Today, we will discuss the strong and weak
convergence in the norm linear space. We have
seen that in calculus. There are so many ways
of defining the convergence. Like we have
ordinary convergence of a sequence, then conditional
convergent sequence, absolute convergence
of the sequence and uniform convergence; these
concepts we have gone through this concept
in case of calculus. Ordinary convergence
is simply x n converges to x mean mod of x
n minus x goes to 0, conditioner or a series
is ordinary convergent.
Conditionally convergent is when the series
is convergent, but the absolute is not convergent,
conditionally convergent of the series and
absolutely convergent is when each term is
replaced by the absolute value. And the series
is convergent and uniform convergent is when
the sequence of the functions is defined over
a certain domain, then we say that the sequence
of the function converges uniformly.
It means the epsilon should not depend on
the point mod of f n x minus f m. It should
remain less than, whatever f n f x and f z,
whatever x and z may be inside that. Epsilon
should not depend on the point. But in case
of the convergence, the points are important.
So, these are the various concepts of convergence
in case of calculus.
We have the same type of concepts in functional
analysis also. We have these concepts in case
of norm space. Apart from this, one more concept
which we have in case of the norm space is
a weak and strong convergence. In every norm
space, one can find out the dual space of
this, and then once you get the dual space,
the f belongs to the dual set of all boundary
linear functional. The convergence with respect
to the boundary linear function plays an important
role in the.
So, that convergence is termed as a weak convergence
and the convergence in the norm, is termed
as a strong convergence.
Convergence in norm, we mean that if a sequence
x n belongs to a norm space x t and we say
that x n converges, it means that there must
be some point x available in the x such that
the difference between x n n x under the norm
should go to 0 as n tends to infinity. That
is the converge norm.
We define first the strong convergence as
the convergence in norm. Strong convergence
means, a sequence x n 
in a norm space, x norm 
is said to be strongly convergent or we can
also say convergence 
in the norm, if there exist or there is an
x belonging to capital X such that the norm
of this x n minus x as n tends to infinity
is 0.
Then, such a sequence we say, the sequence
x n is strongly convergent in n. We denote
this by saying that x n converges to x in
this norm. It means that it is strongly convergent
and x is called a strong limit of the sequence
x n. ok
The concept of the weak convergence is as
follows: A sequence x n 
in a normed space x norm is said to be 
weakly convergent. If there is an x belongs
to capital X such that for every f belongs
to its dual, this is the dual space of x for
every a belongs to x is limit of this f of
x n as n tends to infinity is nothing but
f x.
It is denoted by saying x n converges to x
weakly. So, x n converges to x. x is called
weak limit of the sequence x.
Basically, the meaning of the weakly convergence
is that x n is a point in x. This is in the
normed space, a vector quantity. When f is
the functional defined on x, a linear boundary
functional, we say that x n converges to x,
means that corresponding sequence of scalars
and scalars is obtained by taking the image
of x n and that f.
So, f of x n becomes a scalar when the sequence
of scalars converges. Then, we say such a
sequence x n as weakly convergent.
This limit f of x n equal to f x as n tends
to infinity must hold good for all f belongs
to x dash. Then only we say that the sequence
x n converges to x weakly, means image under
each boundary linear functional goes to corresponding
f of x. Then sequence of scalar goes to this.
Now, in this weak convergence, there are various
applications in analysis and most like a differentiation
equation, general theory of differentiation
equations. We require certain lemmas to go
in deep to the results on weak convergence.
First lemma is that, on the weak convergence,
the proof of these results on the weak convergence
requires rigorously the application of uniform
boundary theorem. So, this gives you results
where the bounded uniform boundary theorem
is used or in an application of uniform boundary
theorem to analysis problem.
What is lemma? Let x n be a weakly convergent
sequence in a normed space x norm, say weakly
convergent.
Suppose, this sequence goes to x weakly, x
n is a weakly convergent sequence, now, it
means that there must be some point x available
in x such that x n converges to x weakly or
f of x n goes to f x for every f. Then the
following results hold: the weak limit 
x of the sequence x n is unique. Just like
that in the case of an ordinary convergence,
if a sequence x n converges to x, alpha n
converges to alpha, limit alpha will be unique.
