Let $E,F$ be Banach spaces and $U$ be an open subset of $E$ containing $0$. Suppose that $f:U\to F$ is a $C^p$ locally homogeneous of degree $p$ function. In Lang's Differnetial and Riemannian Manifolds it is stated that we then have $$f(x)=\frac{1}{p!}D^pf(0) x^{(p)}, $$ where $x^{(p)}$ is the vector $(x,\ldots, x)$, $p$ times.
I gave it a go to prove this, I proceeded by induction. Let $h_x:\mathbb{R}\to F$ be defined by $t\mapsto t x $, then the homogeneity of $f$ can be stated as: for any $x\in E$ and for $t\in \mathbb{R}$ $$(*)\quad \bigr( f\circ h_x \bigr)(t)=t^pf(x)$$ For $p=1$ the result follows immediately from the chain rule. Now, I am not sure how to follow the induction step, to figure out I tried for $p=2$, so we differentiate the two expression in $(*)$ for $f\circ h_x$ we have first $$D\bigr(f\circ h_x \bigr)(t)=Df\bigr(h(t)\bigr)\circ Dh(t)=Df(tx)\circ h_x $$ But I do not get how to compute $D^2\bigr(f\circ h_x \bigr)$. Could you make this clear for me? The left right hand side can easily be twice differenciated it is the RHS that I do not get.
For fixed $x\in U$ let $g_x,\ell_x:\mathbb{R}\to F$ be defined as $g_x(t)=f(tx)$ and $\ell_x(t)=t^pf(x)$. By the homogeneity of $f$ we have $g_x=\ell_x$, we look at the derivative at $t\in \mathbb{R}$ \begin{align} Dg_x(t)&=D\ell_x(t) \tag{1}\\ Dg_x(t)&=pt^{p-1}f(x) \tag{2}\\ \end{align}
This shows that $Dg_x$ is homogeneous of degree $k-1$, by inductive hypothesis we have $$Dg_x(t)=\frac{1}{(p-1)!}D^{p-1}\bigr(Dg_x\bigr)(0)\;t^{(p-1)}$$ This can be rewritten using $(2)$ and the fact that $D^{p-1}\bigr(Dg_x\bigr)(0)=D^pg_x(0)$ $$pt^{p-1}f(x)=\frac{1}{(p-1)!}D^pg_x(0)\; t^{(p-1)} $$ Let $t=1$ and rearrange $$f(x)=\frac{1}{p!}D^pg_x(0)(1)^{(p-1)} \tag{3}$$ Since $g_x$ is the composition of $f$ with the linear map $t\mapsto tx$ we have by the chain rule and that $D^lT=0$ with $l\geq 2$ if T is a linear maps that $D^pg_x(t)(z_1,\ldots,z_p)=D^pf(tx)(z_1x,\ldots ,z_px)$ this argument finishes the proof, by just evaluating $(3)$ at $1$.