A nonzero dual vector is strictly positive on the interior of the cone

Membership in the dual cone $K^{*} = \{y : \langle x, y\rangle \ge 0 \ \text{ for all } x \in K\}$ only guarantees a nonstrict inequality against points of $K$. On the interior of $K$ the inequality is automatically strict, provided the dual vector is nonzero: if $z \in K^{*}$ with $z \ne 0$ and $w \in \operatorname{int} K$, then

The reason is that an interior point can be moved a little against $z$ and stay inside $K$. Suppose $\langle w, z\rangle = 0$. Since $w$ is interior, some ball $B(w, \varepsilon)$ is contained in $K$, so with $\delta = \varepsilon / (2 \lVert z\rVert) > 0$ the point $w - \delta z$ still lies in $K$; its distance to $w$ is $\delta \lVert z\rVert = \varepsilon / 2 < \varepsilon$. But then

contradicting $z \in K^{*}$.

Both hypotheses are needed. For $z = 0$ the pairing vanishes. And on the boundary of $K$ a nonzero element of $K^{*}$ may well pair to zero — that is exactly what a hyperplane supporting $K$ at that point does.

This is the strict form underlying the recorded property that $K^{*}$ is pointed whenever $K$ has nonempty interior. In conic duality it is the step that rules out a degenerate multiplier: a Slater point lies in $\operatorname{int} K$, so pairing it against a nonzero dual vector cannot vanish.