1
00:00:28,480 --> 00:00:42,040
So therefore this set of functions that we
talked about a minute ago is indeed a linear
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00:00:42,040 --> 00:00:51,629
space that is why I have called it a space
here And we will give that space a name So
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00:00:51,629 --> 00:01:11,740
we will call that space V0 So V0 is a set
now I am going to write mathematical notations
4
00:01:11,740 --> 00:01:18,610
V0 is a set of all x of t such that 2 things
happen x of t now you know x is a function
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00:01:18,610 --> 00:01:28,560
so when I write like this what I mean is I
am suppressing the explicit value of the independent
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00:01:28,560 --> 00:01:35,090
variable but I recognise there is an independent
variable here then treating the whole thing
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00:01:35,090 --> 00:01:36,900
as an object
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00:01:36,900 --> 00:01:47,900
It is a function I am treating the whole function
as an object and this object belongs to L2R
9
00:01:47,900 --> 00:01:54,250
recall that L2R is the space of all functions
which are square integrable and this stands
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00:01:54,250 --> 00:02:00,810
for belongs to So such that x belongs to L2R
and x is piecewise constant on all intervals
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00:02:00,810 --> 00:02:14,020
of the form n n Plus1 n integer Now once we
have talked about V0 in fact the reason for
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00:02:14,020 --> 00:02:23,240
giving the subscript 0 here is that we are
talking about 2 to the power of 0 as the size
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00:02:23,240 --> 00:02:28,190
of the interval That is important enough I
think to make a noting
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00:02:28,190 --> 00:02:34,800
So we say V0 because of piecewise constancy
on intervals of size 2 raised to 0 which is
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00:02:34,800 --> 00:02:42,299
one And similarly therefore in fact you know
you could call it 2 raised to the power of
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00:02:42,299 --> 00:02:49,970
0you could call it 2 raised the power of minus0
We will prefer to use 2 raised to the power
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00:02:49,970 --> 00:02:58,660
minus0 because it will be consistent in future
So we could similarly have V1 then Similarly
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00:02:58,660 --> 00:03:06,470
V1 is a set of all x let us define it the
set of all x x belongs to L2R and x is piecewise
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00:03:06,470 --> 00:03:08,930
constant on standard 2 raised to the power
minus1 intervals That is intervals of the
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00:03:08,930 --> 00:03:13,999
form n into 2 raised to minus1 n Plus 1 into
2 raised to minus1 for all n integers
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00:03:13,999 --> 00:03:15,410
In general we have Vm the set of all xt for
completeness we should write down the definition
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00:03:15,410 --> 00:03:17,349
properly and x is piecewise constant on all
open intervals of the form simple enough To
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00:03:17,349 --> 00:03:25,189
fix our ideas let us sketch a couple of examples
So let us take an example of x belonging to
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00:03:25,189 --> 00:03:30,530
V2 It would look something like this you would
have intervals of one fourth here And in fact
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00:03:30,530 --> 00:03:35,290
to be complete we should also include intervals
before 0 and so on
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00:03:35,290 --> 00:03:43,499
And we have piecewise constancy on these and
so on then And please remember x is also in
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00:03:43,499 --> 00:03:51,780
L2R so when you say it is in V2 it is automatically
of course in L2R and that means that if I
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00:03:51,780 --> 00:04:00,560
take the sums squared of all these constants
that sum square is going to be finite that
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00:04:00,560 --> 00:04:16,739
is an important observation The constants
that we assign here must be such that when
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00:04:16,739 --> 00:04:23,330
we sum the square of all of them magnitude
square of all of them that sum must converge
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00:04:23,330 --> 00:04:28,060
This observation is so important that I think
we should make a note of it
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00:04:28,060 --> 00:04:40,000