It means that we cannot take a sequence converging
to two different limit points.
Then, we say that the sequence does not converge.
So, just like a sequence of a scalar, the
limit is unique. It is similar in case of
this weak convergence or weak limit is unique.
Every sub sequence 
of x n converges weakly to x. Just like an
ordinary sequence of a scalar, if a sequence
converges, then, all of its sub-sequences
will also converge. So, similarly, here also
it will converge weakly.
Third is the sequence of norms is bounded.
The proof of this is easy. The weak limit
of x of x n is unique. So, given that x n
converges to x weakly. It means f of x n goes
to f x for every f belongs to its dual. This
is the dual space.
Now, f of x n is a scalar quantity. So, basically,
what you are getting is that this sequence
will go to f x scalar. Suppose, a weak limit
is not the same, what is given is that we
wanted weak limit of x is unique.
Suppose, the weak limit x is not unique, it
means that there exist y such that x n also
converges to y weakly. So, that is the meaning.
f of x n goes to f y for every y, for every
f belongs to its dual.
Now, f x n goes to f x for every f, f n x
goes to f y for f. We want x equal to y. So,
let us consider f x minus f y. f is a bounded
linear functional. We can write f of x minus
y as f is linear. ok
But, f of x minus y f x equal to f y is given.
So, this must be 0 and this is 0 for every
f belongs to its dual space. So, what this
shows is that f of x minus y is 0 for every
f belongs to its dual.
Why is it 0? The reason is that f of x n is
a sequence of a scalars and sequence of a
scalar cannot have two different limits. So,
this implies that the limiting point f x must
be equal to f y.
x n converges to x means f of x n goes to
f x, but f of x n is a sequence of scalars
and we are assuming that this x n is not unique.
We are assuming another y, but basically by
definition of b convergence f, image of x
n under f will go to f of y.
Since f of x n is a sequence of a scalar,
its limiting value f x and f y will not be
different. If it is convergent, it is equal.
Once they are equal, it will be 0. So, this
is 0 for every f belongs to x.
Now, from this, we can say x minus y equal
to 0. If this is 2 for every f, then, this
must be 0 for every y and this implies x equal
to y. Why is it 0?
There is a lemma. This is by the following
result. What is the result? The result is
corollary of the Hahn Banach theorem.
The corollary of the Hahn Banach theorem says,
for every x 
in a normed space x norm, we have norm of
x equal to supremum of mod f x over norm f
belongs to x dash and f is not equal to 0.
Hence, if x 0 is such that f of x 0 is 0,
for all f belongs to its dual, then x 0 will
be 0. So, because of this result, we can say
that f of x minus y 0 implies x minus y is
0.
Now, this result follows from Hahn Banach
theorem. How does it follow from Hahn Banach
theorem? What is Hahn Banach theorem? Hahn
Banach theorem says that if x be a normed
space and x 0, if I picked up any non zero
point here, then corresponding to this x 0,
we can find f or f 0 such that f 0 x 0 is
1.that
The Hahn Banach theorem is: Let x be a normed
space and x 0 be a non 0 element of x, then
there exist a bounded linear functional f
delta. There exists a bounded linear functional
f delta on x such that f delta x 0 is norm
of x 0 and norm of f delta is 1.This is what
we call as the Hahn Banach theorem in case
of the normed space.
Let x be a normed space and x 0 be a non 0
point in this. Then there exists a boundary
linear functional f delta such that image
of this x 0 under f delta is norm of x 0 and
the norm of this is 1.
So, using this, we can say the norm of x we
wanted to show this. So, start with this supremum
mod f x over norm f when f belongs to the
dual and f is not equal to 0. Obviously, this
will be greater than equal to this particular
f delta. So, this is greater than equal to
f delta x, f delta x over norm of f delta.
But, x is fixed. So, norm f delta we can find
out if such f delta where norm is 1. So, this
is equal to f delta x. But f delta x is equal
to norm of x. So, this is norm of x.
