FejÚr kernel). These are: (1) (that's obviously true); (2) (easy integral, if you express it in polar coordinates); and (3) given and there exists such that whenever and (That last condition essentially says that when z is small, is also small except when (x,y) is close to the origin.)
Once you have checked that those three conditions hold, you can calculate as follows:
You now want to show that for small enough z, that last integral can be made arbitrarily small. To do that, split the plane into two regions and , where r is chosen so that is small whenever (that will require assuming that f is continuous). Then the integral over will be small by (2), and the integral over will be small by (3) provided we assume that |f| is integrable over and provided that z is small enoungh.