2
$\begingroup$

Suppose we have a sequence of functions $\{ f_n \}$: $\mathbb{R}^p \to \mathbb{R}^q$ differentiable at $x$.

Let $x_n \to x$, and consider the Taylor expansion

$f_n(x_n) = f_n(x) + \nabla f_n(x)(x_n-x) + R_n$

Under which conditions do we have $R_n = o(|x_n - x|)$ (Peano form for the remainder)?

Additionally, if the second derivative is defined in a neighborhood of $x$, under which conditions do we have $R_n = O(|x_n - x|^2)$ (Mean-value form for the remainder) ?

My intuition is that the remainder form is correct if the family $\{\nabla f_n\}$ is equicontinuous at $x$, but I'm not sure how I would prove this.

$\endgroup$

1 Answer 1

0
$\begingroup$

For the first part, what is needed is "equidifferentiability", as defined in

Moore, R.H., 1966. Differentiability and convergence for compact nonlinear operators. Journal of Mathematical Analysis and Applications, 16(1), pp.65-72.

With equidifferentiability at a point $x_0$, you have $\max_n(|R_n|) = o(|x_n-x_0|)$, which let you do the Taylor expansion required.

Alternatively, we can have equicountinuity of $\{ \nabla f_n(x)\}$ in a $r$-neighborhood of $x_0$. Proof here.

For the second part, if the second derivative $\{ \nabla^2 f_n(x)\}$ exists in a neighborhood $U_x$ of $x_0$ and $max_{x \in U_x}\nabla^2 f_n(x) = O(1)$ as $n \to \infty$, then we have $R_n = O(|x_n-x_0|^2)$ from the mean-value theorem applied to each $q$ component of $f_n$ separately.

Again, this can be strengthened to $R_n = (x_n - x_0)^T\nabla^2 f_n(x_0)(x_n - x_0) + o(|x-x_n|^2)$ if we assume equidifferentiability of the second derivative at a point $x_0$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .