Let ψ0(t) = (
e1/t if t < 0
0 if t ≥ 0 and define ψ : Rn → R by ψ(x) = ψ0(|x|2 − 1). Then ψ ∈ Cc∞(Rn) with R
Rnψ > 0.
Next, for each x ∈ Rn, let ρ(x) = Rψ(x)
Rnψ . Then ρ(x) ≥ 0,
Z
Rn
ρ(x)dx = 1, supp ρ = {x ∈ Rn| kxk ≤ 1} = ¯B1(0)
and ρ is called the standard mollifier. For each ² > 0, if we let ρ²(x) := 1
²n ρ(x
²), then ρ² ∈ C∞(Rn),
Z
Rn
ρ²(x)dx = 1, supp ρ² = ¯B²(0)
Definition. Let Ω be an open subset of Rn, and for each ² > 0, let Ω² = {x ∈ Ω | dist (x, ∂Ω) > ²}.
If f : Ω → R be a locally integrable function, then its mollification f² is a function on Ω² defined to by
f²(x) = ¡ ρ²∗ f¢
(x) = Z
Rn
ρ²(x − z) f (z) dz = Z
Ω
ρ²(x − z) f (z) dz for each x ∈ Ω².
Theorem 1.2.1 Let Ω be an open subset of ⊂ Rn, and let f be a locally integrable function on Ω.
Then
(a) f²∈ C∞(Ω²).
(b) lim
²→0f²(x) = f (x) for a.e. x ∈ Ω².
(c) If f ∈ C0(Ω) then f²converges to f uniformly on each compact subset of Ω. Thus, if ∂αf ∈ C0(Ω) then ∂αf² converges to ∂αf uniformly on each compact subset of Ω.
(d) If 1 ≤ p < ∞ and f ∈ Lploc(Ω), t hen f² converges to f in Lploc(Ω).
Proof.
(a) Fix ² > 0 , x ∈ Ω², i ∈ {1, . . . , n}. Since Ω²is open, there exists 0 < δ < ² such that Bδ(x) ⊂ Ω², and Bδ+²(x) ⊂ Ω²−δ ⊂⊂ Ω., Thus, for any 0 < |h| < δ, x + hei ∈ Ω² and
∂f²
∂xi(x) = lim
h→0
f²(x + hei) − f²(x)
h = lim
h→0
1
²n Z
Ω
1 h
£ρ¡x + hei− y
²
¢− ρ¡x − y
²
¢¤f (y) dy
= lim
h→0
1
²n Z
Ω²−δ
1 h
£ρ¡x + hei− y
²
¢− ρ¡x − y
²
¢¤f (y) dy = Z
Ω
∂ρ²
∂xi(x − y) f (y) dy exists. Similarly, for each x ∈ Ω² and for each multiindex α, one can show that
Dαf²(x) = Z
Ω
Dαρ²(x − y) f (y) dy.
(b) Recall that Lebesgue’s differentiation theorem says that if f : Rn→ R is locally integrable, then for a.e. point x ∈ Rn lim
r→0
1 Vol¡
Br(x)¢ Z
Br(x)
|f (y) − f (x)| dy = 0.
Fix such a point x. Then
lim²→0 |f²(x) − f (x)| = lim
²→0
¯¯
¯¯ Z
B²(x)
ρ²(x − y)£
f (y) − f (x)¤ dy
¯¯
¯¯
≤ lim
²→0
1
²n Z
B²(x)
ρ¡x − y
²
¢ |f (y) − f (x)| dy
≤ lim
²→0
C Vol¡
Br(x)¢ Z
B²(x)
|f (y) − f (x)| dy = 0
(c) Given V ⊂⊂ Ω, we choose W such that
V ⊂⊂ W ⊂⊂ Ω and note that f is uniformly continuous on W. Thus,
limr→0
1 Vol¡
Br(x)¢ Z
Br(x)
|f (y) − f (x)| dy = 0 for all x ∈ Ω and the convergence is uniform on V.
Using this and the proof in the preceding part, we conclude that the convergence of lim²→0 f²(x) = f (x)
is uniform on V.
(d) See e.g. Appendix C in the book “Partial Differential Equations” by Lawrence C. Evans.