So we are saying the sums squared the absolute
squared sum of the piecewise constant values
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00:04:40,000 --> 00:04:53,790
come in all these Vm must be convergent and
this follows from belonging to L2R Let us
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00:04:53,790 --> 00:05:01,450
also take an example of a function belonging
to V minus1 So minus1 means intervals of size
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00:05:01,450 --> 00:05:11,220
of 2 2 raised to the power minus of minus1
so intervals of size 2 and so on there and
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00:05:11,220 --> 00:05:27,230
so on there and we are piecewise constants
there and so on here and so on there So now
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00:05:27,230 --> 00:05:46,990
we get our ideas fixed what we mean by the
spaces VM
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00:05:46,990 --> 00:06:03,250
Now the moment we put down these spaces with
these examples so clearly we see a containment
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00:06:03,250 --> 00:06:07,590
relationship so there is a relation between
these spaces they are not arbitrary they are
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00:06:07,590 --> 00:06:13,580
not just totally disjoint and unrelated In
fact you can notice that if a function belongs
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00:06:13,580 --> 00:06:19,400
to V0 for example which means that it is piecewise
constant on the standard unit intervals it
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00:06:19,400 --> 00:06:25,000
is also going to be piecewise constant on
the standard half intervals And for that matter
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00:06:25,000 --> 00:06:27,390
if a function belongs to V1 which means that
it is piecewise constant on the standard half
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00:06:27,390 --> 00:06:31,940
intervals it is automatically going to be
piecewise constant on the standard one 4th
45
00:06:31,940 --> 00:06:32,940
intervals
46
00:06:32,940 --> 00:06:40,650
To exemplify this let me go back to this example
of x belong into V minus1 that I have here
47
00:06:40,650 --> 00:06:51,280
Notice that this function is piecewise constant
on the standard intervals of size 2 So obviously
48
00:06:51,280 --> 00:07:04,280
should take the standard intervals of size
1 for example 0 to 1 1 to 2 2 to 3 3 to 4
49
00:07:04,280 --> 00:07:13,120
minus2 to minus1 minus1 to 0 and so on the
function is still piecewise constant So therefore
50
00:07:13,120 --> 00:07:22,690
a function that belongs to V minus1 automatically
belongs to V0 a function that belongs to V0
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00:07:22,690 --> 00:07:25,160
automatically belongs to V1
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00:07:25,160 --> 00:07:38,760
And therefore there is a ladder of subspaces
that is implied here What is that ladder the
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00:07:38,760 --> 00:08:07,140
space V0 is contained in V1 the space V1 is
contained in V2 and so on this way and of
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00:08:07,140 --> 00:08:17,791
course the space V minus1 is contained in
V0 the space V minus2 in V minus1 and so on
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00:08:17,791 --> 00:08:24,170
And we expect intuitively as we move in this
direction we should be going towards L2R Of
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00:08:24,170 --> 00:08:39,620
course it is an important question what happens
when we go in this direction That is interesting;
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00:08:39,620 --> 00:08:48,210
we will spend a minute now and reflect on
that
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00:08:48,210 --> 00:09:03,490
So you see what happens when we go leftwards
What I mean by that is of course you have
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00:09:03,490 --> 00:09:12,230
V0 contained in V1 contained in V2 and then
V minus1 contained in I mean contained in
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00:09:12,230 --> 00:09:30,410
V0 yes and so on here and so on there What
happens when we go this way What do we think
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00:09:30,410 --> 00:09:54,220
should happen what are we doing we are taking
piecewise constant function on larger and
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00:09:54,220 --> 00:10:32,480
larger intervals let us write that down so
piecewise constant functions
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00:10:32,480 --> 00:10:34,459
on larger intervals
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00:10:34,459 --> 00:10:54,399
Now you see what is the L2 norm of functions
as you go leftwards What kind of a form will
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00:10:54,399 --> 00:11:04,050
