Yesterday, we have discussed the weak and
strong convergence. Today, we will see convergence
of sequences 
of operators. If X 
and Y are two normed spaces, then an operator
T from X to Y, if it is a boundary linear
operator, then the class of all such boundary
linear operator is denoted by B of X, Y, a
set of all boundary linear operators T from
X to Y. And this class forms a normed space
under the operations that the addition is
defined as T 1 plus T 2 X is the T 1 X plus
T 2 X 
and alpha T X is alpha times of T X and then
norm of T is defined as supremum of norm of
T X over norm of X, where the X belongs to
the domain of T capital X not equal to 0.
Now, this is the norm of Y, this is the norm
of X and this is the norm of B X, Y. This
is what we have. So, basically, when we construct
the normed space with the help of these operators,
then, we require the three norms. One is the
norm in X, norm in Y and these two, together
gives the norm of B X, Y. If we take any sequence
belonging to this class B X, Y, then 
convergence of sequences of operator T n will
depend on the norm whether we are taking norm
of B X, Y or we are taking norm of T X or
whether it goes to in the usual way in the
norm of this. So, with respect to this norm,
we can classify this convergence of sequence
of operators in three categories.
These categories are; uniform operator convergence
in the norm of B X Y, strongly convergence
of T n X in Y 
and third is weak convergence 
of T n X in Y. So, we will have three categories
for the convergence of the sequence, uniform
convergence, strong convergence and weak convergence.
Now, we will first define these concepts and
then find out the relation whether 1 implies
2, 2 implies 2 or vice-versa is also 2.
So, let us see how to define 
convergence of sequence of operators. Let
X and Y be normed spaces. A sequence T n of
operators, T n belonging to B X Y means bounded
linear operator from X to Y is said to be
uniformly operator convergent 
if sequence T n converges in the norm on B
X Y, that is, the norm of T n minus T must
go to 0 when n tends to infinity.
Then second is, this sequence T n is said
to be strongly operator convergent 
if the sequence T n X, that is, the element
norm in Y converges strongly in Y for every
X in X because Y is a normed space. So, norm
of Y is there. Therefore, this sequence converges
strongly in y. So, in normed space, we have
seen two, one is the strongly convergence,
another one is the weak convergence. Strong
convergence means, the sequence convergence
to X under the norm norm of X n minus X goes
to 0 and weak convergence means when f of
X n goes to f of X for every f, then it said
to be the weakly convergence sequence for
f belongs to X Ts. Third is weakly operator
convergent. Sequence T n X, this sequence
converges weakly in Y for every X belonging
to X. So, this is known as the weakly operator
convergence.
Notation wise, we can write a sequence T n
of B X Y, B X Y, a sequence T n belongs to
B X Y and there is an operator T from X to
Y such that one - norm of T n minus T tends
to 0, second is norm of T n X minus T X that
is the norm of Y goes to 0 for all X belonging
to capital X.
And third is modulus of f T X n minus f of
T X tends to 0 for all X, belonging to capital
X and for all f belongs to all f belongs to
the dual space of y respectively respectively.
Then T is called the uniform. This is uniform.
This 
is strong and this is weak. The operator T
is called uniform, strong 
and weak 
operator limits limit of T n respectively.
So now, we have seen the three ways of convergence
of the sequence of beta, uniform operator
convergent, strong operator and weak operator.
Now, there arises question that sequence T
n is a bounded linear operator, but we have
not told anything about T whether this T remains
bounded linear operator, whether T is a point
of X Y. What will be the nature of T in case
of this uniform, in case of this strong or
in case of the weak. So, we will discuss that
whether T n converges to T uniformly, T has
to be bounded linear operator or not and second,
in case of the strong convergence, whether
the limiting point T is bounded or unbounded.
Similarly, it is for the weak convergent.
This is the first case.
Second point is, what is the relation between
these three? It is clear from the definition
that 1 implies 2 and 2 implies 3. Every uniform
operator convergent implies the strong operator
convergent and strong operator convergent
implies this. Why because suppose if uniform
operator is given, it means that norm of T
n minus T goes to 0. So obviously, the reason
is 1 is given suppose, then what is to consider
norm of T n X minus T X of Y. This is equal
to norm of T n minus T X Y. Now, this will
be less than equal to norm of T n minus T
into norm of X. So, that will be T into norm
of X.
Now, one thing which I point out here is that
this is T. I think this is wrong which I did.