1.3. Distributions
(a) Example 1. Let f ∈ L1loc(Rn). Then we can define a distribution f : Cc∞(Rn) → R by hf , φi =
Z
Rn
f (x) φ(x) dx ∀ φ ∈ Cc∞(Rn).
It is easy to check that f ∈ D0(Rn), since for each compact set K, by letting C = R
K |f | and N = 0, we have
|hf , φi| ≤¡Z
K
|f |¢ sup
K
kφk = C X
|α|≤N
sup
K
|φ| ∀ φ ∈ Cc∞(Rn) with supp(φ) ⊆ K.
Similarly, for any multi-index α, one can define Dαf ∈ D0(Rn) by hDαf , φi = (−1)α
Z
Rn
f (x)¡ Dαφ¢
(x) dx ∀ φ ∈ Cc∞(Rn).
Then, for each compact set K, by taking C =R
K |f | and N = |α|, we have
|hDαf , φi| ≤¡Z
K
|f |¢ sup
K
kDαφk = C X
|β|≤N
sup
K
|∂βφ| ∀ φ ∈ Cc∞(Rn) with supp(φ) ⊆ K.
(b) Example 2 (Dirac distribution). Let δ and δy : Cc∞(Rn) → R be defined by hδ , φi =
Z
Rn
δ(x) φ(x) dx = φ(0) hδy, φi =
Z
Rn
δ(x − y) φ(x) dx = φ(y)
for each φ ∈ Cc∞(Rn). Then, for each compact set K, by taking C = 1 and N = 0, we have
|hδy, φi| = |φ(y)| ≤ C X
|α|≤N
sup
K k∂αφk ∀ φ ∈ Cc∞(Rn) with supp(φ) ⊆ K.
(c) For any n ∈ {0, 1, 2, . . .}, let δan+1 : Cc∞(R) → Rbe defined by δn+1a ¡
φ¢
= δn+1(x − a)¡ φ¢
=¡ Dnφ¢
(a).
Then
T :=
X∞ n=0
δnn+1 = X∞ n=0
δn+1(x − n) ∈ D0(R),
since for each compact set K, there exists m ∈ N such that K ⊆ [−m , m] and
|T (φ)| = | Xm n=0
¡Dnφ¢ (n)| ≤
Xm n=0
kDnφk∞ ∀ φ ∈ Cc∞(R) with supp(φ) ⊆ K.
Note that there does not exist a f ∈ L1loc(R) such that
| Xm n=0
¡Dnφ¢
(n)| = |T (φ)| = | Z
[−m,m]
f (x) φ(x) dx|
≤ ¡Z
[−m,m]
|f |¢ sup
[−m,m]
|φ| ∀ φ ∈: Cc∞(R) with supp φ ⊆ [−m, m], ∀m ≥ 0.
(d) (Cauchy principal part integral). Define P (1
x) on Cc∞(R) by hP (1
x) , φ(x)i := lim
²→0+
Z
|x|≥²
1
xφ(x) dx ∀φ ∈ Cc∞(R).
This is well defined since Z
|x|≥²
1
xφ(x) dx = Z ∞
²
φ(x) − φ(−x)
x dx
with
φ(x) − φ(−x)
x ∈ Cc∞((0, ∞)) and
limx→0+ φ(x) − φ(−x)
x = limx→0+ £φ(x) − φ(0)
x + φ(−x) − φ(0) (−x)
¤ = 2φ0(0).
Observe that for x > 0
¯¯
¯¯φ(x) − φ(−x) x
¯¯
¯¯ =
¯¯
¯¯1 x
Z x
−x
φ0(t) dt
¯¯
¯¯ ≤ 1 x
Z x
−x
|φ0(t)| dt ≤ 2 kφ0k∞. This implies that
|hP (1
x, φ(x)i| =
¯¯
¯¯ Z 1
0
φ(x) − φ(−x)
x dx +
Z ∞
1
φ(x) − φ(−x)
x dx
¯¯
¯¯
≤ Z 1
0
2 kφ0k∞dx + Z ∞
1
{|φ(x)| + |φ(−x)|} x dx x2
≤ 2 kφ0k∞+ 2kx φ(x)k∞ Z ∞
1
dx x2
≤ 2 kφ0k∞+ 2mkφ(x)k∞ ∀ φ ∈ Cc∞(R) with supp φ ⊆ [−m, m]
≤ C X
|α|≤1
k∂αφk∞ for some ∀ φ ∈ Cc∞(R) with supp φ ⊆ [−m, m],
where C = max{2 , 2m}. Hence, P (1
x) ∈ D0 1(R).
