Journal of Inequalities in Pure and
Applied Mathematics
http://jipam.vu.edu.au/
Volume 6, Issue 4, Article 113, 2005
ESTIMATION OF THE DERIVATIVE OF THE CONVEX FUNCTION BY MEANS
OF ITS UNIFORM APPROXIMATION
KAKHA SHASHIASHVILI AND MALKHAZ SHASHIASHVILI
T BILISI S TATE U NIVERSITY
FACULTY OF M ECHANICS AND M ATHEMATICS
2, U NIVERSITY S TR ., T BILISI 0143
G EORGIA
mshashiashvili@yahoo.com
Received 15 June, 2005; accepted 26 August, 2005
Communicated by I. Gavrea
A BSTRACT. Suppose given the uniform approximation to the unknown convex function on a
bounded interval. Starting from it the objective is to estimate the derivative of the unknown
convex function. We propose the method of estimation that can be applied to evaluate optimal
hedging strategies for the American contingent claims provided that the value function of the
claim is convex in the state variable.
Key words and phrases: Convex function, Energy estimate, Left-derivative, Lower convex envelope, Hedging strategies the
American contingent claims.
2000 Mathematics Subject Classification. 26A51, 26D10, 90A09.
1. I NTRODUCTION
Consider the continuous convex function f (x) on a bounded interval [a, b] and suppose that
its explicit analytical form is unknown to us, whereas there is the possibility to construct its
continuous uniform approximation fδ , where δ is a small parameter. Our objective consists
in constructing the approximation to the unknown left-derivative f 0 (x−) based on the known
^
function fδ (x). For this purpose we consider the lower convex envelope f δ (x) of fδ (x), that
is the maximal convex function, less than or equal to fδ (x). Geometrically it represents the
thread stretched from below over the graph of the function fδ (x). Now our main idea consists
^
of exploiting the left-derivative f 0δ (x−) as a reasonable approximation to the unknown f 0 (x−).
The justification of this method of estimation is the main topic of this article.
These kinds of problems arise naturally in mathematical finance. Indeed, consider the value
function v(t, x) of the American contingent claim and suppose we have already constructed its
uniform approximation u(t, x), where 0 ≤ t ≤ T , 0 < x ≤ L (for example, by discrete Markov
Chain approximation developed by Kushner [1]). The problem is to find the estimation method
ISSN (electronic): 1443-5756
c 2005 Victoria University. All rights reserved.
185-05
2
K AKHA S HASHIASHVILI AND M ALKHAZ S HASHIASHVILI
∂v
(t, x) that can be used to construct the optimal hedging strategy
of the partial derivative ∂x
(we should note here that the explicit form of the value function v(t, x) is typically unknown
for one in most problems of the option pricing). It is well-known (see, for example El Karoui,
Jeanblanc-Picque, Shreve [2]) that for a variety of practical problems in the pricing of American
contingent claims that utilize one-dimensional diffusion models, the value function v(t, x) is
convex in the state variable and hence we can apply the above mentioned method of estimation.
^
Thus, consider first for fixed t ∈ [0, T ] the lower convex envelope u(t, x) of the function u(t, x)
^
∂v
and then use its left-derivative ∂∂xu (t, x−) instead of the unknown ∂x
(t, x). Then in this way we
construct the approximation to the optimal hedging strategy.
2. T HE E NERGY E STIMATE
FOR
C ONVEX F UNCTIONS
Consider the arbitrary finite convex function f (x) on a bounded interval [a, b]. It is well
known that it is continuous inside the interval, and at the endpoints a and b it has finite limits
f (a+) and f (b−). Moreover it has finite left and right-derivatives f 0 (x−) and f 0 (x+) in the
interval (a, b) (see, for example Schwartz [3, p. 205]).
We will use the following inequality several times (Schwartz [3, p. 205]) concerning convex
function f (x) and its left-derivative f 0 (x−)
f (x2 ) − f (x1 )
≤ f 0 (x2 −)
x2 − x1
for arbitrary x1 , x2 with a < x1 < x2 < b.