it have It is going to have a form like this
summation on n now we see remember the L 2
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00:11:04,050 --> 00:11:10,610
norm is the integral of the absolute square
of the function And please remember the function
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00:11:10,610 --> 00:11:21,910
is piecewise constant so you have one constant
let us call it Cn on the nth interval and
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00:11:21,910 --> 00:11:24,269
the interval is of size 2 raised to the power
minus m So this is essentially you know you
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00:11:24,269 --> 00:11:26,230
are talking about integrating mod Cn square
it is a constant over an interval of minus2
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00:11:26,230 --> 00:11:31,507
raised to the power minus m and please remember
m is negative and m goes towards minus infinity
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00:11:31,507 --> 00:11:33,139
as a go leftwards
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00:11:33,139 --> 00:11:34,490
That is the same thing as 2 raised now you
see 2 raised to the power minus m is 2 raised
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00:11:34,490 --> 00:11:43,589
to the power mod m in the context of negative
m and summation on n mod Cn square Now you
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00:11:43,589 --> 00:11:52,589
see the subtle point is that if this needs
to be finite irrespective of how large m is
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00:11:52,589 --> 00:11:58,589
we have no control on this except that this
part must be finite But then when we say finite
76
00:11:58,589 --> 00:12:00,680
if it is nonzero and if we allow m to grow
without bound this is going to diverge
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00:12:00,680 --> 00:12:02,175
So the only way in which this can converge
no matter how large I mean large in the sense
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00:12:02,175 --> 00:12:03,175
large in magnitude how large in magnitude
m is no matter how large in magnitude m as
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00:12:03,175 --> 00:12:08,819
if this is to converge then this must be 0
A very important conclusion So we are saying
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00:12:08,819 --> 00:12:13,420
that if 2 raised to the power mod m summation
n mod Cn square must converge no matter how
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00:12:13,420 --> 00:12:14,970
large or how negative then we must have summation
over n Cn square tending to 0 So essentially
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00:12:14,970 --> 00:12:15,970
what we are seeing is as we move leftwards
we are going towards the 0 function
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00:12:15,970 --> 00:12:18,889
A point that takes a minute to understand
but is not so difficult as you can see So
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00:12:18,889 --> 00:12:26,540
now we have very clearly an idea of our destination
as you move up this ladder towards plus infinity
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00:12:26,540 --> 00:12:32,310
and as we move down the ladder towards minus
infinity and we can formalise that What we
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00:12:32,310 --> 00:12:36,069
are saying is moving upwards now you know
one has to use proper notations We would have
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00:12:36,069 --> 00:12:38,529
been tempted to say something like limit as
m tends to infinity or plus infinity or something
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00:12:38,529 --> 00:12:41,300
like that but you see it is not really correct
to talk about limits of sets
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00:12:41,300 --> 00:12:45,220
So we need to use notation that is appropriate
in the context of sets namely union So when
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00:12:45,220 --> 00:12:55,480
we take a union of 2 sets and if one set is
contained in the other we are automatically
91
00:12:55,480 --> 00:13:02,939
taking the larger set So moving upwards is
attained by using union in other words we
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00:13:02,939 --> 00:13:14,639
are saying the union of Vm m over all the
integer should almost be L2R now that is where
93
00:13:14,639 --> 00:13:38,550
the little catch is
94
00:13:38,550 --> 00:13:43,360
I mean we would have been happy to write is
equal to L2R but you know we need to make
95
00:13:43,360 --> 00:13:51,769
a little detail here we need to put something
called a closure I will explain what I mean
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00:13:51,769 --> 00:13:57,490
by a closure You know suppose you were to
visualize L2R to be like an object with a
97
00:13:57,490 --> 00:14:05,040
boundary So suppose this were L2R just notionally