So, this part is greater than or equal to
norm of x, but mod of f x is less than equal
to norm of f into norm of x. So, this implies
mod f x over norm of f supremum is taken over
all f belongs to this. f is not equal to 0
will remain less than equal to norm x.
So, combining these two, we get this lemma
is 2, A is 2, A follows. Now, if A is 2, then,
what he says is x 0 is such that f of x 0
is 0 for all x, then x 0 must be 0.Now, if
this part is 0 for all f, then; obviously,
the supremum will be 0. Obviously, norm of
x 0 will be 0. So, it follows immediately
and norm x 0 implies x 0 must be 0 because
it is a norm.
So, this result, f of x minus y equal to 0
implies x minus y 0 follows from this result.
Now, let us see the second part of this. Proof
of second part:
Every subsequence x n converges weakly to
the same limit x. Now, this follows here,
since x n converges to x weakly. This is given.
So, this implies that f of x n converges to
f x for every f belongs to x dash.
But, f of x n is a convergence sequence of
a scalar and every convergence sequence is
scalar, the subsequence is to converge to
the same limit. So, all of its subsequences
will converge to the same limit point. Hence,
this follows.
Now, part c. What is part c is that sequence
norm of x n is bounded.
To show that the sequence norm of x n is bounded,
let us start with this given x n converges
to x weakly. So, this implies f of x n will
converge to f x for every f belongs to the
dual. It means the sequence f x n is a convergent
sequence.
Every convergent sequence is bounded sequence
of scalars. So, it is bounded. Therefore,
there exists a constant c which depends on
f such that mod of f of x n, this will remain
less than equal to constant c that depends
on n for all n. c will not depend on n, it
will depend on f.
Because if f changes, the corresponding constant
will change. So, we get this. Now, let c be
the mapping from x to x double dash which
sends x to g of x.This is a cannonical mapping
which we have already discussed.
So, the cannonical mapping is defined by g
x f equal to f of x. It is linear. This is
already shown to be linear and bounded also
So, let us consider g of x n f mod of this
is equal to f of x n.
Mod of g of x n f is equal to mod of f of
x n, but mod of x is less than equal to C
f. So, this is less than equal to C f for
every f belongs to the dual.
It means that this sequence mod of g x n f
is a bounded sequence for every f belongs
to the dual.
So, this operator g of x n is a bounded operator
pointwise it is pointwise bounded because
g is defined on x dash, and g of x n f is
less than equal to c for every f and c depends
on f.
So, it is a pointwise bounded theorem. Since
g x n is defined from x dash, this is defined
from X dash to Rnx.
It is an element of x double dash, but basically
the domain will be g n of f will be the point
in real. So, it is defined on the real or
c.
Now, this will be a complete normed space,
whether x is complete or not, dual space will
always be complete. So, a sequence of the
operator is defined on a Banach space x. This
is x to y.
This one is R n c. So, this will be uniform
boundedness theorem. The uniform 
boundedness theorem says that if T n be a
sequence of operators from x to y where x
is a Banach space, y may or may not be a Banach
space, just norm space such that norm of T
n x. This is a bounded sequence for every
x belongs to x.
Then, the sequence of the norm is bounded.
Now g n is defined on x dash which is a complete
norm space, Banach space and g is bounded
point y.
So, according to the Banach uniform boundedness
theorem, we say norm of g n will be bounded.
So, from here, using the canonical mapping
concept 
and uniform boundedness theorem, we say that
norm of this sequence g of x n is bounded.
What 
is the canonical mapping? When g of x n is
there, then, norm of g x n is equal to norm
of x n, which is the same as norm of x n by
cannonical mappings because the operator which
we have defined is also bounded.
Hence the theorem is proved. So, this proves
c because we want the norm of x n to be bounded.
So, this is convergent.
Now, in a general norm space, we have seen
the concept of the weak convergence and strong
convergence. Why do not we have this concept
in a real or complex, when x is reduced to
the real set of earlier number or set of complex
number or in general a finite dimensional
space?