It is T n X n. Please make the correction.
T n X minus f of t X. Now, here one thing
is, when 1 is given that is if the sequence
T n converges to T uniformly then T must be
bounded linear operator. Otherwise, this meaning
of norm T n minus T does not carry any meaning
because T n, if it is unbounded, the difference
norm of T n minus T cannot go to 0. So, if
T n is a sequence of the operator that converges
in the norm of T uniformly, their limiting
point must be a bounded linear operator. So,
T n minus T becomes bounded. Therefore, by
definition of the bounded operator, norm of
T X is less than equal to norm T into norm
X. Now, this goes to 0 as n tends to infinity.
Therefore, this will go to 0. So, it implies
it. So, first implies second.
Now, second implies third. What is second?
second 
Second shows that mod of T n X minus f of
T X. This is equal to f of T n X minus T X
mod of this f is giving to be a bounded linear
functional on Y. So, mod of f is less than
equal to norm of f into norm of this. Now,
this is giving to be 0, 10 into 0, this is
bounded. So, it goes to 0 as n tends to infinity.
So, second implies third. So, first implies
second, second implies third is obviously
true.
In general, the converse of this may not be
true. Let us take the counter example first.
Let us see first example. The sequence converges
strongly, second need not imply first. This
is an example for this. It means that a sequence
of the operator T n converges strongly, but
not uniformly. So, Let us take T n as a sequence
of the operator from l 2 to l. Let T n be
a sequence of the operator from l to l defined
by T n X is 00 0 0. Then, xi n plus 1 n plus
2 and so on where X xi 1, xi 2, xi n, xi n
plus 1 and so on. This is an element of l
2.
Now, we know that L2 is also a Hilbert space
and norm of l 2, we know this l 2 X belongs
to l 2 means that is sigma of mod xi I whole
square, I is 1 to infinity. This is finite
convergent and norm on l 2 is given as norm
of X 2 is sigma xi I square power half. This
is the norm on l 2. So, under this norm, we
can say T n X is a point in this.
Now, the operator T n which is defined above,
we claim that this operator T n is linear
and bounded. Why because, T n alpha X plus
beta Y that will be by definition 0 0 0 xi
and alpha times xi I n plus 1 beta times say
eta n plus 1 and so on, where y is equal to
eta 1, eta 2, eta n, eta n plus 1 and so on
belongs to l 2 and X is our xi 1 xi 2 belongs
to l 2. So, alpha X plus beta means, this
will come. Now, this can be written here as
alpha times 0 0 0 xi n plus 1 xi n plus 2
plus beta times 0 0 0 eta n plus 1 and so
on and that will be equal to alpha times T
n X plus beta times T n Y. So, T n is linear.
T n is bounded also. Why it is bounded, because
what is the norm of T n X? This is the norm
of l 2. So, by definition, this will be sigma
mod of xi j j equal to n plus 1 to infinity
under the norm. So, let it be square. Now,
this will be less than equal to norm of X
2 because this is power under root half. So,
you can take this. Let it be removed. Now,
j equal to 1 to infinity. So, T n is bounded.
Therefore, T n is a sequence of the operator
which is linear and bounded.
Further, sequence T n converges to 0 strongly.
This sequence converges to 0 strongly. That
is norm of T n X minus 0 under l 2 must go
to 0 as n tends to infinity. What is the norm
of T n X T n X minus 0?It means, this is equal
to under root sigma j equal to n plus 1 to
infinity mod of xi j square.
Now, this is a series after n th term. So,
it is a remainder of a convergent series.
So, this goes to 0 as n tends to infinity
because this is the remainder of the series
sigma mod xi j whole square j equal to 1 to
infinity which is convergent.
Therefore, it will go here. So, T n X converges
to 0 under this norm. It means T n sequence
is strongly convergent to 0. This is the 0
operator operator, but this T n sequence is
not uniformly convergent 
to 0 operator. It is not uniformly convergent
because what is the 
norm of t n minus 0 operator. Norm of t n
minus 0 means j equal to 1 to infinity. What
is this? By definition, this will be equal
to norm of T n X. Let me write this thing
supremum over taken whole x.
Norm of 
T n is the supremum of norm of T n X over
norm of X, X is belonging to our l, 2 space
not equal to 0. Now, T n X is this. So, this
is less than equal to norm of 1, norm of X,
norm of X that is 1; supremum of this thing
is less than equal to 1.
