Let $V$ be a real vector space and $f: V \to \mathbb{R}$ be a convex function with $f(0) \geq 0$. I'm trying to show that
$$\forall v \in V:\inf_{t > 0} \frac{f(tv)}{t} > -\infty$$
Fix $v\in V$. We want to show that
$$\{\frac{f(tv)}{t}: t > 0\}$$
is bounded below.
I was not able to find a lower bound. I can find a lower bound for $1 \leq t <\infty$, but I'm still looking for a lower bound for $0 < t < 1$.
I think that $f(0) \geq 0$ will be crucial for this.
For $v \in V$ and all $t > 0$ we have $$ 0 \le f(0) = f\left( \frac{t}{t+1}(-v) + \frac{1}{t+1}(tv)\right) \le \frac{t}{t+1} f(-v) + \frac{1}{t+1}(tv) $$ which implies the lower bound $$ \frac{f(tv)}{t} \ge -f(-v) $$ and therefore $$ \inf_{t > 0} \frac{f(tv)}{t} \ge -f(-v) > - \infty \, . $$