Proof of Theorem 1.3.3 A distribution u ∈ D0(X), where 0 ≤ m < ∞, has a unique extension to a linear form on Ccm(X) which is sequentially continuous. Conversely, the restriction of a sequentially continuous linear form on Ccm(X) to Cc∞(X) is a member of D0(X).
Proof. If u is a sequentially continuous linear form on Ccm(X), since Ccm(X) ⊃ Cc∞(X), then, by restricting to Cc∞(X), u ∈ D0(X), i.e. u is a distribution on X.
Suppose that u /∈ D0 m(X), i.e. u is a distribution but not of finite order, then there exists a compact set K ⊂ X such that for each j ∈ {1, 2, . . .}, there exists a φj 6≡ 0 ∈ Cc∞(X) with supp φ ⊂ K and
|hu , φji| > j X
|α|≤j
sup
K |∂αφj| Thus, we have
|hu , φj
jP
|α|≤j supK|∂αφj|i| > 1 and lim
j→∞
φj jP
|α|≤j supK|∂αφj| = 0 in Cc∞(X) ⊂ Ccm(X).
This contradicts to that u is a sequentially continuous linear form on Ccm(X). Hence, u ∈ D0 m(X).
Proof of Theorem 1.5.1 needs to apply the
Lebesgue’s dominated convergence theorem. Let {fn} be a sequence of measurable function on X such that f (x) = lim
n→∞(x) exists for every x ∈ X. If there is a function g ∈ L1loc(X) such that
|fn(x)| ≤ g(x) for each n = 1, 2, . . . , and for all x ∈ X, then f ∈ L1loc(X) lim
n→∞
Z
X
|fn− f | = 0, and lim
n→∞
Z
X
fn= Z
X
f.
Lebesgue’s dominated theorem implies that
n→∞limhfn, φi = hf , φi ∀ φ ∈ Cc∞(Rn) =⇒ lim
n→∞ fn= f in D0(X).
Proof of Theorem 1.5.2: is a direct application of the following
(Principle of uniform boundedness) Let X be a Banach space and let Y be a normed linear space. Let {Tα : X → Y }α∈A be a family of bounded linear operators from X into Y. If for each x ∈ X, the set {Tα(x)} is bounded, then the set {kTαk}α∈A is bounded.
Remark.
(a) The linear operator Tα : X → Y is said to be bounded if there exists Kα ≥ 0 such that kTα(x)kY ≤ KαkxkX for all x ∈ X.
(b) The set {Tα(x)} is said to be bounded if there exists Kx ≥ 0 such that kTα(x)kY ≤ Kx for all α ∈ A .
(c) The operator norm is defined by kTαk = sup
x6=0
kTα(x)kY kxkX . (d) In Theorem 1.5.2, the hypothesis
{uj}j∈N is a family of distributions on X with the property limj→∞ uj(φ) = limj→∞huj, φi exists ∀ φ ∈ Cc∞(X) implies that
{uj}j∈N is a family of bounded linear operators from Cc∞(X) to C with the property that the set {uj(φ) | j ∈ N} is bounded ∀φ ∈ Cc∞(X).
Example. Let {fn} be a sequence of functions on R such that (a) fn(x) ≥ 0 for all x ∈ R.
(b) Z
R
fn(x) dx = 1 for all n ∈ N.
(c) For each a > 0, lim
n→∞
Z
|x|≥a
fn(x) dx = 0.
Then
n→∞lim fn= δ in D0(R), i.e. lim
n→∞hfn, φi = φ(0) ∀φ ∈ Cc∞(R).
Proof. Fix a φ ∈ Cc∞(R) with φ 6≡ 0. For any ² > 0, let η > 0 be chosen such that
|φ(x) − φ(0)| < ²
2 ∀ |x| ≤ η.
