PROBLEM SET 4 FOR 18.102, ‘SPRING’ 2015
DUE SAURDAY 28 FEBRUARY, BY 7AM
RICHARD MELROSE
A couple of these problems were a little too hard given the limited experience
you have to date and overall it was a bit long. Not to worry, most people had the
same experience – but remember you CAN ask me for a hint. In view of this, there
is a little more commentary than I would usually put in the proofs and don’t feel
bad if you lost a few marks!
Problem 4.1
Combining the original definition with Lebesgue’s dominated convergence, show
that f : R −→ C is in L1 (R) if and only if there exists a sequence un ∈ C(R) and
F ∈ L1 (R) such that |un (x)| ≤ F (x) a.e. and un (x) → f (x) a.e.
1
By the initial
P R definition, if f ∈ L (R)
P then there exists a sequence vn ∈ Cc (R)
such that
|vn | < ∞ and un =
vk → f a.e. The same definition shows
n
k≤n
P
that F (x) defined as
|vn (x)| when this converges absolutely, and 0 otherwise, is
n
an element of L1 (R) and |un | ≤ F a.e.; this proves that f ∈ L1 (R) satisfies the
given condition. Conversely it there exist a sequence un as in the statement then
un ∈ L1 (R) by applying Lebesgue dominated convergence, since χ[−N,N ] un → un ,
|χ[−N,N ] un | ≤ F a.e. and we know that χ[−N,N ] un ∈ L1 (R). Applying Lebesgue
dominated convergence again for the sequence un shows that f ∈ L1 (R).
Problem 4.2
Define L∞ (R) as the set of functions g : R −→ C such that there exists C > 0
and vn ∈ C(R) with |vn (x)| ≤ C and vn (x) → g(x) a.e. Show that L∞ is a linear
space, that
kgk∞ = inf{sup |g(x)|; E has measure zero and sup |g(x)| < ∞}
R\E
R\E
∞
∞
is a seminorm on L (R) and that this makes L (R) = L∞ (R)/N into a Banach
space, where N is the space of null functions.
Remark: This one was harder than I intended.
Note that if g ∈ L∞ (R) there exists for each n a set of measure zero En such
that
kgk∞ ≤ sup |g(x)| ≤ kgk∞ + 1/n.
R\En
Taking E =
S
n En , still of measure zero, it follows that
kgk∞ = sup |g(x)|
R\E
so the infimum is attained.
1
2
RICHARD MELROSE
Linearity is clear since if vn is a sequence showing that g ∈ L∞ (R) and wn such
a sequence for h then for a, b ∈ C avn + bwn is a sequence which is bounded and
converges a.e. to af + bg – the union of two sets of measure zero being of measure
zero.
If g ∈ L∞ (R) then |g(x)| ≤ C on R \ E where this is the set on which vn (x)
converges to g. Thus kgk is finite, since there is one set E for which the supremum
is defined. It is a seminorn since kzgk∞ = |z|kgk∞ for z ∈ C as this is true for
z = 0 and for z 6= 0, sup |zg| = |z| sup |g| on any set on which either is defined. To
see the triangle inequality for f, g ∈ L∞ (R) choose > 0 and observe that there
are sets of measure zero E and F on the complements of which |f | ≤ kf k∞ + and
|g| ≤ kgk∞ + , so on R \ (E ∪ F ), kf + gk∞ ≤ kf k∞ + kgk∞ + 2. Or use the fact
that the infimum is attained.
Certainly N ⊂ L∞ (R) since vn = 0 works in the definition and kf k∞ = 0 in that
case. The converse is immediate. It follows that L∞ (R) = L∞ (R)/N is a normed
space.
So, to completeness. If [gn ] is a Cauchy sequence in L∞ (R) then for each n there
is a set of measure zero En such that |gn (x)| ≤ kgn k∞ on R \ E
Sn . From the triangle
inequality it follows that gn (x) is Cauchy for all x ∈
/ E, E = n En since
|gn (x) − gm (x)| ≤ kgn − gm k∞ on R \ E.
