First of all, I'm sorry if I used the wrong tags for this question, please tell me if you want me to change them.
During a reading of a paper, I came across a lemma that is useful for my research (I study the characteristics of a certain function in algebraic geometry) and I tried to prove it without success. Here it is (Lemme 2.7 from B.Lehmann’s paper - "Volume type functions for numerical cycle classes"):
Let $V$ be a finite dimensional $\mathbb Q$-vector space and let $C \subset V$ be a salient full-dimensional closed convex cone (i.e it is a convex cone such that $C - C = V$ and for every $v \in C$, we have $-v \notin C$). Suppose that $f : V \to \mathbb R_{\geq 0}$ is a function satisfying
- $f(e) > 0$ for any $e \in C^{\text{int}}$,
- there is some constant $c > 0$ so that $f(me) = m^cf(e)$ for any $m \in \mathbb Q_{>0}$ and $e \in C$,
- for every $v,e \in C^{\text{int}}$ we have $f(v+e) \geq f(v)$.
Then $f$ is locally uniformly continuous on $C^{\text{int}}$.
All I could observe is that the values taken by elements of the cone C close to $0_V$ must be small by the formula
$$\forall v \in C^{\text{int}}, \forall m \geq 1 : ~~f\left( \dfrac{1}{m}v \right) = \dfrac{1}{m^c} f(v)$$
I tried to formulate a proof in the same style that demonstrates the continuity of a linear application in $0_V$ but I did not have more ideas than that. Any thought is welcome. Thank you!
This answer is somewhat verbose, it might helpt to just draw the corresponding picture for yourself.
We are assuming that $V=\mathbb{Q}^d$ with the subspace topology inherited from the Euclidean topology on $\mathbb{R}^d$ (the OP clarified in the comments that this is the setting of interest to them). Furthermore, we will assume that $d\geq 2$ (the case $d=1$ is trivial as $C$ would be a half-space and $f(x) = \vert x\vert^m f(e)$ for $x\in C^\mathrm{int}$ and $e=1$ if $C=[0,\infty)$ and $e=-1$ if $C=(-\infty,0]$. Clearly $f$ is locally uniformly continuous). In this answer, we denote by $B(x,r)=\{v\in \mathbb{Q}^d \ : \ \vert v-x\vert <r\}$ the ball in $\mathbb{Q}^d$ of radius $r$ centered at $x$.
We want to prove that the map $f$ is locally uniformly bounded on $C^\mathrm{int}$. The only thing which is of quantitative nature in our assumption is the requirement that $$ f(c x) = c^m f(x) $$ for all $c\in \mathbb{Q}_{>0}$ and all $x\in C$. We would like to make use of this to estimate the difference between close points. Namely, if we want to compare $x,y\in C^\mathrm{int}$, then we would like to make a "zigzag" between $x$ and the ray spanned by $y$. More precisely, we would like to find $\varepsilon(x,y)>0$ such that $$ y-(1-\varepsilon(x,y))x, (1+\varepsilon(x,y))x-y\in C^\mathrm{int}. $$ In that case we have by conditions $1$ and $2$ that $$ \left( 1-\varepsilon(x,y) \right)^m f(x) = f((1-\varepsilon(x,y)x) \leq f(y) \leq f((1+\varepsilon(x,y))x) = \left( 1+\varepsilon(x,y) \right)^m f(x). $$ In particular, this implies $$ \vert f(x)-f(y) \vert \leq \left[ \left( 1+\varepsilon(x,y) \right)^m - \left( 1-\varepsilon(x,y) \right)^m \right] f(x). $$
We need to make sure that both $\varepsilon(x,y)$ are locally uniformly controlled and $f(x)$ too.
