[\;|\; a < x + u < b\}.
[/tex]
For all [tex]\delta > 0 [/tex] there exists [tex]U>0[/tex] so that [tex]]a+\delta, b-\delta[\subset E_u[/tex] when [tex]|u| < U[/tex]. So now it makes sense to ask, that under which conditions does the collection [tex]\epsilon_u|_{]a+\delta, b-\delta[}[/tex] approach zero uniformly when [tex]u\to 0[/tex], for all [tex]\delta > 0 [/tex]?
For example, could f being continuously differentiable be enough?