Yeah let us begin with this ah theorem.
Suppose f is an element of this space that
is all square integrable functions minus pi
to pi . let f suffix N of x be given by this
expansion a naught plus summation k equals
1 to , a k cos kx plus b k sine kx, where
a k and b k are the fourier coefficients of
this function f which is a square integrable
function. .
Then f of x N heads to f in L 2 over the internal
minus pi to pi as N goes to infinity, which
means the norm between fN and f the integral
norm this heads to 0 as N goes to infinity.
So this is what the statement of the theorem
is consider as square integrable function,
let fN be the fourier expansion then this
expansion heads to this function in the L
2 sense that means, L 2 norm of f N minus
f heads to 0 heads to 0 as N goes to infinity
of course, if you are little careful about
it you might sense that f this function is
L 2 is in L 2. We have to say that in some
where we can approximate this function by
a piecewise smooth periodic function, if you
somehow get it in that form then we can approximate
that function ah the original function by
a piecewise smooth periodic function and approximate
the piecewise smooth periodic function uniformly
by the ah fourier representation.
There are 2 two parts to this right i mean
all your discontinuities can happen though
it is square integrable, you may have discontinuities
and you will have to extend them carefully
right, so we will we will see these aspects
carefully,
but let us get a feel for the sketch of the
proof the proof sketch involves 2 steps, first
show that any function in L 2 over the interval
minus pi to plus pi can be approximated 
by a piecewise smooth periodic function g.
And then 
approximate g uniformly and we saw the theorem
that if you have if it converges uniformly
therefore, it converges in L 2 by its fourier
series . So we need to get an idea towards
1. We have already established 2 right the
second part we have already established suppose
you it converges uniformly then it ah it converges
in L 2 right. We have established this result
already, so let us get the idea towards 1.
.
An element f belonging to this space which
is a space of all square integrable functions,
may not be continuous even if it is continuous
its periodic extension is often not continuous,
there are plenty of such cases a good example
is f of x equals x over the interval minus
pi to pi right if you if f of x equals we
saw the points of discontinuity.
Now let us just take a sketch of this we have
minus pi pi 2 pi 3 pi and so on. So have something
like this say and this is this is a discontinuity
point is a jump, similarly take a periodic
extension 
you have a same problem . 
So let us form, another function g of x 
that agrees with f of x 
at all segments except 
on segments 
connecting 
the continuous components of f it is very
important.
So let us from another function g of x that
agrees with f of x at all segments except
on segments connecting the continuous components
of f right. So these are this is a continuous
component this is a continuous component at
these points g of x is going to be differing
from f of x that is the whole idea right let
us see what we could do .
So revisiting this like a spy pi is 2 pi its
3 pi now here i am connecting them 
and i am making them continuous ok. So i think
you can see i can i can i can i am i am putting
this i am joining these segments i am and
i am making them continuous 
we are fine if .
We choose is very important 
a differentiable periodic function. g of x
such that the norm between f and g in the
L 2 sense is less than epsilon for a tolerance
epsilon right,this is the first first step
right you try to make it you choose a periodic
function differentiable as well such that
in the L 2 sense f minus g this norm right
in L 2 sense is within epsilon correct i can
choose such ah such a function. Now we this
we have to recall from our last theorem .
from our last theorem it is 1 of the previous
theorems let us rec recall the following,
the fourier series of a piecewise smooth 2
pi periodic function, f of x converges uniformly
to f of x on the interval minus pi to plus
pi. So let gN of x b equals C naught the summation
k equal to 1 to N Ck cos kx plus dk sine kx
. Where cks and dks or fourier coefficients
of g of x right. We know how to extend the
function f of x to g of x right that we choose
a differentiable periodic function and extend
the function original function f which was
just square integrable, but possibly discontinuous
at certain points right.
