# Lec 65 – Basic analysis: Pointwise and uniform continuity of functions

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So, in signal processing we will often have to encounter with limits sequences sequences

of functions, you know series summations and. In fact, each term in the series could be

a function so on and so forth right functional series etcetera.

And, there are also these notions of continuous functions discontinuous functions and we need

some treatment into this. So, the route to these ideas stem from analysis in mathematics,

I will not delve too much into analysis because it is a separate course in itself in the field

of mathematics and you could do a rigorous course for 30 hours as part of your NPTEL,

but I will just briefly give you some ideas about continuity convergence etcetera ah as

part of the discussions here, which is going to be useful in your understanding of some

of the signal processing ideas ok. Let us, first talk about ‘continuity’ and

forms of continuity . Now consider S to be a subset of real numbers R and f is a mapping

from S to R and this is a real valued function defined on S right.

There is some subset S of real numbers and there is a mapping from this subset to real

and this is a real valued function defined on the subset S. So, S for example, if you

think of S as say 0 say 2 this is basically all x belonging to the set of real number

such that 0 is strictly less than, x less than 2 that is it is open on both sides. It

can be infinite like S is 0 infinity, which is basically

x belonging to R such that x is strictly positive right, because 0 is not included it is strictly

positive . So, we may encounter many such type of these

sets. So, let us start with the notion of what continuity is so, I will start with the

definition . For some function to be continuous on this

subset ok . The function f is said to be continuous honest if and only if if and only if for every

x 0 belonging to S for every epsilon, which is greater than 0 there exists a delta, which

is greater than 0, which depends upon x 0 and epsilon for every x belonging to this

subset the following holds that is if x minus x 0 absolute is less than delta, this implies

the absolute value of f of x minus f of x 0 is strictly less than epsilon and it is

not continuous on S if this is greater than or equal to epsilon.

There is a notion of I will go to the next slide there is a notion of uniform uniformly

continuous . So, I can call this concept as uniformly continuous . So, the function f

is uniformly continuous on S, if and only if for every epsilon greater than 0 . There

exists delta greater than 0 ; that means, delta is great is depending only on epsilon

right. And then you choose points x 0 belonging to S and x belong to S choose these 2 points,

if absolute of x minus x 0 is less than delta this implies that absolute of f of x minus

f of x 0 is strictly less than epsilon . Now, you have to sort of be careful how the

definitions are worded very carefully. So, in the case of uniformly continuous you choose

first an epsilon greater than 0. So, I will just write it down in a red here. So, we choose

epsilon greater than 0 and we let some delta, which is basically delta which is a function

of epsilon . Then you choose x 0 belonging to S and x belonging to S you assume, that

the absolute value of x minus x 0 is less than delta and this implies that absolute

value of f of x minus f of x 0 is less than epsilon this is the English way of writing

things this is a little more mathematical way of writing things.

So, I think you have to observe very carefully that in the order I first for every epsilon

greater than 0 I choose delta, which is greater than 0 and this delta depends upon epsilon

and not really on the point right. So, just going back one slide where we define continuity

property on this subset. So, here the difference is as follows. So, here we choose x 0 belonging

to this subset, then we choose epsilon which is greater than 0 and the let delta which

is basically a function of both the point x 0 and epsilon observe this very careful

point, that it has to depend on this x 0 on this point, delta depends on this point. And

then you choose some x belonging to S and assume that x minus x 0 is less than delta

and if you assume this with all these conditions this implies an absolute value of f of x minus

f of x 0 is strictly less than epsilon. So, at least we got an idea of what is continuous

and what is uniformly continuous? Let us look at some examples towards this

understanding ok .

Let this upset be the real line itself choose the mapping to be mx plus c ok . Let us examine

if f is uniformly continuous on on S right . So, what do we know what do we need to do

here. So, first we choose some epsilon which is greater than 0, because we have to examine

uniform continuity. So, we first choose some epsilon, which is

greater than 0 let delta be epsilon upon m I will tell you why we choose this m I mean

at the moment just trust me and if things become very clear choose x 0 belonging to

R choose x belonging to R assume x minus x 0 is less than delta .

Now, consider mod f of x minus f of x 0 this is mod mx plus c minus m x 0 plus c which

is so, I will assume that m is just to make things a little easier assume m is greater

than 0 and m 0 equal to 0. So, that this is taken care of now this is m times mod x minus

x 0 and this quantity is if this is basically m times delta and this is strictly this is

quantity strictly less than delta right. So, I can say this is quantity strictly less than

m times delta and delta is epsilon upon m. So, therefore, this quantity modulus of fx

minus fx 0 is strictly less than epsilon right and that is reason why I chose delta to be

epsilon upon m. So, now, you can see the choice of delta equals epsilon upon m why it has

been carefully chosen right and this is satisfied therefore, f this is func this function, which

is a mapping from S to R is uniformly continuous ok .

Let us, take another example suppose S is given by x belonging to the set of real numbers

such that 0 is less than x less than say some 2 and f of x equals x square . Let us examine

if f is uniformly continuous on S . So, it is straight forward let us just work through

these steps here right. So, what we have to do is we have to choose

some epsilon, which is greater than 0 let delta be epsilon upon 4 I will tell you why

this thing appears like this, then we choose x 0 belonging to S and x belonging to S ok.

Now 0 is less than x 0 less than 2 and 0 is also less than x less than 2 therefore, if

I add these 2 intervals 0 is less than x 0 plus x this is less than 4 ok.

Now, let us assume the absolute value of x minus x 0 is less than delta, now consider

mod f of x minus f of x 0 in this case is x square minus x 0 square, you can factorize

this as x plus x 0 into mod x minus x 0 because x plus x 0 is basically definitely greater

than 0 right. We know this quantity therefore, I had I did not put the absolute here and

this is basically 4 times I would say this is basically 4 times mod x minus x 0 and this

is basically less than 4 times delta. And therefore, this is if I choose delta to be

epsilon upon 4 this basically satisfies the condition that absolute value of f of x minus

f of x 0 is strictly less than epsilon ok. So, this is again

uniformly continuous well there is a small catch here, because the range is from 0 less

than x less than 2 precisely x plus x 0 will be strictly less than 4. So, this equality

here would have to be an inequality, which is strictly

less than and this is something to note . In our examples on if x is mx plus c and if

x is x square that is be considered linear function we considered a quadratic function

we landed up with a situation where mod f of x 1 minus f of x 2 is less than or equal

to some m times mod x 1 minus x 2 right we landed up with a situation like this, for

all x 1 x 2 belonging to S right. Inequality of this form is called Lipschitz inequality

and the constant m is called the corresponding Lipschitz constant ok.

So, this is a an important note I think we we understood what this this idea of continuity

is what the idea of uniform continuity is I gave you examples of uniform continuity?

I will give you some homework exercises that you would want to work . First examine, if f of x equals x square is

uniformly continuous on the interval S, which is basically 0 to infinity and it is basically

an open interval. So, this is one one exercise that I would want you to try and the second

would be examine if f of x given by one if x is positive 0, if x is less than or equal

to 0 is continuous. Of course, you know the answer to the second problem, because there

is a jump here right I mean so, like a step function there is a jump here and you know

it is a point of discontinuity, but using the definitions, that we learned just now

on continuous functions and uniform uniformly continuous functions, I would want you to

examine if this is continuous or not ok. So, I think with this we recover the essentiality

of continuous property, then the next we will talk about convergence and then we will discuss

some of the certain issues related to convergence of sequences and then convergence of functions

which is essential for your signal processing understanding.

Thank you.