It won't destroy the perturbation. There is a low dimensional manifold along which the cost function will decrease/increase. The adversarial perturbation lies o

It won't destroy the perturbation. There is a low dimensional manifold along which the cost function will decrease/increase. The adversarial perturbation lies on that manifold, but random white noise (which is a random high dimensional vector) will have close to zero length with high probability when projected onto the manifold and hence won't affect the cost function.