Thus the limit g(x) = limn gn (x) exists a.e. and if we set g(x) = 0 on the set of
measure zero E then in fact supx∈E
/ |gn (x)−g(x)| → 0. This shows that kgn −gk∞ →
0 and we are done, provided we show that g ∈ L∞ (R).
The trick here is that this is really the ‘measurability’ of g – and I have not
discussed this is lectures. What we can see directly is that for each N, χ[−N,N ] g ∈
L1 (R). Indeed, from the definition of h ∈ L∞ (R) it follows that χ[−N,N ] h ∈ L1 (R)
by applying Lebesgue dominated convergence to the sequence χ[−N,N ] vn which
converges to it a.e. is in L1 (R) and is bounded by Cχ[−N,N ] ∈ L1 (R). Applying
this to the sequence gn above, we see that χ[−N,N ] gn ∈ L1 (R) and by Lebesgue
dominated convergence again it follows that χ[−N,N ] g ∈ L1 (R).
To finish things off we ‘only’ need to prove:Lemma 1. If g : R \ C is such that supR\E |g| < ∞ for some set of measure zero
E and also χ[−N,N ] g ∈ L1 (R) for all N then g ∈ L∞ (R).
It would have been better if I had used something closer to this as the definition.
Proof. So now, from the first problem above we know that for each N there is
a sequence uN,n ∈ C(R) which converges a.e. to χ[−N,N ] g and is bounded by an
L1 function. However, we can replace this sequence by a new one which is 0 if
kgk∞ = 0 or otherwise
(
uN,n (x)
if |uN,n (x)| ≤ kgk∞
UN,n (x) =
uN,n (x)/|uN,n (x)| if |uN,n (x)| > kgk∞ .
This is again a sequence of continuous functions which converges a.e. to χ[−N,N ] g
and which is now bounded by kgk∞ . So the remaining problem is to get one such
sequence which converges to g a.e. on the whole line. To do this we can use a
‘partition of unity’ – this is one point where it was unreasonable of me to expect
PROBLEMS 4
you to be able get there. Look at the explicit function


if |x| ≤ 12
1
3
(1)
φ(x) = 2 − |x| if 12 ≤ |x| ≤


0
if |x| ≥ 32 .
3
3
2
This is just a flat top with legs and is non-negative and continuous. Its nice proper
is that
X
φ(x − k) = 1
k∈Z
where the sum exists because no more than two terms are non-zero at any one point
and is easily checked. So now, set
X
un (x) =
φ(x − k)U|k|+2,n (x).
k
From the discussion above, φ(x − k)U|k|+2,n (x) → φ(x − k)g(x) a.e. – since
U|k|+2,n (x) → g(x) a.e. where φ(x − k) 6= 0. It follows that un ∈ C(R) is a bounded
sequence in the supremum norm which converges a.e. to g so g ∈ L∞ (R).
When I get around to reading your solutions, maybe I will find a better method
than the one I have given!
Problem 4.3
Show that if g ∈ L∞ (R) and f ∈ L1 (R) then gf ∈ L1 (R) and that this defines a
map
L∞ × L1 (R) −→ L1 (R)
which satisfies kgf kL1 ≤ kgkL∞ kf kL1 .
If un is a sequence as in the definition above of g ∈ L∞ (R) and vn is a sequence
as in Problem 1 for f ∈ L1 (R) then un vn is such a sequence for gf since it converges
to it a.e. and is bounded Ra.e. by CFRif C is a bound for the |un |. Again by Lebesgue
dominated convergence, un vn → gf and by restricting to [−N, N ] and passing
to the limit again using dominated convergence we can see that
Z
Z
Z
Z
Z
(2)
| un vn | ≤ C |vn | → C |f | =⇒ |gf | ≤ C |f |.
