Now, I should have basically covered uniform
convergence and point wise convergence, but
I deliberately avoided this ah. But if you
go through analysis lectures basically we
will start with sequences and then you will
start with what point wise convergence is
and that leading into uniform convergence
and implications and so, on. But, here we
will just stress upon only certain ideas from
analysis which are useful for us to discuss
the convergence aspects of Fourier series
ok.
One of the important concepts is uniform convergence
so, let me define what this means . 
Given a sequence of functions f small n subscript
small n x , 
for a given tolerance epsilon greater than
0 some positive tolerance . There exists capital
N such that absolute value of F N of x minus
F of x is less than epsilon for every x and
all small n greater than or equal to capital
N .
And this is basically what you reform convergences,
I give you some function, I give you a tolerance
epsilon, I give you some number capital N,
it could be a million, it could be a billion
some number. Then I take f of x Fourier series
expansion, I take f n of x truncated to n
partial sum and this is within epsilon for
every value of x and all small n greater than
or equal to capital N right. That is; that
means, if I take the sum it it it basically
is is bounded, this is what I mean here is
this what I mean here. And if the the tolerance
depends upon the point, then it is point wise
conversions.
So, uniform is a more ah sort of a more general
statement right. So, you can think think about
epsilon which which you know which I can give
you a tolerance. And the tolerance if it is
dependent upon the point; that means, every
if if you evaluate the function and you look
at the approximation the tolerance if it depends
upon the point; that means, some point x naught
I give you some tolerant tolerance epsilon
epsilon ah which depends on x naught, for
some other it is absolutely depending on x1
so, on and so, forth.
Then it is for a very specific set of points.
Here it is not dependent on any point, I give
you some tolerance for all points x this is
bounded with an epsilon ok, the absolute value
is bounded within epsilon . So, there is another
definition which is I will state a function
is piecewise smooth , 
if it is continuous and it is derivative is
defined everywhere except importantly except
possibly 
at a discrete set of points . An example of
this is the saw tooth stall saw too ah saw
tooth waveform ok.
A function is piecewise smooth if it is continuous
and it is derivative is defined everywhere
except possibly at a discrete set of points
at some points. So, saw tooth waveform is
one of them I mean it is imagine this 
is a saw tooth form . Now there is a result
so, what does it mean? Ah the smoothness property
is very important because if it is it depends
upon the number of derivatives. If it is nth
order smooth let us take the function f of
x equals x, first derivative is a constant
x, the second derivative is 0. It is vanishing
take f of x equals x square, second derivative
it exists right second derivative first derivative
is 2 2 x, second derivative is 2 constant
right. 2 derivatives exist after which it
just vanishes.
So, if a function is nth or smooth if all
the n derivatives exists and you take the
next derivative it just vanishes next higher
order derivative it just vanishes. So, when
we talk about smooth functions it should be
really differentiable many many many times.
Now you think about the Gaussian function
, how many derivatives does it have? Many.
So, this Gaussian kernel it is used useful
in many many applications in signal processing
because it is smooth, keep on taking the derivatives
it is smooth to that extent right. This is
one of the important ah ideas in signal processing.
And I think in the wavelet also the hour for
example is not a nth order smooth ah function,
just take the immediate derivative it just
there is trouble at the points. And of course,
we can investigate the time frequency uncertainty
at those points, ah it is ah it is it is it
is an interesting exercise. But, I think what
you have to bear in mind is since it is not
smooth if you want a smoother approximation,
you may need conditions in creating your wavelet
functions etcetera. So, this is one of the
important considerations in signal processing
ok.
So, a little bit of digression, but I think
you have to get the idea of what smoothness
is.
Now, we have a theorem , 
the Fourier series of a piecewise smooth 2
pi periodic function converges uniformly to
f of x on the interval minus pi to pi . I
mean f of x equals x is not smooth everywhere
right. I mean it is exhibiting discontinuities
when you try to periodically extend it does
not Gibbs phenomena we saw that. Now, imagine
that we have periodic is it piecewise smooth
2 pi periodic function and the Fourier series
of that function converges uniformly to f
of x on the interval minus pi to pi you have
to prove this result .
So, we should prove this theorem with the
assumption 
that f is everywhere twice differentiable
, we will see the condition why this twice
differentiability needs to be ah satisfied
ok. So, let us start with the function, let
f of x be summation over n a n cosnx plus
b n sin n x . Now, I take a derivative 2 times
, this one thing nice about this trigonometric
functions cos becomes minus sign when you
take a derivative right, sign when you take
a derivative it becomes cos.
So, taking twice derivative is not so difficult
for us. So, if this is giving us a n double
dash cosnx plus b n double dash sign n x , then
we can relate a n double dash and b n double
dash in terms of a n and b n ok. So, here
a n double dash is minus a n times n square
and b n double dash is minus these suffix
n times n square. Not too difficult just take
a derivative n pulls out take another derivative
n pulls out, cos becomes minus sin sin becomes
causes a negative sign, there it is a straight
forward .
