This goes by then theorem Cauchy Criterion
for so this is a theorem, let us write it
as theorem, Cauchy’s Criterion, fn converges
to f uniformly if and only if, fn is Cauchy,
Cauchy in B X, R. So, that is, so we have
seen Uniform Convergence, we have seen when
a sequence does not converge uniformly a criterion
for that. We have seen now a criterion which
says, when does a sequence converge uniformly,
namely it should be converging in the B X,
R and that is same as saying it should be
Cauchy there. So, we know we can test sequences
being converging uniformly or not.
So, let us finally look at whether this serves
our purpose of saying that, when pointwise
convergence limit was not continuous functions,
converging pointwise the limit was not continuous
convergence uniform, differentiability did
not imply the derivatives converge. And integral
also we had an example that, fn’s are integrable
converging pointwise did not imply that limit
is integrable. And these are important things
as far as the convergence problems are concerned.
We were important problems in analysis because,
whenever you want to go beyond algebra, algebra
is 1, 2 and 3, only finite number of operations,
when you, you can look at only polynomials.
To generate functions which are nonpolynomial,
nonrational functions, you have to go to analysis
that is why we had to do all that exponential
logarithm, trigonometric all those functions
were cannot be obtained using algebra alone,
you have to go to limits. So, things that
are preserved.
So, let us write theorem 1, fn converges to
f uniformly implies and each fn continuous
at some point x not implies f also continuous
at the point x naught. So, let us write a
proof, that continuity is preserved in uniform
convergence. So let, we want to analyse so
for convergence what we have to analyse, f
of x minus f of x naught, this difference
right. We want to show f is continuous, we
have to make this thing small whenever x is
close to x naught. So, fn converges to f uniformly,
so let us how to somehow bring fn inside it
right.
Because something is given about fn so let
us bring in, so this is fx minus fn of x plus,
I have subtracted so let me write fn of x
minus fn of x naught plus fn of x naught minus
f of x naught. I am adding and subtracting,
what I am f of x, I subtracted fn of x, I
add fn and then I subtract fn at x naught
add fn at x naught and the final term is there
as it is. So, I have added and subtracted
use triangle inequality.
Why I am doing that, I am forced because something
is given to me about fn to use that fact I
have to bring fn, then only you can use it.
And now, if I look at this quantity, the first
one is less than or equal to norm of f minus
fn, the first term, this term. And this term,
let us keep it at it is if you like, so it
is mod of fn x minus fn at x naught plus and
this quantity again is a same point. So, less
than or equal to fn minus f norm infinity.
And now, I know that this quantity is small
because fn is converging uniformly to f, both
these quantities are small this and this.
And what is this quantity middle one, that
is fn of x minus fn at x naught. Same function
at the point x naught but fn is given to be
continuous, fn is given to be continuous.
So, this quantity is also small so we can
just simply write, so using the fact that
each fn is continuous, using the fact that
each fn is continuous. And fn minus f goes
to 0, so this equation star implies f is continuous.
I am not writing that epsilon every time kind
of a thing. So, if you want to write all that
thing, you can write given epsilon bigger
than 0.
I can find some stage n, so that this quantity
is small, this quantity is small less than
epsilon. And by continuity, for fn after that
stage, for that stage this will be less than
epsilon, by continuity. So, I can find a delta
such that you want me to write everything,
let me write so that you feel happy. So, what
does this mean, so here is what I am saying.
Let, epsilon greater than 0 be given. So,
there is how formal proof will be written.
Choose n naught such that, mod of f minus
fn is less than epsilon for every n bigger
than n naught.
That is by using the fact that, fn is converging
to f uniformly, also for, so let us take this
fn n bigger than n naught. Let us take n naught
itself, fn naught being continuous, there
exist a delta bigger than 0 such that, mod
x minus x naught less than delta implies fn
naught x minus fn naught of x naught is less
than epsilon. So, this is 1, this is 2 then
in this equation, star then in star put this
values. So, this is less than epsilon, this
is less than epsilon, this is less than epsilon
and this is less than epsilon, when x is closed
to x naught by delta.
So, you get, so in star x minus x naught less
than delta implies fx minus fx naught will
be less than 3 epsilon, does not matter. If
you want it very nice you get off cut down
everything in, so that is an idea of the proof.
So, you should understand what you are doing
saying this thing I can manage, I can make
it small by using the fact that fn converges
to f uniformly. I can make this thing is small,
using the fact that fn is continuous at the
point x naught and I can make this small again
by uniform continuity.
So, the middle term is small whenever x is
closed to x naught by continuity, this will
be small. So, that is what we have written,
given epsilon there is a delta and so on,
so that proves the thing. So, this is, that
means continuity is preserved in uniform convergence.
So, let us prove that, so theorem 2 says,
if fn belonging to B X, R are Riemann integrable
and fn converges to f uniformly, then f, I
should Riemann integration so I should not
write X now. Because Riemann integration is
only for some interval, so we are taking X
is equal to the interval a, b. Integrability
of functions is defined Riemann integrable
only for closed bounded intervals. So, here
x is taken as, so we are looking at all bounded
functions on a, b to R, which are Riemann
integrable.
And fn converges to f uniformly, f is a function
on a, b to R, then is also Riemann integrable
and integral of fn a to b dx converges to
integral f dx, a to b. So, one has to put
a slightly stronger condition, pointwise we
saw that even the function may not be integrable
at all, limit may not be integrable. In fact,
you can give examples of functions where fn’s
are Riemann integrable, fn’s converge pointwise
and the derivative function is integrable
but the integrals do not converge. So, many
such examples are possible we are not going
into all those things.
What we are saying is uniform convergence
at least is a one criteria which allows you
to pass on under limits, integrability is
preserved. I think will prove it next time
because the proof requires a bit of more partitions
and so on. And we will also analyse what happens
to differentiability, will show that differentiability,
also is not preserved in the pointwise convergence.
And in fact, one has to put very strong conditions
for differentiability also, even for uniform
convergence. So, these are slightly differ
theorems. And this, this thing that vendors
integral of fn converge to integral of f for
Riemann integrable functions.
As I had mentioned is a beginning of the story
of Liebig integration. In general, it does
not happen, it happens for uniform convergence.
What other class can happen that is, so we
will stop here.
