# Derivative being linear

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?

## 1 Answer

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.

• 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 – Displayname Feb 14 at 11:12
• by notation I just mean what is $R^R$ – Displayname Feb 14 at 11:13
• @Displayname: In your case, the derivative at $a$ is the linear map which takes the real number $h$ to $(2a+2)h$. – Hans Lundmark Feb 14 at 11:14
• @Displayname $\mathbb{R}^{\mathbb{R}}$ is just a notation for the set of real functions. – Botond Feb 14 at 11:14
• @Displayname See the examples here: en.wikipedia.org/wiki/Function_space – Botond Feb 14 at 11:17