Now from the theorem above we can uniformly
approximate g of x by some g N of x right
and how do we do that,.
by choosing a large positive N naught. We
can set i mean with the fourier expansion
g of x minus gN of x the absolute value of
this is less than epsilon for all x in the
interval minus pi to plus pi this is very
important because it is uniform.
Now the norm of g N minus g square in the
L 2 sense is integral minus pi to plus pi
absolute value of g of x minus g N of x square
dx, and this is minus pi to plus pi this is
epsilon square dx which is 2 pi epsilon square
that is for N greater than N naught. Now therefore
the norm of g minus g N is basically bounded
by i mean if you choose this to be equal to
fN of these little tolerance norm square is
within tolerance to pipe system, square this
norm is if you choose some epsilon you can
meet it with epsilon times root 2 pi right
straightforward .
Now let us consider the norm of f minus g
N that is we had our original function f which
is in L two, but which possibly had discontinuity
points etcetera right this norm i can write
it as follows f minus g N is basically i add
and subtract g into it is familiar trick that
1 does. Now we apply the triangle inequality
this is f minus g norm plus norm g minus g
N this is because triangle inequality and
this we said is less than i just put a less
than here less than epsilon this is less than
epsilon times root of 2 pi for N greater than
N naught .
Now gN of x is a linear combination of sine
kx and cos kx for values k equals one. So
on till capital N , so therefore gN of x belongs
to the space V N, now its easy to interpret
this because from an earlier lemma f suffix
N is a closest element from the space V N
to f in the L 2 sense right.
So from .
our earlier lemma earlier result f suffix
N is the closest element from this space VN
to this function f in the L 2 sense . So therefore,
if you look at the norm f minus f N f suffix
capital N this is less than or equal to norm
f minus gN this is less than epsilon times
1 plus root of 2 pi, which means given a tolerance
epsilon, we can get as closely as we can to
the original function belonging to L 2 using
the fourier expansion.
So this is an important result ok i think
this is sort of summarizing the various modes
of ah convergence and basically trying to
tell that if you are given a function which
is possibly suppose you are not able to achieve
point wise convergence or ah uniform ah convergence
basically look at the weaker form of convergence
which is basically in the L 2 sense and for
you to get towards L two. So take a function
f which is belonging to L 2 it may have discontinuities,
but extend that function using a differentiable
function right and connecting those points
of discontinuities through ah a differentiable
function and periodically extend that right,
and you can figure out such a g which is close
to f within a tolerance epsilon and then use
uniform. If you can do a uniform approximation
then you can show that that result will converge
in the L 2 sense and then we are home for
good right and how do we approximate the function
g itself is a construction process which is
not discussed in the proof of this result,
but if you can do that then the . it is valid
ok.
So this gives us an idea that fourier expansion
is quite generic enough we can figure out
we we studied the convergence aspects at a
point of discontinuity at points where it
was continuous it was no problem, but we had
subtle issues where we had to link the additional
ah we had to bring the additional a kernel
and look at some subtle aspects ah where those
points were a differentiable right, and then
we looked at the L 2 convergence aspects and
i think we have sort of ah summarized ah the
convergence aspects of the fourier series
expansion and this is a very useful tool for
expanding signals in the fourier space right,
and i think you can think about a similar
expansion in the wavelet sense you have all
these aspects.
That you have to deal with i mean i remember
in 1 of the theorems which i left it as a
as a theorem without a proof that you can
expand this in terms of the direct some spaces
in the limiting form, and when you have to
look at the limiting form then you have to
look at all these subtle aspects at points
of continuities at points of discontinuities
etcetera.You can sort of take a lot of analysis
in this work and think about in the wavelet
expansion as well right.
I think you now get a very clear picture as
to how to expand functions using bases and
what are the subtle aspects ah particularly.
You are in trouble if you have jump points
or discontinuity points and i think this is
where you have to spend little little bit
of thought to ah carefully see in what sense
it converges ok. So with this we are done
with this module on fourier expansion ah our
next lecture would be on kl ok, so we can
stop here.
