Prove that (Clarke's) generalized directional derivative for locally Lipschitz function is subadditive

Question

I'm reading the proof on link(doi: 10.1.1.145.6632), lemma 2.6, on subadditivity of (Clarke's) generalized directional derivative for locally Lipschitz functions. The generalized directional derivative for $f:\mathbb{R}^n \to \mathbb{R}$ and $x_0 \in \mathbb{R}^n$ is defined as

$$f^{o}\left(x_{0} ; d\right):=\limsup _{h \rightarrow 0,\lambda \rightarrow 0} \frac{f\left(x_{0}+h+\lambda d\right)-f\left(x_{0}+h\right)}{\lambda}$$

In the proof of subadditivity of $f$ where $f$ is locally Lipschitz, i.e., prove that

$$f^{o}\left(x_{0} ; d+d'\right) \leq f^{o}\left(x_{0} ; d\right) + f^{o}\left(x_{0} ; d'\right)$$

I'm confused by why

$$\limsup _{h \rightarrow 0, \lambda \rightarrow 0} \frac{f\left(x_{0}+h+\lambda d+\lambda d^{\prime}\right)-f\left(x_{0}+h+\lambda d^{\prime}\right)}{\lambda}$$

is equal to $f^{o}\left(x_{0} ; d\right)$. I have tried to prove it using the Lipschitzness of $f$, but couldn't figure out (Lipschitzness will give something like $K \lambda \|d\|$ but the $\lambda$ will be cancelled by the denominator, so I might be missing something). Thanks!

Answer 1

Let $h'=h+\lambda d'$, then when $h\to 0,\lambda\to 0$ we have $h'\to 0$. Thus, \begin{align} &\limsup _{h \rightarrow 0, \lambda \rightarrow 0} \frac{f\left(x_{0}+h+\lambda d+\lambda d^{\prime}\right)-f\left(x_{0}+h+\lambda d^{\prime}\right)}{\lambda} \\ = &\limsup _{h' \rightarrow 0, \lambda \rightarrow 0} \frac{f\left(x_{0}+h'+\lambda d\right)-f\left(x_{0}+h'\right)}{\lambda}\\ = &f^{o}\left(x_{0} ; d\right). \end{align}