Let $u^{}$ a function in $\operatorname{Lip}_{k}(\Omega)$, that is the set of lipschitz continuous functions from $\Omega \subset \mathbb{R}^n$ to $\mathbb{R}$ with lip constant $\le k$. Denote by $\left[u^{}\right]$ the lipschitz constant of $u$.
Let $v \in \operatorname{Lip}(\Omega)$ and for $t \in[0,1]$ define $$ v_{t}=u^{}+t\left(v-u^{}\right) $$ Then I would have to get that $\left[v_t\right] < k$ but I cannot reach this.
What I tried is the following, by subadditivity of Lipschitz constants and homogeneity and by knowing $\left[u^{}\right] < k$:
$$\left[v_t\right] \le \left[u^{}\right] + t\left[v-u^{}\right] < k + t\left[v-u^{}\right]$$
Then I should be able to conclude from the last passage that it is strictly less than $k$. If $t$ is fixed in $(0,1)$ and the lipschitz constant $\left[v-u^{}\right]$ is always non-negative, how can I conclude that $k + t\left[v-u^{}\right] < k$, getting $\left[v_t\right]<k$? Thanks!
Restate $v_t = u + t(v - u)$ as $v_t = t v + (1 - t)u$
Then for $w \in \Omega$ we have $v_t(w) = tv(w) + (1-t)u(w)$
So, $|v_t(w)| = |tv(w) + (1-t)u(w)| \le t|v(w)| + (1-t)|u(w)| \le tk|w| + (1 - t)k|w| = k|w|$
I.e. $\left[v_t\right] \le k$.