and this is the boundary of L2R so to speak
98
00:14:05,040 --> 00:14:16,379
so it is a space you know Now what we are
saying is as we go in union that is union
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00:14:16,379 --> 00:14:22,860
m Overall integers of Vm it would cover all
the all the inside cover all the interior
100
00:14:22,860 --> 00:14:25,759
But then it might leave out some peripheral
things on the boundary so it may also cover
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00:14:25,759 --> 00:14:29,110
some part of the boundary Now of course do
not ask me at this stage what we mean by boundary
102
00:14:29,110 --> 00:14:30,110
and interiors save that you know you are talking
about situations you know boundary now informally
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00:14:30,110 --> 00:14:31,110
when you say boundary you are talking about
functions where moving in a certain direction
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00:14:31,110 --> 00:14:32,110
does not remain in L2R moving in the other
one does
105
00:14:32,110 --> 00:14:35,162
So you know this boundary and the interior
at the moment needs to be understood only
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00:14:35,162 --> 00:14:38,450
informally But what we are saying is as far
as this union goes it can take you almost
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00:14:38,450 --> 00:14:39,709
all over L2R it covers all the interior it
may also cover quite a sizeable part of the
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00:14:39,709 --> 00:14:41,899
boundary but it might leave some patches of
the boundary untouched And therefore when
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00:14:41,899 --> 00:14:46,509
we do a closure we are covering up those patches
What we just did was covering up those patches
110
00:14:46,509 --> 00:14:47,910
so closure means cover-up boundary patches
111
00:14:47,910 --> 00:14:49,330
Now this is a small detail and we need not
spend too much of time in reflecting about
112
00:14:49,330 --> 00:14:54,089
this idea of closure and so on But to be mathematically
accurate we do need to note that it is after
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00:14:54,089 --> 00:14:55,949
closure that the union overall m integers
of Vm becomes L2R Otherwise it is almost L2R
114
00:14:55,949 --> 00:15:01,279
which means that when you take this union
that is when you make piecewise constant approximation
115
00:15:01,279 --> 00:15:05,980
is on smaller and smaller and smaller intervals
You can go as close as you desire to a function
116
00:15:05,980 --> 00:15:09,089
in L2R so you can reduce the L2 norm of the
function to 0
117
00:15:09,089 --> 00:15:15,809
So if you look at it that is what we mean
by that is what mean that is what is implied
118
00:15:15,809 --> 00:15:21,910
by boundary You know you can you can go as
close as you like to a certain function you
119
00:15:21,910 --> 00:15:22,910
can make the L2 norm 0 but still it would
not quite reach there So you know you could
120
00:15:22,910 --> 00:15:24,899
just visualize and you might just be a teeny-weeny
bit inside that boundary but not quite on
121
00:15:24,899 --> 00:15:27,180
the boundary And how teeny-weeny as small
as you like they touch where the union takes
122
00:15:27,180 --> 00:15:28,259
you that is the subtle idea of closure
123
00:15:28,259 --> 00:15:31,720
Anyway as I said we do not need to spend too
much of time in talking about this closure
124
00:15:31,720 --> 00:15:35,152
but we should be aware of this idea because
when we need literature on wavelets or for
125
00:15:35,152 --> 00:15:36,152
that matter when we really wish to put down
the axioms of multiresolution analysis properly
126
00:15:36,152 --> 00:15:41,802
we must be aware that this closure is required
So much so anyway Now let us take the 2nd
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00:15:41,802 --> 00:15:42,802
of our inferences here moving downwards so
to speak So how do we move downwards
128
00:15:42,802 --> 00:15:43,802
Just as union takes you upwards Intersection
takes you downwards So if you take an Intersection
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00:15:43,802 --> 00:15:44,879
on all m belonging to Z on VM there I do not
need to worry about closure and anything of
130
00:15:44,879 --> 00:15:46,839
that kind I can simply put down this is a
trivial subspace essentially the subspace
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00:15:46,839 --> 00:15:49,089
of L2R with only the 0 function included This
is called the trivial subspace Now again I
132
00:15:49,089 --> 00:15:52,730
must make an observation here to clarify The
trivial subspace is not the same as the null
133
00:15:52,730 --> 00:15:54,209
space The trivial subspace has only the trivial
0 element in it the null subspace does not
134
00:15:54,209 --> 00:15:56,209
have any element So that is a subtle distinction
and we must bear in mind that we are talking
135
00:15:56,209 --> 00:15:56,550
about the trivial subspace