1

https://en.wikipedia.org/wiki/Summability_kernel

The third condition of the definition involving $\delta$, I don't understand what it is trying to say. Why would the integral go to 0, when condition 1 says otherwise?

MinYoung Kim
  • 1,015

1 Answers1

1

What it is saying is that the mass concentrates around 0. For large enough $n$, basically all the function is in the region $|t|<\delta$, i.e.

$$1 = \int_{\mathbb T} k_n(t) \ dt \approx \int_{|t|<\delta} k_n(t) \ dt$$

Check the example of the Fejer kernels. The picture from that page: enter image description here As $n$ increases, the function becomes taller, but also most of the integral comes from a small region around 0, in a way that the total area is fixed (condition 1)

Calvin Khor
  • 34,903
  • ohhh, so the integral bound is saying as $n \rightarrow 0$, for the region $\delta < |x| < a$, there is not a lot of "mass" in that region (as it is now centered around x = 0). Makes sense now, thanks! – MinYoung Kim Jul 10 '19 at 11:34
  • @MinYoungKim you're welcome – Calvin Khor Jul 10 '19 at 11:34