Furthermore, by the hypothesis, there exists N ∈ N such that if n ≥ N, then Z
|x|≥η
fn(x) dx < ² 4 kφk∞
. Thus, we have
¯¯
¯¯ Z
R
fn(x) φ(x) dx − φ(0)
¯¯
¯¯ =
¯¯
¯¯ Z
R
fn(x)¡
φ(x) − φ(0)¢ dx
¯¯
¯¯
≤ Z η
−η
fn(x) |φ(x) − φ(0)| dx + Z
|x|≥η
fn(x) |φ(x) − φ(0)| dx
≤ ² 2
Z ∞
−∞
fn(x) dx + Z
|x|≥η
fn(x)¡
2 kφk∞¢ dx
< ²
2 + ²
4 kφk∞
¡2 kφk∞¢
= ² Hence, for all n > N, we find that
¯¯
¯¯ Z
R
fn(x) φ(x) dx − φ(0)
¯¯
¯¯ < ².
This implies that lim
n→∞ fn = δ in D0(R).
Remark. If we replace (c) by the requirement that (c0) lim
n→∞
Z
|x−y|≥a
fn(x) dx = 0 for each a > 0.
Then we have
n→∞lim fn= δy in D0(R).
Example. For each ² > 0, let f²(x) = ²
(x − y)2+ ²2. Then lim
²→0+ f² = π δy in D0(R).
Proof. Clearly, f²(x) ≥ 0 for all x ∈ R.
For (b), it is easy to see, for each ² > 0, that Z
R
f²(x) dx = Z
R
²
(x − y)2+ ²2 dx = Z
R
1
u2+ 1du = tan−1u|−∞∞ = π.
For (c), we check, for each a > 0,
²→0lim+ Z
|x|≥a
f²(x) dx = 2 lim
²→0+
Z ∞
a
²
x2 + ²2 dx = lim
²→0+
¡2 tan−1x
²|∞a ¢
= lim
²→0+
£2¡π
2 − tan−1a
²
¢¤= 0.
Thus, lim
²→0+
1
π f² = δy in D0(R).
Example. For each ² > 0, let f²(x) = 1
x − y + i². Then lim
²→0+ f² = P¡ 1 x − y
¢− iπ δy in D0(R).
Proof. Clearly, we have f²(x) = 1
x − y + i² = x − y − i²
(x − y)2+ ²2 = x − y
(x − y)2+ ²2 − i² (x − y)2+ ²2 Thus, lim
²→0+ f² = P¡ 1 x − y
¢− iπ δy in D0(R).
Proof of Theorem 2.1.2: For each y ∈ X ⊆ Rn, since X is open, we may choose δ > 0 such that B¯4δ(y) ⊂ X.
Thus, for each 0 < ² < δ, and for each x ∈ Bδ(y), f²(x), g²(x) ∈ C∞( ¯Bδ(y)), where f²(x) = ¡
ρ²∗ f¢ (x) =
Z
Rn
ρ²(x − z) f (z) dz = Z
B2δ(y)
ρ²(x − z) f (z) dz g²(x) = ¡
ρ²∗ g¢ (x) =
Z
Rn
ρ²(x − z) g(z) dz = Z
B2δ(y)
ρ²(x − z) g(z) dz.
By hypothesis ∂if = g in D0(X), we have
hf , ∂iφi = −hg , φi for all φ ∈ C∞(X).
It is obvious that ρ²(x − z) ∈ C∞(X), and ∂ρ²(x − z)
∂xi = −∂ρ²(x − z)
∂zi , thus, we have
∂f²
∂xi
(x) = −hf (z) , ∂ρ²(x − z)
∂zi
i = hg(z) , ρ²(x − z)i = g²(x) for each x ∈ ¯B2δ(y).
For each x ∈ ¯Bδ(y), since f² ∈ C∞( ¯B2δ(y)), the Mean Value Theorem implies that for any |h| < δ, there exists xh ∈ ¯B2δ(y) lying on the line segment joining x and x + hei such that
¡f²1(x + hei) − f²(x + h)¢
−¡
f²1(x) − f²(x)¢
= h¡∂f²1
∂xi(xh) − ∂f²
∂xi(xh)¢ . This implies that
¯¯
¯¯f²1(x + hei) − f²1(x)
h −f²(x + hei) − f²(x) h
¯¯
¯¯ =
¯¯
¯¯∂f²1
∂xi
(xh) − ∂f²
∂xi
(xh)
¯¯
¯¯ = |g²1(xh) − g²(xh)| .
