Tensor ranks and symmetric tensor ranks are the same for points with low symmetric tensor rank

Fix integers n ≥ 1, d ≥ 2. Let V be an (n + 1)-dimensional vector space over a field with characteristic zero. Fix a symmetric tensor \({T\in S^d(V)\subset V^{\otimes d}}\). Here we prove that the tensor rank of T is equal to its symmetric tensor rank if the latter is at most (d + 1)/2.