k = 1

In the special case k = 1, the Wirtinger inequality is a special case of the Cauchy–Schwarz inequality:

${\displaystyle \omega (v_{1},v_{2})=g(J(v_{1}),v_{2})\leq \|J(v_{1})\|_{g}\|v_{2}\|_{g}=1.}$

According to the equality case of the Cauchy–Schwarz inequality, equality occurs if and only if J(v₁) and v₂ are collinear, which is equivalent to the span of v₁, v₂ being closed under J.

k > 1

Let v₁, ..., v_2k be fixed, and let T denote their span. Then there is an orthonormal basis e₁, ..., e_2k of T with dual basis w₁, ..., w_2k such that

${\displaystyle \iota ^{\ast }\omega =\sum _{j=1}^{k}\omega (e_{2j-1},e_{2j})w_{2j-1}\wedge w_{2j},}$

where ι denotes the inclusion map from T into V.^[2] This implies

${\displaystyle \underbrace {\iota ^{\ast }\omega \wedge \cdots \wedge \iota ^{\ast }\omega } _{k{\text{ times}}}=k!\prod _{i=1}^{k}\omega (e_{2i-1},e_{2i})w_{1}\wedge \cdots \wedge w_{2k},}$

which in turn implies

${\displaystyle (\underbrace {\omega \wedge \cdots \wedge \omega } _{k{\text{ times}}})(e_{1},\ldots ,e_{2k})=k!\prod _{i=1}^{k}\omega (e_{2i-1},e_{2i})\leq k!,}$

where the inequality follows from the previously-established k = 1 case. If equality holds, then according to the k = 1 equality case, it must be the case that ω(e_{2i − 1}, e_2i) = ±1 for each i. This is equivalent to either ω(e_{2i − 1}, e_2i) = 1 or ω(e_2i, e_{2i − 1}) = 1, which in either case (from the k = 1 case) implies that the span of e_{2i − 1}, e_2i is closed under J, and hence that the span of e₁, ..., e_2k is closed under J.

Finally, the dependence of the quantity

${\displaystyle (\underbrace {\omega \wedge \cdots \wedge \omega } _{k{\text{ times}}})(v_{1},\ldots ,v_{2k})}$

on v₁, ..., v_2k is only on the quantity v₁ ∧ ⋅⋅⋅ ∧ v_2k, and from the orthonormality condition on v₁, ..., v_2k, this wedge product is well-determined up to a sign. This relates the above work with e₁, ..., e_2k to the desired statement in terms of v₁, ..., v_2k.