I have been reading “A course on rough paths’’ by Peter Friz and Martin Hairer, and at one point a gnome is defined with the following properties. I am finding it hard to prove subadditivity. I presume the hint is incorrect, as the function $N(\cdot)$ does not seem to be subadditive. Would anybody be able to help me prove this result?
- Identifying $1, b, c$ with the elements $(1,0,0),(0, b, 0),(0,0, c) \in T^{(2)}(V)$, we may write $(1, b, c)=1+b+c$. The resulting calculus is familiar from formal power series in non-commuting indeterminates. For instance, the usual power series $(1+x)^{-1}=$ $1-x+x^2-\ldots$ leads to, omitting tensors of order 3 and higher, $$ \begin{aligned} (1+b+c)^{-1} &=1-(b+c)+(b+c) \otimes(b+c) \\ &=1-b-c+b \otimes b, \end{aligned} $$ allowing us to recover (2.9). We also introduce the dilation operator $\delta_\lambda$ on $T^{(2)}(V)$, with $\lambda \in \mathbf{R}$, which acts by multiplication with $\lambda^n$ on the $n$th tensor level $V^{\otimes n}$, namely $$ \delta_\lambda:(a, b, c) \mapsto\left(a, \lambda b, \lambda^2 c\right) . $$ Having identified $T_1^{(2)}(V)$ as the natural state space of (step-2) rough paths, we now equip it with a homogeneous, symmetric and subadditive norm. For $\mathbf{x}=(1, b, c)$, $$ \|\mathbf{x}\| \stackrel{\text { def }}{=} \frac{1}{2}\left(N(\mathbf{x})+N\left(\mathbf{x}^{-1}\right)\right) \quad \text { with } N(\mathbf{x})=\max \{|b|, \sqrt{2|c|}\}, $$ noting $\left\|\delta_\lambda \mathbf{x}\right\|=|\lambda|\|\mathbf{x}\|$, homogeneity with respect to dilation, and $\left\|\mathbf{x} \otimes \mathbf{x}^{\prime}\right\| \leq$ $\|\mathbf{x}\|+\left\|\mathbf{x}^{\prime}\right\|$, a consequence of subaddivity for $N(\cdot)$ which requires a short argument left to the reader. It is clear that $$ \left(\mathbf{x}, \mathbf{x}^{\prime}\right) \mapsto\left\|\mathbf{x}^{-1} \otimes \mathbf{x}^{\prime}\right\| \stackrel{\text { def }}{=} d\left(\mathbf{x}, \mathbf{x}^{\prime}\right) $$*
Writing this up, I realised that $N(\cdot)$ is indeed subadditive. The trick in the proof is to realise that the authors define a compatible metric on the tensor product to be such that $|a\otimes b|\le |a||b|$, then it is just a matter of algebra.
Once the $N$ function is proven to be subadditive, it is trivial that the norm $|||\cdot|||$ is subadditive.