The thing is in case of finite dimensional
space the weak convergence and strong convergence
are equivalent concepts. It means that strong
will imply weak and weak will imply strong.
But if x is not a finite dimensional space,
then these two concepts differ. Strong always
implies the weak, but weak may not imply the
strong convergence.
We will see that in case of the finite dimensional
space, the strong convergence and weak convergence
are identical. So, that is the next target
of
We have the theorem. Let x n be a sequence
in a normed space x norm space norm, then
the following result hold. Strong convergence
implies weak convergence with the same limit.
The converse of a 
is not generally true.
And c part is, if the dimension of x is finite,
then, weak convergence 
implies 
strong convergence. Let us see the proof.
Strong convergence always implies the weak
convergence. So, it is given x n converges
to x strongly means under this norm, that
is norm of x n minus x goes to 0 as n tends
to infinity.
We want the weak convergence. So, start with
this mod of f x n minus f x
Now, f is a bounded linear functional. So,
this can be written in this form, further
f is bounded. So, we can say this is less
than equal to norm f norm x n minus x as f
is bounded. This is true for every f belongs
to the dual.
Now, x n goes to x is given. So, it will not
tend to 0 as n tends to infinity, because
this is a bounded and this will go to 0.
So, f x n converges to n. Therefore, f x n
converges to x weakly, because this is true
for every x.
Now, part b, the converse of this is not true
in general. It means in a general normed space,
weak convergence need not imply the strong
convergence. So, we have to take a counter
example where the sequence converges weakly,
but it is not strongly. I take x to be a Hilbert
space. Hilbert space is also a normed space,
but every Hilbert space is not a normed space.
We can introduce the norm is 
an inner product norm of x is the inner product
x x under root. So, we can find out the.
So, let us take the Hilbert space and let
e n be an orthonormal sequence in a Hilbert
space H. Now, it is given that weakly convergent.
Let f is an element belonging to the dual,
that is f is a bounded linear functional on
a Hilbert space H. So, it is representation
by Riesz theorem.
By Riesz Representation Theorem, every bounded
linear functional f can be represented in
terms of the inner product.
By- bounded linear functional, x can be represented
in terms of the inner product x z where z
is uniquely determined by f and norm of f
equal to norm of z. x belongs to H. H and
f is this by Riesz Representation Theorem.
So, we get f of e n f of e n will be inner
product of e n z.
Now, use the Bessel’s Inequality. What the
Bessel’s Inequality says is that sigma of
the inner product x n y n mod square is less
than equal to sigma norm of x into norm of
y. That is what is Bessel’s Inequality.
Let e k be a orthonormal sequence inner product
space, then for every x belongs to x, that
is if e k be orthonormal sequence in an inner
product space x, then for every x belonging
to capital X, the sigma of inner product modulus
of inner product x e k whole square k is equal
to 1 to infinity is bounded by norm of x square.
This is what inner product at the Bessel’s
Inequality says.
Apply the Bessel’s Inequality here. What
we get from there is, n is equal to 1 to infinity
modulus of inner product of e and z. This
whole square is dominated by norm of z square,
but this is finite. So, this series converges.
Therefore, the n th term must go to 0. So,
inner product of e and z must go to 0 as n
tends to infinity, but what is this, f of
e n goes to 0 as n tends to infinity.
So, e n 
converges to 0 weakly because this is 2 for
every f belongs to h dash dual of this. So,
e n converges to 0 weakly, but e n does not
converge strongly to 0.
Sir..
This is because norm of e n minus e m whole
square, this is inner product e n minus e
n e n minus e m and this will be 2.
So, it is not Cauchy. Therefore, it will not
converge. So, e n does not go to 0.
Therefore, every weak convergent need not
imply the strong convergence.
Now, third part: the proof of this, if dimension
of this is finite, then weak convergence and
strong convergence are the same.
So, let the dimension of X be n and e 1, e
2, e n be the basis elements for x.
Let x n, n comma x are the elements of x.
x n can be expressed in a linear combination
of e 1 e 2 n.