Letting x2 tend to b we get
(2.1)
f 0 (x1 −) ≤
f 0 (x1 −) ≤
f (b−) − f (x1 )
,
b − x1
and similarly letting x1 tend to a, we have
f (x2 ) − f (a+)
≤ f 0 (x2 −).
x2 − a
From here we get
f (x) − f (a+)
f (b−) − f (x)
≤ f 0 (x−) ≤
for a < x < b.
x−a
b−x
Multiplying the latter inequality by (x − a)(b − x) we obtain the following crucial estimate
(2.2)
(b − x) f (x) − f (a+) ≤ (x − a)(b − x)f 0 (x−)
≤ (x − a) f (b−) − f (x) for a < x < b.
From this estimate we see, that the function
w1 (x) = (x − a)(b − x)f 0 (x−)
is bounded on the interval (a, b), moreover
w1 (a+) = 0,
w1 (b−) = 0,
hence it is natural to extend this function at the endpoints a and b by the relations
w1 (a) = 0,
w1 (b) = 0.
Thus we obtain the function w1 (x) defined on the closed interval [a, b], which is left-continuous
with right-hand limits and is bounded on the interval [a, b]. Similarly for another finite convex
function ϕ(x) defined on [a, b] we may denote
w2 (x) = (x − a)(b − x)ϕ0 (x−) for a ≤ x ≤ b
J. Inequal. Pure and Appl. Math., 6(4) Art. 113, 2005
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E STIMATION OF THE D ERIVATIVE OF THE C ONVEX F UNCTION
3
Finally introduce the following function
w(x) = w1 (x) − w2 (x)
(2.3)
= (x − a)(b − x) f 0 (x−) − ϕ0 (x−) for a ≤ x ≤ b
and note that it is bounded, left-continuous (with right-hand limits) and also
w(a) = 0,
w(b) = 0.
Rb
Now our objective in this section is to bound the Riemann integral a w2 (x)dx, it holds the
key to our method of estimation of the derivative of the arbitrary convex function. This bound
is given in the theorem below.
Theorem 2.1. For arbitrary two finite convex functions f (x) and ϕ(x) defined on a closed
interval [a, b] the following energy estimate is valid
Z b
2
(2.4)
(x − a)2 (b − x)2 f 0 (x−) − ϕ0 (x−) dx
a
≤
8 √
· 3 · sup f (x) − ϕ(x) sup f (x) + ϕ(x)(b − a)3
9
x∈(a,b)
x∈(a,b)
4
+
3
!2
sup f (x) − ϕ(x)
(b − a)3 .
x∈(a,b)
Proof. The proof is lengthy and therefore divided in two stages. At the first stage we verify the
validity of the statement for smooth (twice continuously differentiable) convex functions. At
the second stage we approximate arbitrary finite convex functions inside the interval [a, b] by
smooth ones in an appropriate manner and afterwards pass on limit in the previously obtained
estimate.
Thus, at first we assume that the convex functions f (x) and ϕ(x) are twice continuously
differentiable on [a, b] in which case we obviously have
f 00 (x) ≥ 0,
ϕ00 (x) ≥ 0,
a < x < b.
Introduce the functions
u(x) = f (x) − ϕ(x),
v(x) = (x − a)2 (b − x)2 f (x) − ϕ(x) .
Consider the following integral and use in it the integration by parts formula
Z b
b Z b
0
0
0
u (x)v (x)dx = u (x)v(x) −
v(x)u00 (x) dx
a
a
a
Z b
=−
(x − a)2 (b − x)2 f (x) − ϕ(x) f 00 (x) − ϕ00 (x) dx,
a
as v(a) = v(b) = 0.
From here we get the estimate
Z b
Z b
0
0
(x − a)2 (b − x)2 |f 00 (x) − ϕ00 (x)| dx
u (x)v (x) dx ≤ sup |f (x) − ϕ(x)|
a
x∈[a,b]
a
However, as pointed above f 00 (x) ≥ 0, ϕ00 (x) ≥ 0, hence
00
f (x) − ϕ00 (x) ≤ f 00 (x) + ϕ00 (x)
J. Inequal. Pure and Appl. Math., 6(4) Art. 113, 2005
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4
K AKHA S HASHIASHVILI AND M ALKHAZ S HASHIASHVILI
and from the previous estimate we obtain the bound
Z b
Z
b
0
0
u (x)v (x) dx ≤ sup f (x) − ϕ(x)
(x − a)2 (b − x)2 f 00 (x) + ϕ00 (x) dx.