Let us fix some $e\in C^\mathrm{int}$ and some $0<\delta <\vert e\vert/8$ such that $B(e,4\delta)\subset C^\mathrm{int}$. In particular, this means that for every $x\in B(e,\delta)$ the cone $C_x = \mathbb{Q}_{>0} B(x,2\delta)$ generated by $B(x,2\delta)$ is contained in $C^\mathrm{int}$. There exists $\theta \in (0,\pi/2)$ such that all $C_x$ contain a cone of the form $$ \{ v\in \mathbb{Q}^d \ : \ 0<\langle v,x \rangle \leq\cos(\theta) \vert x \vert \} \subseteq C_x \subseteq C^\mathrm{int}.$$ In particular, this means that if we were in $\mathbb{R}^d$ (instead of $\mathbb{Q}^d$), then we could just use (highschool) geometry to compute $\varepsilon(x,y)$ for $y\in B(x,\delta)$. Let's do this first.
We can rotate everything to reduce to the case where $x=(\vert x \vert, 0, \dots, 0)\in \mathbb{R}^d$ and $y=(y_1, y_2, 0, \dots, 0)\in \mathbb{R}^d$ and $y_1, y_2\geq 0$ (so really we are in $\mathbb{R}^2$). We are asking where we need to start on the ray generated by $(1,0, \dots, 0)$ to reach $y$ by adding only a vector in $C_{(\vert x \vert,0,\dots, 0)}$. The cone $C_{(\vert x \vert,0,\dots, 0)}$ has an opening angle of $\theta$ around $\mathbb{R}_{\geq 0}(1, 0, \dots, 0)$, thus, starting at $( y_1-y_2 \cot(\theta), 0, \dots, 0)$ would do the job. Similarly, if we start at $y$ and need to reach the ray generated by $(1,0, \dots, 0)$ by adding only a vector in $C_{(\vert x \vert,0,\dots, 0)}$, we can choose $$ (y_1+y_2 \cot(\theta), 0, \dots, 0). $$ We have $$ \vert \vert x \vert - (y_1\pm\cot(\theta)y_2)\vert \leq \vert x-y\vert (1+\cot(\theta)).$$ Here we have used that $0\leq y_1\leq \vert x-y\vert$.
All of those computations assumed that we are in $\mathbb{R}^d$. In order to account for the fact that not all those numbers would be in the reals, we just introduce an additional factor of $2$ (if we'd really care about constants, we could drop it by just approximating suitably and noting that everything is well-behaved). Then we one sees that $$ \varepsilon(x,y) = 2 \vert x-y\vert (1+\cot(\theta)) $$ is an admissible choice (recall that $\theta$ is constant in the choosen neighborhood).
So far we have shown that there exists $\theta\in (0, \pi/4)$ such that for all $x,y\in B(e,2\delta)$ we get $$ \vert f(x)-f(y) \vert \leq \left( \left[ 1+2 \vert x-y\vert (1+\cot(\theta)) \right]^m - \left[ 1-2 \vert x-y\vert (1+\cot(\theta)) \right]^m \right) f(x). $$ Using the same argument as before, comparing $x$ to $e$, we get for all $x,y\in B(e,2\delta)$ $$ \vert f(x)-f(y) \vert \leq \left( \left[ 1+2 \vert x-y\vert (1+\cot(\theta)) \right]^m - \left[ 1-2 \vert x-y\vert (1+\cot(\theta)) \right]^m \right) \left( 1+ 2\delta(1+\cot(\theta)) \right) f(e). $$ This proves that $f$ is locally uniformly continuous on $C^\mathrm{int}$ as there exists a constant $C_{\theta,m}$ depending only on $\theta,m$ such that for all $x,y\in \mathbb{R}^d$ holds $$ \left[ 1+2 \vert x-y\vert (1+\cot(\theta)) \right]^m - \left[ 1-2 \vert x-y\vert (1+\cot(\theta)) \right]^m \leq C_{\theta,m} \vert x-y\vert.$$
Note that this proof gives us a slightly stronger result. Namely, we get that $f$ is locally lipschitz on $C^\mathrm{int}$.