Thus, for each η > 0, since f² and g² converge uniformly on the compact set ¯B2δ(y) to f and g, respectively, there exists η1 > 0 such that
(1) if 0 < ² , ²1 < η1 then
¯¯
¯¯f²1(x + hei) − f²1(x)
h −f²(x + hei) − f²(x) h
¯¯
¯¯ = |g²1(xh) − g²(xh)| < η which implies that
¯¯
¯¯f (x + hei) − f (x)
h − f²(x + hei) − f²(x) h
¯¯
¯¯ = |g²1(xh) − g²(xh)| < η
Fix an 0 < ² < η1, since ∂f²
∂xi(x) = g²(x), there exists η2 > 0 such that
(2) if 0 < |h| < η2 then
¯¯
¯¯f²(x + hei) − f²(x)
h − g²(x)
¯¯
¯¯ < η
With this 0 < ² < η1, the uniform convergence of g² implies that
(3) |g²(x) − g(x)| ≤ η
Hence, by combining (1) − (3), we have if 0 < |h| < η2 then
¯¯
¯¯f (x + hei) − f (x)
h − g(x)
¯¯
¯¯ < 3 η Letting h → 0, we have proved that
∂f
∂xi(x) = lim
h→0
f (x + hei) − f (x)
h = g(x)
exists.
Definition. The Borel sets of R is the smallest family of subsets of R with the following properties:
(a) The family is closed under complements.
(b) The family is closed under countable unions.
(c) The family contains each open interval
If {Bα | α ∈ A } is a collection of families satisfying all three properties in the definition, then B := ∩α∈ABα is the smallest such family satisfying these three properties.
Definition. Let B denote the collection of Borel sets of R, and let F be a family of all countable unions of disjoint open intervals, i.e. F is just the family of open sets in R, and let µ¡
∪∞i=1(ai, bi)¢ :=
X∞ i=1
(bi− ai) (which may be ∞.) For each Borel set B ∈ B, define
µ(B) = inf
I∈F , B⊂Iµ(I).
Then the function µ satisfies the following properties.
(a) µ(∅) = 0.
(b) If {Ai}∞i=1 is a collection of mutually disjoint Borel sets, then µ¡
∪∞i=1Ai¢
= X∞
i=1
µ(Ai).
(c) µ(B) = inf{µ(I) | B ⊂ I , I is open } . (d) µ(B) = sup{µ(C) | C ⊂ B , C is closed } .
Definition. A function f is called a Borel function iff f−1¡ (a, b)¢
is a Borel set for all a, b. [It is often convenient to allow our functions to take the values ±∞ on small sets in which case we require f−1¡
{±∞}¢
to be Borel.]
Proposition.
(a) f : R → R is Borel iff for each B ∈ B, f−1(B) ∈ B.
(b) A complex, locally Borel measure µ defined on an open set X ⊆ Rn determines a distribution by
hµ, φi = Z
φ dµ, ∀ φ ∈ Cc∞(X). (1.3.9)
Riesz Theorem. For every bounded linear functional F on a Hilbert space H, here exists a uniquely determined element f ∈ H such that
F (x) = hx, f i ∀ x ∈ H, and kF k = kf k.
Since every distribution u of order 0 is a bounded linear form on Cc∞(X), by Theorem 1.2.1(d), it can be extended to be a bounded linear functional on the Hilbert space H = L2loc(X). By Riesz Theorem, there exists f ∈ L2loc(X) such that u is a distribution of the form (1.3.9), i.e.
hu, φi = hf, φi = Z
φ f dx = Z
φ dµ ∀φ ∈ Cc∞(X).
(Calculus) Theorem. Let {fn}n∈N: I = [a, b] → R. Suppose that (a) ∃ x0 ∈∈ I at which lim
n to∞ fn(x0) exists.
(b) fn0(x) exists for all x ∈ I.
(c) fn0 converges uniformly on I to a function g.
Then fn converges uniformly on I to a function f and f0(x) = g(x) for each x ∈ I.
(Calculus) Theorem. Suppose that the functions f = f (x, t) , ft = ft(x, t) : [a, ∞) × [α, β] → R are continuous in (x, t).
Suppose that the functions F and G on J = [α, β] defined by F (t) =
Z ∞
a
f (x, t) dx = lim
b→∞
Z b
a
f (x, t) dx exists for all t ∈ J, and G(t) =
Z ∞
a
ft(x, t) dx = lim
b→∞
Z b
a
ft(x, t) dx is uniformly convergent on J.