x∈[a,b]
a
a
Next we have to transform the integral
Z b
(x − a)2 (b − x)2 f 00 (x) + ϕ00 (x) dx
a
b
= (x − a) (b − x) f (x) + ϕ (x) a
Z b
0
−
(x − a)2 (b − x)2 f 0 (x) + ϕ0 (x) dx
a
Z b
=−
2(x − a)(b − x)(−2x + a + b) f 0 (x) + ϕ0 (x) dx
2
2
0
0
a
b
= −2(x − a)(b − x)(−2x + a + b) f (x) + ϕ(x) a
Z b
0
+
2(x − a)(b − x)(−2x + a + b) f (x) + ϕ(x) dx.
a
Therefore
Z b
(x − a)2 (b − x)2 f 00 (x) + ϕ00 (x) dx
a
≤ sup f (x) + ϕ(x)
x∈[a,b]
Z b
0 2(x − a)(b − x)(−2x + a + b) dx
a
Evaluating the last integral we get
Z b
0 4 √
2(x − a)(b − x)(−2x + a + b) dx = · 3 · (b − a)3 .
9
a
Whence we come to the estimate
Z b
4 √
0
0
(2.5) u (x)v (x) dx ≤ · 3 · sup f (x) − ϕ(x) · sup f (x) + ϕ(x) · (b − a)3 .
9
x∈[a,b]
x∈[a,b]
a
On the other hand
Z b
Z b
0
0
0
u (x)v (x) dx =
f 0 (x) − ϕ0 (x) (x − a)2 (b − x)2 f (x) − ϕ(x) dx
a
a
Z b
= 2 (x−a)(b−x)(−2x+a+b) f (x)−ϕ(x) f 0 (x)−ϕ0 (x) dx
a
Z b
2
+
(x − a)2 (b − x)2 f 0 (x) − ϕ0 (x) dx.
a
Therefore we get the equality
Z b
2
(2.6)
(x − a)2 (b − x)2 f 0 (x) − ϕ0 (x) dx
a
Z b
Z b
0
0
=
u (x)v (x) dx −
(x − a)(b − x) (f 0 (x) − ϕ0 (x))
a
a
× 2(−2x + a + b) (f (x) − ϕ(x)) dx.
J. Inequal. Pure and Appl. Math., 6(4) Art. 113, 2005
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E STIMATION OF THE D ERIVATIVE OF THE C ONVEX F UNCTION
5
Bound now the last term
Z b
0
0
(x − a)(b − x) f (x) − ϕ (x) 2(−2x + a + b) f (x) − ϕ(x) dx
a
Z b
2
1
≤
(x − a)2 (b − x)2 f 0 (x) − ϕ0 (x) dx
2 a
Z b
2
+2
(−2x + a + b)2 f (x) − ϕ(x) dx
a
Z b
2
1
(x − a)2 (b − x)2 f 0 (x) − ϕ0 (x) dx
≤
2 a
!2
1
+ 2 sup f (x) − ϕ(x)
(b − a)3 ,
3
x∈[a,b]
as
b
Z
a
1
(−2x + a + b)2 dx = (b − a)3 .
3
Therefore we obtain
Z b
0
0
(2.7) (x − a)(b − x) f (x) − ϕ (x) · 2(−2x + a + b) f (x) − ϕ(x) dx
a
1
≤
2
Z
b
2
(x − a)2 (b − x)2 f 0 (x) − ϕ0 (x) dx
a
2
+
3
!2
sup |f (x) − ϕ(x)|
· (b − a)3 .
x∈[a,b]
Finally, if we use the bounds (2.5) and (2.7) in the equality (2.6), we come to the following
estimate
Z b
2
(2.8)
(x − a)2 (b − x)2 (f 0 (x) − ϕ0 (x)) dx
a
≤
8 √
· 3 · sup |f (x) − ϕ(x)| · sup |f (x) + ϕ(x)| · (b − a)3
9
x∈[a,b]
x∈[a,b]
!2
4
+
sup |f (x) − ϕ(x)| · (b − a)3 .
3 x∈[a,b]
Let us pass to the second stage of the proof. Consider two arbitrary finite convex functions f (x)
and ϕ(x) on the closed interval [a, b]. We have to construct the sequences of smooth convex
functions fn (x) and ϕn (x) approximating, respectively, the functions f (x) and ϕ(x) inside the
interval [a, b] in an appropriate manner.
