[Math] Definition of Frechet derivative

banach-spacesfunctional-analysis

Let $f: U \to Y$ be a map from $U \subset X$ open, $X,Y$ Banach spaces. Then the Frechet derivative at $x_0 \in X$ is defined as follows:

$f$ is called Frechet differentiable at $x_0$ if there exists a linear map $A: X \to Y$ such that the following limit is $0$:

$$ \lim_{x \to x_0}{\|f(x) – f(x_0) – A(x-x_0) \|\over \|x – x_0\|}$$

What is the role of the open subset and why is it necessary that $A$ be defined on the whole space? Concretely, is it possible to define the Frechet derivative equivalently as follows:

Let $f: X \to Y$ be a map between Banach spaces. Then $f$ is Frechet differentiable at $x_0$ if there exists a linear map $A: X \to Y$ such that the following limit equals $0$:
$$ \lim_{x \to x_0}{\|f(x) – f(x_0) – A(x-x_0) \|\over \|x – x_0\|}$$

Or equivalently:

Let $X,Y$ be a Banach spaces and $U\subset X$ an open subset. Then $f: X \to Y$ is (Frechet) differentiable on $U$ if for $x_0 \in U$ there exists a linear map $A: U \to Y$ such that the following limit equals $0$:
$$ \lim_{x \to x_0}{\|f(x) – f(x_0) – A(x-x_0) \|\over \|x – x_0\|}$$

Best Answer

  1. To allow functions from an open subspace of $X$ makes the definition more general, and thus applicable for cases where the given function could not be extended in a differentiable way to the whole space (simplest exaple is $x\mapsto 1/x$ which is defined only on an open subset of $X=\Bbb R$, and is differentiable there).
  2. If a linear function $U\to Y$ is defined on an open subset $U$ of the Banach space $X$, then it automatically extends to the whole space by linearity.
  3. However, your question might make sense: among infinite dimensional spaces there are several operators that are defined only on a dense subspace of the domain, and we might consider such 'unbounded' operators as well in the definition.
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