Now further, norm of T n X T n is greater
than equal to norm of T n X over norm of X
for particular X. If I take X to be 1 0 0
0, this is an element of l 2. So, what is
this norm? t n X will be 1, norm of X will
be 1. So, it means norm of T n minus 0, this
is coming to be norm of T n which is 1. So,
it cannot go to 0 as n tends to infinity.
Therefore, this sequence does not converge
uniformly. So, this is an example where the
sequence of the operator converges strongly,
but not uniformly. Now, second example when
the sequence of the operator convergent weakly,
but not strongly.
Let us 
take the example. So, define an operator T
n from l 2 to l 2 such that T 
n X T n X equal to 0 0 0. This is up to n
terms; and then xi 1, xi 2 and so on. X 
belongs to e 2.
It is a set operator and n digits are separate.
Now, this operator is clearly linear. T n
is linear and bounded. I think there is nothing
to prove here. We claim thus that this sequence
T n is weakly convergent operator convergent
to 0, but not strongly. This operator converges
weakly to 0 operator, but not a strong convergent.
We know that every 
bounded linear functional 
f on l 2 can be represented as 
in terms of the inner product f X has inner
product of X Z is uniquely determined by f
and norm of z equal to norm of f and it will
be in l 2. So, the inner product will be I
j is equal to 1 to infinity xi j zeta j conjugate,
where z is zeta j is an element in L 2. This
is by Riesz representation theorem. This can
be written as j equal to 1 to n and so on.
Let us take f of a. So, consider f of t n
X. This is by definition inner product of
T n X Z, but T n X is shift operator. We shifted
the point first position to n plus 1 th position.
So, it means the first few n terms will be
0 basically. So, we can start with this n
plus 1 j n plus 1 to infinity.
Then, xi 1 will be multiplied by zeta n plus
1. So, we can write this thing as j n plus
zeta zeta j minus n and xi j minus n into
zeta j bar because when j equal to n plus
1 this xi 1 is multiplied by zeta n plus 1
conjugate. So, this can be written as sigma
k equal to 1 to infinity xi k zeta n plus
k conjugate.
Now, apply the Cauchy Schwarz inequality.
So, by Cauchy Schwarz inequality, modulus
of f T n X square will be equal to modulus
of inner product of t n X Z square and that
will be less than equal to sigma of mod xi
k square. k is 1 to infinity into sigma say
here, m to m is n plus 1 to infinity mod of
zeta m whole square. By Schwarz inequality,
modulus of this thing is less than equal to
mod of this, all is called raise to the power
half into mod of this. So, half is out now.
This will be less than equal to… Now, we
claim that, this goes to 0 as n tends to infinity,
why because, this is finite, this part is
the norm of X and this part is the remainder
of a convergent series.
So, it will go to 0 as the remainder 
of the convergent series 
sigma mod zeta m is 1 to infinity, remainder
of this convergent series. This convergent
series tends to 0. So, it will go to 0. Therefore,
this implies f of T n X tends to 0 as n tends
to infinity and this is true for every f belongs
to the dual of l 2 dash; that is l 2 dual
of l p is l q.
Therefore, this sequence T n X will converge
to 0 weakly. Therefore, sequence T n is weakly
operator convergent to 0 operators, but this
T n does not converge strongly. Why because
norm of T n is 1, norm of T n X is not strongly
convergent, because if we choose X to be 1
0 0 0, which is in l 2; what is the norm of
T n X minus T m X. This is norm of l 2. is
It is not under root 2, because this difference
will be 1 comma. It will go to root 2, does
not tend to 0 as m n goes to infinity. So,
this sequence T n is not a Cauchy sequence.
Therefore, it cannot be convergent.
This is not a convergent sequence, because
if it convergent, it has to be Cauchy. So,
we contradict the theme, therefore, this term.
So, this is an example where it is weakly
convergent to 0, but not a strongly convergent.
Same thing will be continued if I replace
operator by a functional, because, when we
say T n is a sequence of the operator from
X to Y. Then, both Y and X are normed space.
When we replace Y by r over c, then this sequence
operator shows to be a functional. So, same
type of criteria is there. Let us see convergence
of sequences of functional. Let f n be a sequence
of 
linear functionals defined on X. 
Since they are functional, obviously, Y is
r or c because it maps the point X to the
real or any scalar quantity complex number
and Y n r and c they are complete metric space.
