.
So in todays ah lecture let us consider convergence
at a point of discontinuity right we proved
in the last class that there is convergence
at points where it is continuous. And we arrived
at conditions ah linking the dish slab kernel,
and the function g right that that satisfies
at the points of continuity right.
I mean we have that precise ah equation I
can just pinpoint to you. So, if you said
this g of 0 to two times f dash of x right
then they can extend this function g of u
across u equal 0, as a continuous function
and this is a very important step to show
that there is convergence.
Now, let us look at convergence at a point
of discontinuity . So, the first in the foremost
thing that comes into your mind is 
not all functions 
or continuous and periodic. And the other
part is if we 
consider periodic extensions 
it may not necessarily result in a continuous
function.
There are plenty of such cases right what
we did f of x equals x was one such case,
we start with the following definition we
know from basic calculus limit f of x minus
h h tending to 0 plus is f of x minus 0, this
is the left limit.
Similarly we have limit x tending to 0 plus
f of x plus h is f of x plus 0 this is the
right limit . When the function is continuous
the left and the right limits must be the
same and it has to evaluate to the function
at that particular point right, but it may
not be the case, but we will see the importance
of these left and right limits. Similarly
we can recall the differentiability conditions.
A function f is left differentiable at x,
if f dash of x minus 0 is limit, h tending
to 0 minus f of x plus h minus f of x divided
by h this is the left differentiable condition.
Similarly, a function f is right differentiable
at x if a f dash of x plus 0, is given by
the limit h going to 0 plus f of x plus h
minus f of x divided by h. So, you can think
about if you are going like this, imagine
your computing the tangent and you are going
like this then this is basically derivative
of x minus 0. On the other hand you have this
function like this and you are approaching
like this this is x plus 0 right. So, this
is a sort of air the pictorial notion that
you have to get when you actually evaluate
these this limits.
Now, for f of x equals 
minus x over minus pi less than or equal to
x less than pi and periodic extensions f of
pi plus 0 f of pi plus 0 is minus pi f of
pi minus 0 is pi. If we take f dash of pi
minus 0 is 1 f dash of pi plus 0 this is 1.
Now if you consider f of x is x for 0 less
than or equal to x less than or equal to pi
by 2 and pi minus x between pi upon 2 between
x lying between pi upon 2 and pi f dash of
x is not differentiable at pi upon 2. You
can see the slope how things are changing
when you are approaching just around pi by
2 right, whether you are just going to the
left side and right side you take the derivative
compute the left and right derivatives you
you will conclude that it is not the limits
do not agree.
Now, once we get this kind of feel I mean
you have a function it is not necessarily
continuous, but we take an extension of the
function and it there are discontinued continuity
is because of the periodic extension. What
can we say about the convergence at at such
a point or such points of discontinuities
ok.
So, this leads us into a theorem the statement
of the theorem is as follows. Suppose f of
x is periodic, and piecewise continuous 
suppose x is a point where f is left and right
differentiable , 
but not continuous .
So, look at the subtle point there is a left
derivative left derivative the right derivative,
but it is not continuous. The fourier series
of f at x converges to f of x plus 0 plus
f of x minus 0 upon 2 i it is basically the
mean of the function on either side mean a
mean of the function ah evaluated at the left
and the right values ok. So, let us see the
proof of this ah theorem .
Let us slightly deviate 
from step 4 of our previous theorem . So,
if you recall step 4 I think we said the cardinal
integrates to 1 right from minus pi to pi
that is the result. So, we will slightly deviate
from that point. So, integrals it is a symmetric
kernel.
So, integral 0 to pi p n of u d u is integral
minus pi to 0 P n of you d u and this is 1
upon 2, let us call this equation 1 right.
Just for your recall P n of u is 1 upon 2
pi sin n plus half u put this brace here upon
sin u divided by 2 right these vibe are getting
this result it is an even function. Therefore,
hence this is equation.
Now what will we need? What do we need? To
prove the theorem, we need to show integral
minus pi to plus pi f of u plus x times P
n of u d u heads as N goes to infinity to
this equation which is f of x plus 0 plus
f of x minus 0 upon 2, so we already have
a hint here. So, break this limit from minus
pi to plus pi as from a minus pi to 0 and
then 0 to pi ok. And then it is easy for you
to see through this.
Now, integral 0 to pi f of u plus x P n of
u d u this heads to f of x plus 0 upon 2 as
n goes to infinity. Just taking one one portion
right; and this is basically integral this
is basically half of the the whole kernel.
And therefore, this is just basically x plus
0.
Now, this is done symmetrically for the limit
minus pi to 0 f of u plus x P n of u d u as
N goes to infinity heads to f of x minus 0
upon 2 right. Why you would do it exactly
the same way right you just put this x plus
0 here mine you know exactly the way we did
and show that that is basically heading to
0 right, using the same arguments that we
did .
Using the definition of P N of u 1 upon 2
pi integral 0 to pi f of x plus u minus f
of x plus 0 upon sin u by 2 this is basically
the g of u function form sin of n plus one
half u d u this hits to 0. And that is how
we conclude that this is if you take this
function out, you have this equation here
which is I call this A, I call this B. This
is where you get A from.
Now again 1 upon 2 pi integral 0 to pi 
the sets to 0. So, therefore, this theorem
is is established right, so ah you could put
the limit here ah basically through minus
pi to plus y just the symmetric. So, you take
a limit from minus pi to 0 and here you take
a limit from 0 to pi this establishes this
result ok. Now one of the key questions that
would be bugging your mind is what is the
measure on the set of all such discontinuities
right?
So, when we have finite or countably finite
such 
this continuities such that 
the measure on that set basically it is to
0. If this happens 
we are we are we are ok with the convergence
proof right, this is very important result.
And I will not be discussing a lot of the
aspects regarding measures and and these analysis
ideas because as out of the scope of this
course, but I strongly encourage you to ah
listen through ah the the mooc courses on
analysis in math and you will see all these
details coming up ok basically start from
set theory and that is the reason why you
dont want to have uncountably infinite number
of discontinuities.
So, every point you are looking into is discontinuous
or the measure on that set is basically not
heading to 0, then you have trouble because
it does not get to 0; these these these these
reasons you do not be able to establish ok.
So, this is a an important, but subtle detail.