For this purpose we will use the following smoothing function
(
1
c · e x(x−2) for 0 < x < 2
,
(2.9)
ρ(x) =
0,
otherwise
where the factor c is chosen to satisfy the equality
Z 2
ρ(x) dx = 1.
0
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6
K AKHA S HASHIASHVILI AND M ALKHAZ S HASHIASHVILI
Define
Z
(2.10)
b
n · ρ n · (x − y) · f (y) dy,
fn (x) =
a
Z
b
n · ρ n · (x − y) · ϕ(y) dy,
ϕn (x) =
a
where n = 1, 2, . . . and x ∈ (−∞, +∞).
For arbitrary fixed δ > 0 consider the restriction of functions fn (x) and ϕn (x) on the interval
[a + δ, b − δ] and let n ≥ 4δ . Then n · (x − a) ≥ 4 and n · (x − b) ≤ 0 for x ∈ [a + δ, b − δ].
Perform in (2.10) the change of variable z = n · (x − y), then we’ll have
Z n·(x−a)
z
fn (x) =
ρ(z) · f x −
dz,
n
n·(x−b)
Z n·(x−a)
z
ϕn (x) =
ρ(z) · ϕ x −
dz.
n
n·(x−b)
But the function ρ(z) is equal to zero outside the interval (0, 2) and hence we obtain
Z 2
z
(2.11)
fn (x) =
ρ(z) · f x −
dz,
n
0
Z 2
z
ϕn (x) =
ρ(z) · ϕ x −
dz,
n
0
if n ≥ 4δ . From the definition (2.10) it is obvious, that the functions fn (x) and ϕn (x) are
infinitely differentiable, while the convexity of these functions simply follows from the representation (2.11).
Next we show the uniform convergence of the sequence fn (x) to f (x) on the interval [a +
δ, b − δ] (similarly for ϕn (x) to ϕ(x)). For this purpose we use the uniform continuity of the
function f (x) on the interval [a + 2δ , b − δ]. For fix ε > 0 there exists δb such that we have
δ
f (x2 )−f (x1 ) ≤ ε if |x2 −x1 | < δb and x1 , x2 ∈ a + , b − δ .
2
Take n ≥ max{ 4δb , 4δ }. Then for 0 ≤ z ≤ 2 and x ∈ [a + δ, b − δ] we get
(
)
z
δb δ
z
δ
δ
≤ min
,
, x− ≥a+δ− =a+ .
n
2 2
n
2
2
Hence
z
4 4
− f (x) ≤ ε for n ≥ max
,
f x −
n
δb δ
and consequently
Z
|fn (x) − f (x)| = 2
z
ρ(z) · f x −
− f (x) dz ≤ ε
n
0
4 4
max{ δb , δ }.
for x ∈ [a + δ, b − δ] and n ≥
Thus we’ve shown the uniform convergence of the sequence fn (x) to f (x) on the interval
[a + δ, b − δ].
Now we need to differentiate the relations (2.11). We’ll use again the basic inequality (2.1)
on convex functions. Take therein
z
z
x1 = x −
− h, x2 = x − ,
n
n
J. Inequal. Pure and Appl. Math., 6(4) Art. 113, 2005
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E STIMATION OF THE D ERIVATIVE OF THE C ONVEX F UNCTION
7
where 0 < h < 4δ .
We will have
f (x − z ) − f (x − z − h)
z
n
n
(2.12)
f
x− −h − ≤
n
h
z ≤ f0 x −
−
n
if x ∈ [a + δ, b − δ], 0 ≤ z ≤ 2, 0 < h < 4δ and n ≥ 4δ .
Taking into account that the left-derivative of the convex function is nondecreasing and that
z
δ
z
x− −h≥a+ , x− ≤b−δ
n
4
n
we get
f (x − nz ) − f (x − nz − h)
δ
0
(2.13)
f
a+
− ≤
≤ f 0 ((b − δ)−).
4
h
It follows from here that the family of functions
0
Φn,x
h (z) =
f (x − nz ) − f (x −
h
z
n
− h)
is uniformly bounded by the constant
δ
0
0
c = max f
a+
− , f ((b − δ)−)
4
if x ∈ [a + δ, b − δ], 0 ≤ z ≤ 2, 0 < h < 4δ and n ≥ 4δ .