Then F is differentiable on J and F0(t) = G(t) for each t ∈ J.
Dirichlet’s Test Let f be continuous on [a, ∞) × [α, β] and suppose that there exists a constant A such that |Rc
af (x, t)dx| ≤ A for c ≥ a, t ∈ J = [α, β]. Suppose that for each t ∈ J, the function φ(x, t) is monotone decreasing for x ≥ a, and converges to 0 as x → ∞ uniformly for t ∈ J. Then the integral F (t) =R∞
a f (x, t)φ(x, t)dx converges uniformly on J.
Examples.
(a) Let f (x, t) = cos tx
1 + x2 for x ∈ [0, ∞) and t ∈ (−∞, ∞). Then R∞
0 f (x, t) converges uniformly for t ∈ R by Dominated Convergence Theorem.
(b) Let f (x, t) = e−xxt for x ∈ [0, ∞) and t ∈ [0, ∞). For any β > 0, the integral R∞
0 f (x, t) converges uniformly for t ∈ [0, β] by Dominated Convergence Theorem. Similarly, the Laplace transform of xn, n = 0, 1, 2, . . . , defined by L {xn}(t) = R∞
0 xne−txdx also converges uniformly for t ≥ γ > 0 to n!
tn+1. For t ≥ 1, define the gamma function Γ by Γ(t) = R∞
0 xt−1e−xdx. Then it is uniformly convergent on an interval containing t. Note that Γ(t + 1) = tΓ(t) and hence Γ(n + 1) = n! for any n ∈ N.
(c) Let f (x, t) = e−txsin x for x ∈ [0, ∞) and t ≥ γ > 0. Then the integral F (t) = R∞
0 e−txsin xdx is converges uniformly for t ≥ γ > 0 by Dominated Convergence Theorem and it is called the laplace transform of sin x, denoted by L {sin x}(t). Note that an elementary calculation shows that L {sin x}(t) = 1
1 + t2. (d) Let f (x, t, u) = e−txsin ux
x for x ∈ [0, ∞) and t, u ∈ [0, ∞). By taking φ = e−tx/x and by applying the Dirichlet’s test, one can show that R∞
γ f (x, t, u) converges uniformly for t ≥ γ ≥ 0. Note that if F (t, u) = L {sin ux
x }(t) =R∞
0 e−txsin ux
x dx, then ∂F
∂u(t, u) =R∞
0 e−txcos uxdx = t t2+ u2 and F (t, u) = tan−1 u
t. By setting u = 1 and by letting t → 0+, we obtain thatR∞
0
sin x
x dx = π 2. (e) Let G(t) =R∞
0 e−x2−t2/x2dx for t > 0. Then G0(t) = −2G(t) and G(t) =
√π 2 e−2t. (f) Let F (t) =R∞
0 e−x2cos txdx for t ∈ R. Then F0(t) = −t
2F (t) and F (t) =
√π 2 e−t2/4. 2.4. Primitives in D0(R): Let v ∈ D0(R).
To solve
∂u = v in D0(R) is equivalent to find a u ∈ D0(R) such that
hu , ∂φi = −hv , φi, for all φ ∈ Cc∞(R).
Define a (continuous) map µ : Cc∞(R) → Cc∞(R) by µφ(x) =
Z ∞
x
¡φ(t) − h1 , φi φ0(t)¢
dt, where φ0 ∈ Cc∞(R) with Z
R
φ0 = 1.
Let u be a linear form on Cc∞(R) defined by
hu , φi = hv , µφi + hC , φi, where C = hu , φ0i.
For each compact set [−a, a] ⊂ R, since there exists b ≥ 0 such that [−a, a] ∪ supp φ0 ⊆ [−b, b], it is easy to show that
(a) µφ ∈ Cc∞(R). In fact the supp µφ ⊆ [−b, b] for all φ ∈ Cc∞([−a, a]).
(b) sup |µφ| ≤ 2a¡
1 + 2b sup |φ0|¢
sup |φ|.
(c) sup |∂kµφ| ≤ sup |∂k−1φ| + 2a sup |∂k−1φ0| sup |φ|, for each k ≥ 1.
