Suppose $f:\mathbb R ^n \rightarrow \mathbb R$ is convex and not constant. How can I show that $f(tz)\rightarrow \infty$ as $t\rightarrow \infty$ for some $z\in \mathbb R ^n$?
This seems very intuitive because $f$ is defined on the entire space, but I just could not give any rigorous proof. Any ideas?
Choose some $z_0 \in \Bbb R^n$ with $f(z_0) \ne f(0)$. Then $$ h: \Bbb R \to \Bbb R , h(t) = f(tz_0) $$ is convex with $h(1) \ne h(0)$.
For $t > 1$ and for $t < 0$ the graph of $h$ is above the line joining $(0, h(0))$ and $(1, h(1))$: $$ \tag{1} h(t) \ge h(0) + t \bigl (h(1) -h(0) \bigr ) \, $$ therefore $$ \lim_{t \to \infty} h(t) = \infty \text{ or } \lim_{t \to -\infty} h(t) = \infty \, . $$ It follows that $ \lim_{t \to \infty} f(tz) = \infty $ holds for $z = z_0$ or for $z = -z_0$.
Details for $(1)$: A convex function $h$ satisfies $$ \tag{2} h(x) \le \frac{x_2-x}{x_2-x_1} h(x_1) + \frac{x-x_1}{x_2-x_1} h(x_2) $$ for $x_1 < x < x_2$, i.e. the graph of $h$ lies below the secant joining $(x_1, h(x_1))$ and $(x_2, h(x_2))$.
Now if $x_1 < x_2 < x$ then $(2)$ inequality holds with $x$ and $x_2$ exchanged: $$ h(x_2) \le \frac{x-x_2}{x-x_1} h(x_1) + \frac{x_2-x_1}{x-x_1} h(x) $$ and a simple rearrangement gives $$ \tag{3} h(x) \ge \frac{x_2-x}{x_2-x_1} h(x_1) + \frac{x-x_1}{x_2-x_1} h(x_2) \, , $$ i.e. $(2)$ holds with the reversed inequality sign. In the same way we can derive $(3)$ in the case $x < x_1 < x_2$.
Therefore the graph of $h$ lies above the line joining $(x_1, h(x_1))$ and $(x_2, h(x_2))$ if $x$ is outside of the interval $[x_1, x_2]$.