Very straight forward question, so I'm studying differentiation between (infinite) normed vector spaces and when considering the very basic example of $f(x)=x^2+2x$ from reals to real we have the usual derivative being $f'(x)=2x+2$. But this isn't a linear map from $\mathbb{R}$ to $ \mathbb{R}$ but rather an affine transformation. The frechet derivative has to be a bounded linear map, but this isn't a linear map, what's going on?


The differential operator is $D: \mathbb{R}^{\mathbb{R}} \to \mathbb{R}^{\mathbb{R}}$, and it's linear, because $\forall f,g \in \mathbb{R}^{\mathbb{R}}$ and $a \in \mathbb{R}$, we have that $D(af+g)=aDf+Dg$. It does not matter if $Df$ is linear function or not.

We call $f: \mathbb{R}^n \to \mathbb{R}^m$ Fréchet differentiable at $a\in (\text{dom}(f))'$ if there exists a linear transformation $L:\mathbb{R}^n \to \mathbb{R}^m$ so that $f$ has the following form: $$f(x)=f(a)+L(x-a)+o(x)$$ And we call $L$ the derivative of $f$ at $a$. In the case of $\mathbb{R}\to\mathbb{R}$ functions, $L$ is just a scalar, and multiplication by a scalar is linear.

  • $\begingroup$ Ahhh yes I forgot I'm evaluating it at $x$ so that's not the frechet derivative, could you clarify what your notation is? I've seen it used before but overlooked it, from what I understand $D$ is a function $ D: \mathbb{R} \rightarrow \mathcal{L}( \mathbb{R}:\mathbb{R}) $ from reals to the set of bounded linear maps from reals to reals $\endgroup$ – Displayname Feb 14 at 11:12
  • $\begingroup$ by notation I just mean what is $R^R$ $\endgroup$ – Displayname Feb 14 at 11:13
  • 2
    $\begingroup$ @Displayname: In your case, the derivative at $a$ is the linear map which takes the real number $h$ to $(2a+2)h$. $\endgroup$ – Hans Lundmark Feb 14 at 11:14
  • $\begingroup$ @Displayname $\mathbb{R}^{\mathbb{R}}$ is just a notation for the set of real functions. $\endgroup$ – Botond Feb 14 at 11:14
  • $\begingroup$ @Displayname See the examples here: en.wikipedia.org/wiki/Function_space $\endgroup$ – Botond Feb 14 at 11:17

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