(d) µφ0 = 0 and µ∂φ(x) = −φ(x).
(e) u ∈ D0(R) and u is a solution of ∂u = v in D0(R), i.e. hu , ∂φi = −hv , φi for all φ ∈ Cc∞(R).
Example 1. Show that h 1
x2 , φ(x)i = Z ∞
0
φ(x) + φ(−x) − 2φ(0)
x2 dx ∀φ ∈ Cc∞(R).
Proof. For each φ ∈ Cc∞(R), we have h 1
x2 , φ(x)i = h− d dx
¡ 1 x
¢, φ(x)i
= h1
x, φ0(x)i
= lim
²→0+
·Z ∞
²
φ0(x) − φ0(−x)
x dx
¸
= lim
²→0+
·φ(x) x
¯¯
¯¯
∞
²
+ Z ∞
²
φ(x)
x2 dx + φ(−x) x
¯¯
¯¯
∞
²
+ Z ∞
²
φ(−x) x2 dx
¸
= lim
²→0+
·
−1
²
¡φ(²) + φ(−²)¢ +
Z ∞
²
φ(x) + φ(−x)
x2 dx
¸
= lim
²→0+
·1
²
¡2φ(0) − φ(²) + φ(−²)¢ +
Z ∞
²
φ(x) + φ(−x) − 2φ(0)
x2 dx
¸
= Z ∞
0
φ(x) + φ(−x) − 2φ(0)
x2 dx
where we have use Mean Value Theorem and the fact φ ∈ Cc∞(R) to conclude that
x→0lim+
φ(0) − φ(−²) + φ(0) − φ(²)
² = 0
in the last equality.
Remark. For each k = 0, 1, 2, . . . and for each φ ∈ Cc∞(R), by using the Taylor’s theorem, observe that
x→0lim+
φ(x) − φ(−x)
x = 2φ0(0)
x→0lim+
φ(x) − φ(−x) − 2Pk−1
j=0
φ(2j+1)(0) x2j+1 (2j + 1)!
x2k+1 = φ(2k+1)(0)
(2k + 1)! for k ≥ 1,
x→0lim+
φ(x) + φ(−x) − 2 Pk−1
j=0
φ(2j)(0) x2j (2j)!
x2k = 2
(2k)!φ(2k)(0) for k ≥ 1 and
h 1
x2k+1, φ(x)i = Z ∞
0
φ(x) − φ(−x) − 2Pk−1
j=0
φ(2j+1)(0) x2j+1 (2j + 1)!
x2k+1 dx for k ≥ 1
h 1
x2k , φ(x)i = Z ∞
0
φ(x) + φ(−x) − 2 Pk−1
j=0
φ(2j)(0) x2j (2j)!
x2k dx for k ≥ 1
Example 2. Show that
ft(x) = sin xt
πx → δ in D0(R) as t → ∞.
Proof. Since ft(x) satisfies the following
(a) Z
R
sin xt
πx dx = 2 Z ∞
0
sin xt
πxt d(xt) = 1 for all t > 0, (b) For each a > 0, lim
t→∞ 2 Z ∞
a
sin xt
πxt d(xt) = lim
t→∞ 2 Z ∞
ta
sin u
πu du = 0, Then for ² > 0, and for each t > 0, let 0 < a < π
2t be a fixed number such that
¯¯
¯¯sin xt πx
¡φ(x) − φ(0)¢¯
¯¯
¯ < ²
2, for all |x| ≤ a and
t→∞lim |hft(x) , φ(x) − φ(0)i| ≤ Z a
−a
ft(x) |φ(x) − φ(0)| dx + lim
t→∞
¯¯
¯¯ Z
|x|≥a
ft(x)¡
φ(x) − φ(0)¢ dx
¯¯
¯¯
< ²
2+ lim
t→∞
¯¯
¯¯ Z
|x|≥a
ft(x)¡
φ(x) − φ(0)¢ dx
¯¯
¯¯ .
By the hypothesis (b) and note that |φ(x) − φ(0)| ≤ 2kφk∞, we have
t→∞lim
¯¯
¯¯ Z
|x|≥a
ft(x)¡
φ(x) − φ(0)¢ dx
¯¯
¯¯ = 0.
Thus, ft(x) → δ in D0(R) as t → ∞.
