Notice that has classical derivatives on and on , so any weak derivative of must coincide, on these intervals, with the strong derivative, so the only possible choice of derivative is if and for . Now take , we calculate

By the definition of weak derivative, the sum of these two quantities must be zero, so we arrive at . Since was arbitrary we get our contradiction. Notice that the contradiction works only because we allow the support of to be outside the sets where we know the function is differentiable.