Now, consider 
summation n equals 1 to infinity 
absolute value of a n plus absolute value
of p n . Now, this is equal to summation n
equals 1 to infinity absolute value of a and
double dash upon n square plus absolute value
of b n double dash upon n square right because,
we have the link between a n and b n. Now,
if f 2 dash is continuous 
then a double dash and b double dash stay
bounded by a quantity let us say this is some
M .
So, therefore, the will I use A somewhere
earlier? Ah maybe I can call this A set of
equation, this is I call A . 
B can be written or simplified as this is
bounded, this is less than or equal to summation
n equals 1 to infinity say M upon n square.
Let us say this is M and n if you want or
you can bring in 1 quantity which is M which
is max of M comma N you could do that as well,
but let us just get it into this form.
So, this is basically M plus N times sum ah
n equals 1 to infinity 1 upon n square and
we know this is finite and bounded and given
by pi square upon 6, it is not too difficult
to prove this result. So, now, if M and N
are finite, this is finite. So, everything
is finite therefore, you can say that this
is not heading to infinity and this is ah
this is basically ah bounded ok. So, is this
is this clear? Now, to ensure 
uniform convergence we need another another
result and we will prove that in in a in a
lemma ok.
So, this is one one part of it, if you look
at the statement of this theorem it says it
converges uniformly to f of x, piecewise smooth.
So, first we said is we want to start with
assumption it is twice differentiable and
if it is twice differentiable we said if this
is this function is continuous if f 2 dash
is continuous. Then their coefficients are
bounded by some quantities M and N and then
we replace that bound through an inequality
we say that the sum of our absolute value
of the coefficients is bounded. That is what
we have ah established that that is; that
means, it is finite it is strictly less than
infinity ok.
Let us get into the technical details of another
lemma which is useful towards the main result.
Suppose f of x is a naught plus summa k equals
1 to infinity k suffix k cos kx plus b suffix
k sin kx with k equals 1 to infinity absolute
value of a k plus absolute value of b k, this
being strictly less than infinity. See, this
is this condition is being satisfied in the
previous result right, we took the absolute
value of the coefficients and that is finite.
If this result is true then Fourier series
converges uniformly an absolutely to the function
.
So, basically now in the first part of the
previous result we have not proved it completely,
we have just established this result we need
to start with this condition here then the
correct that the coefficients are bounded
and then establish the uniform convergence
property. So, therefore, this lemma is nested
inside the theorem ok. I could have done a
lemma first and I could have proved something,
but I want you to get a feed of the flow of
the theorem. So, therefore, I am nesting the
lemma literally inside the theorem so that
you can understand how this works. So, we
will start with the proof .
Let us start with our familiar inequality
which is a triangular inequality, you take
absolute value of a k a cos kx plus b k sin
kx . This is less than or equal to absolute
value of a k plus absolute value of a b k
because, cosins and sin the absolute value
of these functions is less than or equal to
1 ok. This is the first thing that you would
do.
Now, let S suffix N of x be a naught plus
summa k equals 1 to capital N a suffix k cos
kx plus piece of xk sin kx .
Let us look at this partial sum. Now, f of
x minus S N of x is summation k equals N plus
1 to infinity a k cos kx plus b k sin kx.
Right, if you just remove this from k equals
1 to N, you you are left with the pile from
N plus 1 to infinity that you need to take
a sum.
Now absolute value of f of x minus S suffix
N of x , this is now less than or equal to
summation k equals N plus 1 to infinity absolute
value of a k plus absolute value of b k . But,
with our twice 
differentiability condition summation k equals
N plus 1 to infinity mod a k plus mod b k
is less than infinity right? We prove that
result because, that constant was heading
to pi square way.
Now, this is easy for us so for a given epsilon
greater than 0, there exists some capital
N naught which is greater than 0 . So, that
for N greater than N naught this implies summation
N plus 1 to infinity k equals N plus 1 to
infinity absolute value of a k plus absolute
value of b k , this is less than epsilon irrespective
of x .
This is very very important. Irrespective
of x this is strictly less than epsilon because,
you could choose your M and N and all these
things carefully. So, this establishes uniform
convergence. So, I would just close the proof
of this lemma here, I would also close the
proof of this theorem saying use lemma to
conclude the theorem ok.
So, this is a very important result, I think
it is a significant theorem it starts here
the theorem starts here and it it ends after
the lemma is proved and then as a consequence
of the proof of the lemma. So, the Fourier
series of a piecewise smooth 2 pi periodic
function converges uniformly to f of x on
the internal minus pi to plus pi. So, we will
stop here.
