PROBLEM SET 4, 18.155
DUE MIDNIGHT FRIDAY 7 OCTOBER, 2011
Consider the space B(Rn ) of complex Borel measures, defined as the
dual space of C00 (Rn ), the space or bounded continous functions on Rn
which ‘vanish at infinity’:
Given > 0, φ ∈ C00 (Rn ) ∃ R > 0 s.t. |φ(x)| < in |x| > R.
Recall that this is a Banach space with respect to the supremum norm
and hence B(Rn ) is a Banach space with respect to the dual norm. In
the second last question you only need to show weak convergence.
(1) Show that L1 (Rn ) is mapped injectively into B(Rn ) by integration,
Z
µf (φ) = f (x)φ(x), f ∈ L1 (Rn ), φ ∈ C00 (Rn ).
(2)
[A] Show that convolution with an element ψ ∈ Cc0 (Rn ), defined by
µ ∗ ψ(y) = µ(ψ(y − ·)), y ∈ Rn , µ ∈ B(Rn )
0
(Rn ), the space of bounded continuous
gives an element of C∞
functions.
[B] Show that µ ∗ ψ can also be interpreted as an element
of B(Rn ) by proving that for φ continuous of compact support
Z
(µ ∗ ψ(y))φ(y)dy = µ(ψ̌ ∗ φ), ψ̌(x) = ψ(−x).
Hint: You can use without proof the fact that the Riemann
integral of a continuous function of compact support is given
by the limit of the sum over the points at 2−k Zn ⊂ Rn – the
integral lattice.
(3) If χ ∈ Cc∞ (Rn ) has χ(0) = 1 then for a given µ ∈ B(Rn ),
x
µk (φ) = µ(χk φ), φ ∈ C00 (Rn ), χk (x) = χ( )
k
n
is a sequence in B(R ) which (norm) converges to µ.
Hint: Weak convergence should be fairly clear; to get norm
convergence one can use a weak formulation of the norm for
measures. First split a measure into real and imaginary parts, so
1
2
PROBLEM SET 4, 18.155 DUE MIDNIGHT FRIDAY 7 OCTOBER, 2011
we can assume that µ is real on real function. Then, somewhat
less trivially, split into positive and negative parts, where
µ+ (ψ) = sup µ(φ), 0 ≤ ψ, φ ∈ C00 (Rn ).
0≤φ≤ψ
A little work (it is in the old notes somewhere) is required to
show that this extends to be linear where µ+ (ψ) = µ+ (ψ+ ) −
µ+ (ψ− ) in terms of splitting continuous functions into positive
and negative parts. Finally then for such a positive measure
(non-negative on non-negative functions) one can check that
the dual norm
kµ+ k = lim µ+ (χk )
k→∞
(a bounded increasing sequence). Here, χ is chosen to have
0χ ≤ 1 and = 1 near 0 χ(x/j)χ(x/k) = χ(x/k) when j >> k
so (χ(x/k)µ)(χ(x/j)) = µ(χ(x/k)) and it follows from the limit
formula that
kχ(x/k)µk = µ(χ(x/k))
(using positivity). The right side increases to k
muk so
|µ − χ(χ(x/k)µk = kµk − µ(χ(x/k)) → 0.
The case of general χ with χ(x) = 1 near
R zero then follows.
(4) Show that if γ ∈ Cc∞ is non-negative and Rn γ = 1 then
µ ∗ γk (ψ) → µ(ψ), where γk (x) = k n γ(xk), µ ∈ B(Rn )
for each ψ ∈ C00 (Rn ).
0
(5) Show that restriction to S(Rn ) ⊂ C∞
(Rn ) gives an injection of
B(Rn ) into S 0 (Rn ).