How does positive homogeneity of degree $1$ yield linearity? The problem I came across is as follows:
Suppose $E, F$ are Banach spaces over $\mathbb{K}$. Show that if $f\in C^1(E,F)$ (continuously differntiable) is positively homogeneous of degree $1$, then $f\in \mathcal{L}(E,F)$ (bounded linear operator).
I could now get for $x\in E$ such that $\Vert x\Vert = 1$, then $D_x f(0)=f(x)$. But using this couldn't yield the desired '$f(\lambda s+\mu t)=\lambda f(s)+\mu f(t)$'. So what to do next?
 A: By homogeneity we have $f(0)=0$. Denote with $D\in \mathcal{L} \left( E,F \right)$ the derivative of $f$ at $0$.
Then for each $x\in E$, with $x\neq 0$, we have
$$
\begin{align}
0 & = \lim_{\lambda\rightarrow0^+}\frac{\left\Vert f\left(\lambda x\right)-f\left(0\right)-D\left(\lambda x\right)\right\Vert }{\left\Vert \lambda x\right\Vert } \\
& = \lim_{\lambda\rightarrow0^+}\frac{\left\Vert \lambda f\left(x\right)-\lambda Dx\right\Vert }{\lambda\left\Vert x\right\Vert } \\
& = \lim_{\lambda\rightarrow0^+}\frac{\left\Vert f\left(x\right)-Dx\right\Vert }{\left\Vert x\right\Vert } \\
& = \frac{\left\Vert f\left(x\right)-Dx\right\Vert }{\left\Vert x\right\Vert }
\end{align}
$$
so that $\left\Vert f\left(x\right)-Dx\right\Vert =0$ that implies $f\left(x\right)=Dx$.
This equality holds even for $x=0$. Hence $f=D\in\mathcal{L}\left(E,F\right)$.
A: Hint: Show that $Df$ is homogeneous of degree $0$.
Full answer: We have for $r > 0$ and $x \in E$ that $f(rx) = rf(x)$. Hence $Df(rx)r = rDf(x)$, so $Df(rx) = Df(x)$. Hence $Df$ is homogeneous of degree 0, i.e. $Df$ is constant. The fundamental theorem of calculus shows that $$f(x) = \int_{0}^{1}Df(tx)x\,dt = Df(0)x.$$
