According to wikipedia, given a locally convex topological vector space $X$ and a functional $f: X \to \mathbb R$ (let's assume $f$ is a real value function here to make things simpler), and a vector $x_0 \in X$, if the mapping\begin{equation}g: v \mapsto \lim_{\tau \to 0}\frac{f(x_0 + \tau v)}{\tau}\end{equation} is defined for every $v \in X$ (i.e. the limit exists for every $v$), we call $g$ the Gauteaux derivitive of $f$ at point $x_0$. Wikipedia says Gauteaux derivitive is homogeneous but not necessarily linear.
I have several questions:
- How to prove the homogeneity of Gauteaux derivitive?
- Can we strengthen the assumption on the space $X$ to make Gauteaux derivitive linear? (say assume $X$ is an Euclidean space)
- If we weaken the assumption, say assume $X$ is barely a vector space with scalar field being $\mathbb R$, will the homogeneity still holds? (becauses it seems we don't really need a locally convex TVS to define the limit)
Homogeneous means $g(\lambda u)=\lambda g(u)$.
Clearly, for $\lambda\ne 0$, \begin{align} g(\lambda u) & =\lim_{\tau\to 0}\frac{f(x_0+\tau\lambda u)-f(x_0)}{\tau} \\ & = \lim_{\tau\to 0}\lambda\frac{f(x_0+\tau\lambda u)-f(x_0)}{\lambda\tau} \\ & =\lambda\lim_{\tau\to 0}\frac{f(x_0+\tau u)-f(x_0)}{\tau} \\ & =\lambda g(u). \end{align}
Even in Euclidean spaces Gateaux does not imply linear: $$ f(x,y)=\frac{xy^2}{x^2+y^2}, $$ and $f(0,0)=0$. This function has directional derivatives at $(0,0)$, but it is not differentiable.