2
$\begingroup$

If $f:\mathbb{R}\to\mathbb{X}$ is a function from the real numbers to any normed vector space (finite or infinite dimension), and $f$ is Gateaux differentiable, is $f$ necessarily Frechet differentiable?

1 Answers 1

1

Yes. In short, what differentiates (pardon the pun) Gateaux and Frechet is that derivatives in Frechet converge uniformly in the direction in the domain, while Gateaux asks only that the directional derivatives converge. Since there is only one `direction' in the domain $\mathbb R$, these notions coincide in your case.

  • 0
    Does this hold when the domain is changed to the complex numbers as well? – 2017-02-10
  • 0
    @AlexError It depends what you mean. Typically complex-derivatives involve limits of the form $\lim_{h \to 0} h^{-1} (f(x +h) - f(x))$, where $h \to 0$ in $\mathbb C$. For to work one must insist that $\mathbb X$ be a complex vector space. In this case, yes, Gateaux and Frechet are equivalent. Is that what you mean? – 2017-02-11
  • 0
    Yes, that's what I mean. Is there a way of proving the equivalence of Gateaux and Frechet differentiation on a vector valued function directly from the limit definitions? – 2017-02-11
  • 0
    @AlexError The crucial observation is that any linear operator $\mathbb R \to \mathbb X$ is of the form $x \mapsto x v$ for some fixed $v \in \mathbb X$. The Frechet derivative is such a linear operator-- the vector corresponding to that operator is precisely the Gateaux derivative. – 2017-02-11