We write from the representation (2.11)
Z 2
f (x − nz ) − f (x − nz − h)
fn (x) − fn (x − h)
=
ρ(z) ·
dz.
h
h
0
Now letting h to zero and using the bounded convergence theorem we come to the following
formula
Z 2
z 0
0
− dz
(2.14)
fn (x) =
ρ(z) · f
x−
n
0
for x ∈ [a + δ, b − δ] and n ≥ 4δ .
From this formula it is easy to see, that for fixed x ∈ [a + δ, b − δ] the sequence fn0 (x)
converges to the left-derivative f 0 (x−). Indeed consider the difference
Z 2
z 0
0
0
0
fn (x) − f (x−) =
ρ(z) · f
x−
− − f (x−) dz,
n
0
where we assume that n ≥ 4δ and choose arbitrary ε > 0. As the left-derivative f 0 (x−) is
left-continuous we can find N (ε) such that (for 0 ≤ z ≤ 2)
0
z
0
f
≤ ε if only n > N (ε).
x
−
−
−
f
(x−)
n
Hence we get
Z
0
fn (x) − f 0 (x−) ≤
2
ρ(z) · ε dz = ε if n > max
0
4
, N (ε) .
δ
Thus for any fixed x ∈ [a + δ, b − δ] we have
lim fn0 (x) = f 0 (x−),
n→∞
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8
K AKHA S HASHIASHVILI AND M ALKHAZ S HASHIASHVILI
and similarly for functions ϕ0n (x), ϕ0 (x−)
lim ϕ0n (x) = ϕ0 (x−).
n→∞
Next write the estimate (2.8) for functions fn (x), ϕn (x) restricted to the interval [a + δ, b − δ]
Z
b−δ
(2.15)
2
2
2
x − (a + δ) · (b − δ) − x · fn0 (x) − ϕ0n (x) dx
a+δ
≤
8 √
· 3 · sup fn (x) − ϕn (x)
9
x∈[a+δ,b−δ]
× sup fn (x) + ϕn (x) · (b − a − 2 · δ)3
x∈[a+δ,b−δ]
4
+ ·
3
!2
sup
fn (x) − ϕn (x)
· (b − a − 2 · δ)3 .
x∈[a+δ,b−δ]
For x ∈ [a + δ, b − δ], 0 ≤ z ≤ 2 and n ≥ 4δ we have
δ
z 0
f
a+
− ≤ f0 x −
− ≤ f 0 ((b − δ)−),
2
n
multiplying this inequality by ρ(z) and integrating by z over (0, 2) from the equality (2.14) we
obtain
δ
0
f
a+
− ≤ fn0 (x) ≤ f 0 ((b − δ)−).
2
Similarly for the functions ϕ0n (x)
δ
0
ϕ
a+
− ≤ ϕ0n (x) ≤ ϕ0 ((b − δ)−).
2
Hence the sequences of the functions fn0 (x) and ϕ0n (x) are uniformly bounded on the interval
[a + δ, b − δ] for n ≥ 4δ . Thus we can apply the bounded convergence theorem in the left-hand
side of the inequality (2.15) passing to limit when n → ∞ and we get
Z
b−δ
(2.16)
2
2
2
x − (a + δ) · (b − δ) − x · f 0 (x−) − ϕ0 (x−) dx
a+δ
≤
8 √
· 3 · sup f (x) − ϕ(x)
9
x∈[a+δ,b−δ]
× sup f (x) + ϕ(x) · (b − a − 2 · δ)3
x∈[a+δ,b−δ]
4
+ ·
3
!2
sup
f (x) − ϕ(x)
· (b − a − 2 · δ)3 .
x∈[a+δ,b−δ]
Finally it remains to pass onto limit when δ → 0 in the inequality (2.16). Introduce the
following function

χ(a+δ,b−δ] (x) · x−a−δ 2 · b−δ−x 2 for a < x < b
x−a
b−x
uδ (x) =
,
0
otherwise
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E STIMATION OF THE D ERIVATIVE OF THE C ONVEX F UNCTION
9
where χ(a+δ,b−δ] (x) is the characteristic function of the interval (a + δ, b − δ]. Evidently
(
1, a < x < b
.