Now, if kgk∞ > 0, we can replace the original sequence by
(
un (x)
if |un (x)| ≤ kgk∞
Un (x) =
un (x)/kgk∞ if |un (x)| > kgk∞
to get a continuous sequence with C = kgk∞ and the same convergence property
a.e. to g so the estimate follows. The map descends from L∞ × L1 to L∞ × L1 since
the integral vanishes if either factor is a null function.
Problem 4.4
Define a set U ⊂ R to be (Lebesgue) measurable if its characteristic function
(
1 x∈U
χU (x) =
0 x∈
/U
4
RICHARD MELROSE
is in L∞ (R). Letting M be the collection of measurable sets, show
(1) R ∈ M
(2) U ∈ M =⇒ R \ U ∈ M S
∞
(3) Uj ∈ M for j ∈ N then j=1 Uj ∈ M
(4) If U ⊂ R is open then U ∈ M
(1) The sequence un = 1 satisfies the definition to show 1 ∈ L∞ (R).
(2) Since χR\U = 1 − χU this follows from linearity of L∞ (R).
(3) If f and g are real functions in L∞ (R) then max(f, g) ∈ L∞ (R) since the
maximum of the real parts of two approximating sequences is bounded
S and
converges to the maximum a.e. It follows that a finite union Vn = j≤n Uj
of measurable sets is measurable since its characteristic function is the
maximum of the χUj . Now, the χVn form an increasing sequence converging
S
pointwise to χV , V = n Un and we see that χ[−N,N ] χVn is in L1 (R) with
norm bounded by 2N. Thus by montonicity, χ[−N,N ] χV ∈ L1 (R) for all N
and so from the Lemma above, χV ∈ L∞ (R).
(4) Any open set in R is a countable union of intervals. For an interval (a, b) we
know that χ[−N,N ] χ(a,b) ∈ L1 (R) for every N so from the Lemma χ(a,b) ∈
L∞ (R) and hence (a, b) is measurable. The measurability of an general
open set follows from the preceding result.
Problem 4.5
If U ⊂ R is measurable and f ∈ L1 (R) show that
Z
Z
f = χU f ∈ C
U
is well-defined. Prove that if f ∈ L1 (R) then
(R
f
(0,x)
R
If (x) =
− (x,0) f
x≥0
x<0
is a bounded continuous function on R.
By definition above, χU ∈ L∞ (R) and the earlier result shows that χU f ∈ L1 (R)
so the integral over U is well-defined. Thus If (x) is defined since the open intervals
(0, x) or (x, 0) if x < 0 (which is what it was supposed to be) are measurable.
Certainly If (0) = 0. Now, if x ≥ x0 > 0 then
Z
Z
(3)
If (x) − If (x0 ) = (χ(0,x) − χ(0,x0 ) )f = χ(x0 ,x) f
so using the inequality proved earlier,
(4)
|If (x) − If (x0 )| ≤
Z
χ(x0 ,x) |f |.
As xn ↓ x0 along any sequence χ(x0 ,xn ) f → 0 a.e. and is always bounded by |f | so
by Lebesgue dominated convergence, If (x) → If (x0 ). This proves the continuity of
If (x) at each x0 > 0 and the same argument works for x0 = 0 and x0 < 0.
If I had the signs right, If (x) would have been the indefinite integral of f and
you would have seen that an L1 function has a continuous indefinite integral.
PROBLEMS 4
5
Problem 4.6 – Extra
Recall (from Rudin’s book for instance) that if F : [a, b] −→ [A, B] is an increasing continuously differentiable map, in the strong sense that F 0 (x) > 0, between
finite intervals then for any continuous function f : [A, B] −→ C, (Rudin shows it
for Riemann integrable functions)
Z b
Z B
f (F (x))F 0 (x)dx.
f (y)dy =
(5)
A
a
Prove the correspondng identity for every f ∈ L1 ((A, B)), which in particular
requires the right side to make sense.
Problem 4.7 – Extra
Show that if f ∈ L1 (R) and If in Problem 4.5 vanishes identically then f ∈ N .
Department of Mathematics, Massachusetts Institute of Technology
E-mail address: rbm@math.mit.edu