406
7 Variational Principles
(7) Let T be a linear operator of X to Y , if for every n 1 and arbitrary points
x n , x ∈ D (n = 1, 2, · · · ), when x n → x, there must be T x n → T x, namely
when x n − x → 0, there must be T x n − T x → 0, then T is called the
continuous linear operator on D.
Let the linear operator T : D → Y be a bounded operator, if there exists a positive
constant K , such that T x Y K x X , here • D and • Y are the norms of D
and Y respectively, then T is called the bounded linear operator. Correspondingly,
when the range of values of T is a set of numbers, T is called the bounded linear
functional. The linear operator that is not bounded is called the unbounded linear
operator.
Theorem 7.4.1 Let X and Y be both the formed linear space, D is the subset of X ,
T is the linear operator of D to Y , if T is continuous at a point x 0 ∈ D, then T is
continuous in all of D.
Proof Taking arbitrary x, x n ∈ D (n = 1, 2, · · · ), and lim
n→∞
x n = x, since T is the
linear operator, there is
T x n − T x = T (x n − x) = T (x n − x + x 0 ) − T x 0
When n → ∞, from x n − x + x 0 → x 0 and the continuity at the point x 0 of T ,
T (x n − x + x 0 ) → T x 0 , namely T x n → T x. Therefore T is continuous at point x.
Since x is arbitrary in D, T is continuous in all of D. Quod erat demonstrandum.
Theorem 7.4.2 Let X and Y be both the formed linear space, D is the sunset of X ,
T is the linear operator of D to Y , then the sufficient and necessary conditions that
T is continuous are that T is bounded.
The sufficiency. Let T be bounded, then there exists M > 0, such that for an
arbitrary x ∈ D, there is T x Mx. Taking an arbitrary number sequence
{x n } ∈ X (n = 1, 2, · · · ), such that lim
n→∞
x n = x, then n → ∞, there is
T x n − T x = T (x n − x) Mx n − x → 0
Therefore, T is continuous at an arbitrary point x in D.
The necessity. Proof method 1. Using reduction to absurdity to prove. Let T be
continuous but unbounded, then for every natural number n, there must exist x n ∈ D
and x n = 0 (n = 1, 2, · · · ), such that T x n nx n . Putting y n =
x n
nx n , then there
is y n =
1
n
, when n → ∞, y n → 0, from the continuity of T , there is T y n → 0.
On the other hand
T y n =
T x n
nx n
=
T x n
nx n
1
This is contradictory to continuity, therefore T is bounded.
Proof method 2. Let T be continuous in D, for point x = 0, there exists δ > 0,
when x < δ, there is T x < 1. Taking an arbitrary x = 0, there is
7 Variational Principles
(7) Let T be a linear operator of X to Y , if for every n 1 and arbitrary points
x n , x ∈ D (n = 1, 2, · · · ), when x n → x, there must be T x n → T x, namely
when x n − x → 0, there must be T x n − T x → 0, then T is called the
continuous linear operator on D.
Let the linear operator T : D → Y be a bounded operator, if there exists a positive
constant K , such that T x Y K x X , here • D and • Y are the norms of D
and Y respectively, then T is called the bounded linear operator. Correspondingly,
when the range of values of T is a set of numbers, T is called the bounded linear
functional. The linear operator that is not bounded is called the unbounded linear
operator.
Theorem 7.4.1 Let X and Y be both the formed linear space, D is the subset of X ,
T is the linear operator of D to Y , if T is continuous at a point x 0 ∈ D, then T is
continuous in all of D.
Proof Taking arbitrary x, x n ∈ D (n = 1, 2, · · · ), and lim
n→∞
x n = x, since T is the
linear operator, there is
T x n − T x = T (x n − x) = T (x n − x + x 0 ) − T x 0
When n → ∞, from x n − x + x 0 → x 0 and the continuity at the point x 0 of T ,
T (x n − x + x 0 ) → T x 0 , namely T x n → T x. Therefore T is continuous at point x.
Since x is arbitrary in D, T is continuous in all of D. Quod erat demonstrandum.
Theorem 7.4.2 Let X and Y be both the formed linear space, D is the sunset of X ,
T is the linear operator of D to Y , then the sufficient and necessary conditions that
T is continuous are that T is bounded.
The sufficiency. Let T be bounded, then there exists M > 0, such that for an
arbitrary x ∈ D, there is T x Mx. Taking an arbitrary number sequence
{x n } ∈ X (n = 1, 2, · · · ), such that lim
n→∞
x n = x, then n → ∞, there is
T x n − T x = T (x n − x) Mx n − x → 0
Therefore, T is continuous at an arbitrary point x in D.
The necessity. Proof method 1. Using reduction to absurdity to prove. Let T be
continuous but unbounded, then for every natural number n, there must exist x n ∈ D
and x n = 0 (n = 1, 2, · · · ), such that T x n nx n . Putting y n =
x n
nx n , then there
is y n =
1
n
, when n → ∞, y n → 0, from the continuity of T , there is T y n → 0.
On the other hand
T y n =
T x n
nx n
=
T x n
nx n
1
This is contradictory to continuity, therefore T is bounded.
Proof method 2. Let T be continuous in D, for point x = 0, there exists δ > 0,
when x < δ, there is T x < 1. Taking an arbitrary x = 0, there is