The Real And Complex Number Systems

n
For the part 1 2
0
that
x sint
t
dt :Since sint
0 t
dt converges, there exists a positive M 1 such
0
n 1 2
x sin t
t
dt ≤ M 1 for all x ∈ I and for all n.
For the part 0
x t−2sin t 2
2t sin t 2
sinn 1 2 tdt
: Consider
0
x t − 2sin t 2
2t sin t 2
sin n 1 2
t dt
≤
0
x t − 2sin t 2
2t sin t 2
dt since t − 2sin t 2 0onI
≤
0
t − 2sin t 2
2t sin t 2
dt : M 2 since lim
t→0
t − 2sin t 2
2t sin t 2
Hence,
|F n x| ≤ M 1 M 2 for all x ∈ I and for all n.
2
So, F n x is uniformly bounded on I. It means that F n x is uniformly bounded on R.
In addition, since
F n x
0
n 1 2
fixed x ∈ I, wehave
x sin t
t
dt
0
x t − 2sin t 2
2t sin t 2
sin t
0 t
dt exists.
and by Riemann-Lebesgue Lemma, in the text book, pp 313,
0.
sin n 1 2 t dt − x 2 ,
x t − 2sin
lim n→
t 2
0 2t sin t sin n 1 t dt 0.
2
2
So, we have proved that
lim n→
F n x sin t
0 t
dt −
2 x where x ∈ 0, .
Hence, F n x is pointwise convergent on I. It means that F n x is pointwise
convergent on R.
Remark: (1) For definition of being boundedly convergent on a set S, the reader can
see the text book, pp 227.
(2) In the proof, we also shown the value of Dirichlet Integral
sin t
0 t
dt 2
by letting x .
(3) There is another proof on uniform bound. We write it as a reference.
Proof: The domain that we consider is still 0, . Let 0, and consider two cases as
follows.
(a) x ≥ 0 : Using summation by parts,