0 ≤ uδ (x) ≤ 1, lim uδ (x) =
δ↓0
0, otherwise
We remind now the definition (2.3) of the function w(x) and the fact that it is a bounded
function on the closed interval [a, b] and we rewrite the inequality (2.16) in terms of the functions
w(x) and uδ (x).
Z b
8 √
(2.17)
uδ (x) · w2 (x) dx ≤ · 3 · sup f (x) − ϕ(x)
9
x∈[a+δ,b−δ]
a
× sup f (x) + ϕ(x) · (b − a − 2 · δ)3
x∈[a+δ,b−δ]
!2
4
+ ·
3
sup
f (x) − ϕ(x)
· (b − a − 2 · δ)3 .
x∈[a+δ,b−δ]
We use again the bounded convergence theorem in this inequality when δ ↓ 0 and at last get
the desired energy estimate (2.4).
3. T HE M AIN R ESULT
The following proposition is the basic result of this article though its proof is a simple consequence of the previous theorem
Theorem 3.1. Let f (x) be the unknown continuous convex function defined on the bounded
interval [a, b] and suppose we have at hand its some continuous uniform approximation fδ (x).
^
Consider the lower convex envelope f δ (x) of the function fδ (x). Then for the unknown left^
derivative f 0 (x−) the following estimate through f 0δ (x−) does hold
Z b
2
^
2
2
0
0
(3.1)
(x − a) · (b − x) · f (x−) − f δ (x−) dx
a
8 √
≤ · 3 · sup fδ (x) − f (x)
9
x∈[a,b]
!
sup |f (x)| + sup |fδ (x)|
x∈[a,b]
· (b − a)3
x∈[a,b]
4
+
3
!2
sup fδ (x) − f (x)
· (b − a)3 .
x∈[a,b]
Proof. Introduce the notation
sup fδ (x) − f (x) = cδ .
x∈[a,b]
It is clear that
f (x) − cδ ≤ fδ (x),
fδ (x) − cδ ≤ f (x), if x ∈ [a, b].
Therefore we get that the convex function f (x) − cδ is less or equal than fδ (x) and hence
^
f (x) − cδ ≤ f δ (x), x ∈ [a, b].
On the other hand we have
^
f δ (x) − cδ ≤ fδ (x) − cδ ≤ f (x), x ∈ [a, b],
J. Inequal. Pure and Appl. Math., 6(4) Art. 113, 2005
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10
K AKHA S HASHIASHVILI
therefore
AND
M ALKHAZ S HASHIASHVILI
^
f δ (x) − f (x) ≤ cδ , x ∈ [a, b],
that is
^
sup f δ (x) − f (x) ≤ sup fδ (x) − f (x).
(3.2)
x∈[a,b]
x∈[a,b]
Denote sup |fδ (x)| = e
cδ , then obviously
x∈[a,b]
−e
cδ ≤ fδ (x) ≤ e
cδ , x ∈ [a, b]
and hence
^
−e
cδ ≤ f δ (x) ≤ fδ (x) ≤ e
cδ , x ∈ [a, b],
that is
^
sup f δ (x) ≤ sup fδ (x).
(3.3)
x∈[a,b]
x∈[a,b]
^
Take now f δ (x) instead of convex function ϕ(x) in the formulation of Theorem 2.1 and use
the inequalities (3.2) – (3.3) in the right-hand side of the estimate (2.4), then we directly come
to the estimate (3.1).
Remark 3.2. As the left and the right-hand derivatives of convex function coincide everywhere
except on the countable set, Theorems 2.1 and 3.1 are obviously true for the right-derivatives
instead of the left-ones.
R EFERENCES
[1] H.J. KUSHNER, Probability methods for approximations in stochastic control and for elliptic equations. Mathematics in Science and Engineering, Vol. 129, Academic Press, Harcourt Brace Jovanovich, Publishers, New York–London, 1977.
[2] N. EL KAROUI, M. JEANBLANC-PICQUE AND S.E. SHREVE, Robustness of the Black and
Scholes formula, Math. Finance, 8(2) (1998), 93–126.
[3] L. SCHWARTZ, Mathematical Analysis, Vol. 1. (Russian) Mir, Moscow, 1972.
J. Inequal. Pure and Appl. Math., 6(4) Art. 113, 2005
http://jipam.vu.edu.au/