There exist scalars alpha 1 n, alpha 2 n,
alpha n n such that the linear combination
of this is x n.
Let the dimension of this be m.
Now x can also be written as alpha 1 e 1 alpha
2 e 2 and alpha n e n in terms of the basis
elements.
Now, it is given x n x m converges to x strongly,
then; obviously, x m will converge to x weakly.
It is by result, every strong convergence
imply weakly.
So, now, let x m converges to x weakly. We
want this converge to this. So, that is given
as f of x n goes to f x for every f belongs
to the dual of it.
For every f belongs to dual, now in particular,
the f 1, f 2 or f n’s, these are also the
elements of x dash where f i e k 
is the chronical delta k which is 1 if I 
is equal to k, otherwise 0. f 1, f 2, f n
will form dual basis. So, these are the elements
of x dash.So, in particular, this f 1, f 2,
f n will also be dual. So, image of this x
n under f j will be equal to alpha j n x m.
As you apply the f j, f j e j will be 1 and
rest will be zero. So, alpha j m and this
is true for all and what is the f j x is alpha
j.
Now, it is given that f j x m converges to
f j x. This is given.
This implies alpha j m goes to alpha j as
m tends to infinity. Therefore, norm of x
m minus x, consider this which is equal to
norm of sigma j equal to 1 to n alpha j m
minus alpha j e j.
Now, this will remain less than equal to sigma
alpha j mod of this into norm of e j.
Now, sigma j equal to 1 norm of e j is finite
and this part goes to 0. So, this will go
to 0. This is finite. So, entire thing will
go to 0 as m tends to infinity.
Therefore, x m will converge to x under this
norm strongly. So, x m converges x weakly
implies x m converges to x strongly and that
is proved
So, in case of the finite dimensional, this
is 2, but it does not mean that infinite dimensional
that we convergent never implies. They are
all examples of an infinite dimensional normed
space l1 has a property that weak convergence
also implies strong convergence. .
There are other spaces also parallel to l
1, where this is true, but in general, it
is not true. In general infinite dimension,
weak convergence need not imply strong convergence
in an infinite dimensional normed space.
But, there are certain spaces even in finite
dimensional where it comes, but we require
the proof of this theorem. So, just this is
an example that I have given in this.
Now, research is going on to find out the
criteria, the sufficient condition when the
sequence converges weakly under that norm
because always the infinite dimension space
is not necessary the weak convergence implies
a strong, but we can impose a restriction
on the sequences so that the sequence will
imply the weak convergence.
What is the criteria for weak convergence?
I will tell these results. The examples: in
a Hilbert space 
x n converges to x weakly if and only if inner
product x and z goes to x z for all z in the
space.
I think this follows immediately by Riesz
representation theorem because f of x n is
x n converges to because x n converges to
x weakly means f of x n goes to f x for every
f belongs to the dual.
But f of x n by Riesz theorem, this is the
inner product of x n z. This will go to the
x z by Riesz.
Therefore, in case of this space l p, the
criteria is that in the space l p where 1
is less than p, less than infinity, x n converges
to x weakly if and only if and only if the
sequence 
norm of x n is bounded. And second part is,
for every fixed j, we have x i j n goes to
x i j as n tends to infinity, where x n is
a sequence x i j n and x is a sequence x i
j.
V is for every fixed j, x i j n converges
to x i j, it means if f is a point in l p
dash dual of l p dash is l q.
So, belongs to the dual of this then f of
x n will go to f x by means of this v convergence.
Now, this will converge weakly if the coordinate
y is convergence of x n is also clear.
X j n, because j you fix it, means x i 1 n
will go to x i 1 x i 2 n will go to x i 2
n. So, if x n converges to x coordinate y
as well as the norm x n is bounded, then,
the sequence will converge weakly.
We are not going for detail in the proof of
this, but these are the results. So, this
type of the continues, means, you take the
space, find out the sufficient condition because
these are basically the necessary and sufficient
in both conditions. Sometimes, we are unable
to get both types. So, we get only the sufficient
part.
It is required condition when the sequence
converges weakly and that. Thank you very
much. .
.
.
