PROBLEM SET 2, 18.155
DUE MIDNIGHT FRIDAY 23 SEPTEMBER, 2011
The ‘Hints’ can be ignored, if anyone finds them intrusive I will move
them to the end.
(1) Let Cc∞ (Rn ) ⊂ S(Rn ) be the subspace of compactly supported
smooth fucntions – those that vanish for |x| > R for some R
(depending on the element of course). Show that this is a dense
inclusion. ‘Hint:’ Use the bump function constructed in Lecture
3 and show that χ( xk )φ(x) → φ(x) in S(Rn ).
(2) Prove that S(Rn ) is a Montel space which means that it has
an analogue of the Heine-Borel property. Namely, (you have to
show that) if D ⊂ S(Rn ) is closed and ‘bounded’ in the sense
that for each N there exists CN such that kφkN ≤ CN for all
φ ∈ D, then D is compact. ‘Hint:’ You may apply the AscoliArzela theorem if you check that a set bounded with respect to
the norm k · k1 is equicontinuous on Rn !
(3) Give a proof of one of the basic steps in the definition of Lebesgue
measure. Namely, by a rectangle in Rn I will mean a set
R = {x ∈ Rn ; aj ≤ xj < bj }
where bj ≥ aj – so the product of semi-open intervals. It has
volume Πj (bj − aj ). Show that if such aSrectangle is written
as
P a countable union of rectanges R = l Rl then vol(R) ≤
vol(Rj ) with equality if Rj ∩ Rk = ∅ for j 6= k. ‘Hint;’ Check
j
finite unions of rectangles can be written as finite unions of disjoint rectangles and hence that the volume is increasing under
inclusion for finite unions. This gives an estimate one way in
the countable case. To get the other estimate, fatten up the Rj ,
a (very) little bit depending on j, to open intervals which cover
the closure of R and use Heine-Borel then the finite case again.
(4) A subset of Rn has measure zero if for any > 0 it is contained
in a countable union of rectangles (as above) with the sum of
the volumes of the rectangles less than . Show that a countable
union of sets of measure zero has measure zero.
(5) Give a proof (I will give one in lectures this week) of the following result. If φj ∈ S(Rn ) is a sequence which is ‘absolutely
1
2
PROBLEM SET 2, 18.155 DUE MIDNIGHT FRIDAY 23 SEPTEMBER, 2011
summable’ in the sense
that
XZ
|φj+1 − φj | < ∞
j≥1
n
then {x ∈ R ;
P
j≥1
|φj+1 (x) − φj (x)| = ∞} has measure zero.