So, if they are complete in that case, r and
c is also dimension of for one. So, these
are the finite dimensional space.
If they are finite dimensional space, then
weak and strong convergence is the same. Therefore,
the concept of the three types of convergence
which we have earlier in case of the operators,
this concept is uniform, strong and weakly
operator convergent. So, second and third
will coincide. We will not get these two different
things. What we get is uniform convergence
and one of them is weakly convergent, because
weak and strong convergence are identical.
Convergence converges in the limit strongly
and weakly. So, here we will instead of weakly,
we say weakly star convergent . So, now, we
define the concepts as…
Now, we define the concepts as: strong and
weak star convergence of sequence 
of functional. Let f n be a sequence of 
bounded linear 
functional defined on a normed space X. 
We say a strong 
convergence of f n means that there is an
f belonging 
to the dual space of X such that norm of f
n minus f tends to 0 as n tends to infinity.
What is the meaning of norm here? The norm
means norm of f is supremum mod f X over X
belongs to the domain of this. So, X belongs
to domain X non 0. So, this goes to 0 in the
norm and weak star convergence 
of f n means that there is an f belongs to
X dash such that f n X goes to f X for all
X. Remember here, for all X belonging to capital
X, that is a modulus of f n X minus goes to
0 as n tends to infinity for all X belongs
to x. We denote it by f n converges to f weak
star. There w l, here w star, weak star convergence.
So, correspondingly we have this. The first
question which we raise is that whether the
limiting operator t remains bounded or not.
We wanted to discuss the behaviour of limiting
operator T under various type of convergence.
This 
is for general operator not for functional.
So, the very first question T n is a sequence
of bounded linear operator. T n belongs to
B X Y and T is an operator from X to Y. What
is the behaviour of this? this is a bounded
linear operator and T is any operator. Now,
the question is that if T n converge T, whether
T is bounded linear or not. So, first is if
T n converges to T uniformly, then that is
norm of T n minus T goes to 0 as n tends to
infinity. Then T has to be bounded linear
operator. Otherwise, difference of this has
no meaning. It cannot go to 0.
In case of uniform convergence of T n, the
limiting operator 
T has to be linear and bounded. So, this is
the case 1. Case 2: If T n converges to T
strongly 
or T n converges to T weakly, then the limiting
operator 
T may or may not be bounded. It will be linear,
but may not be bounded.
For example, suppose, we have the space X
of all 
sequences X is equal to xi I in l 2 with only
finitely many non zero terms, that is X is
the set of those sequence is X xi 1, xi 2
say xi n and rest are zeros like this with
only finitely many non zero terms and rest
are zeros. Now, this set X which is a subset
of l, this is basically set of X belongs to
l 2. So, this will be a subset of l 2 scalars
of all such sequences. This X is not complete
because if I take a sequence 1, one by 2,
1 by 3, 1 by n and then 0 0 0 then sorry any
sequence 1 2 3 first n terms and then belongs
to l 2.
When n tends to infinity, then large numbers
of these terms are non zero. So, basically,
it will be a point or may be point of l 2,
but may not be point in X because it will
not be in X because X contains only those
sequences which have only finite number of
nonzero term. Rest are 0. So, for example,
this sequence if I take X to be 1 2 say 0
0 0, this is in l 2 finite, but this is X
n.
When n tends to infinity, then X does not
belong to capital X. 
It is not a complete metric space. Now, we
will define a sequence of bounded linear operators
T n X as xi 1, 2 xi 2, 3 xi 3 and then n xi
n, after that xi n plus 1 and so on. It means
that first n terms are multiplied like this
1, 2, 3, n and rest are 0.
T n X has these terms j xi j if j is less
than equal to n and equal to xi j if j is
greater than n.
We claim that this sequence T n converges
strongly 
to the 
unbounded operator T defined by T X as eta
j, where the eta j is j times xi j. Obviously,
this T n X, when n tends to infinity, it is
of the form j type this xi j. This is an unbounded
operator, it is a linear operator. So, it
will be unbounded. So, T n X converges to
it strongly. Why this t n X converges strongly
to the unbounded over strongly means norm
of T n X minus this. So, what is this norm
of T n X minus this T X is basically the 0
0 0 0 norm of this. So, it is basically tending
to 0 as n tends to infinity. That is why it
converges. So, it converges strongly of this,
but is not unbounded. So, clearly T is unbounded.
Thank you.
