fa18_sub_eng.srt

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Hello and welcome back to

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functional analysis.

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And as always many, many

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thanks to all the nice people

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that support me on Steady

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or paypal.

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We've already reached part

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18 in our course.

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And today we want to talk

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about compact operators.

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As the name suggests, this

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has something to do with

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the compact sets we've already

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studied.

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Therefore, recall that the

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idea of compactness has

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been to extend the notion

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of finite a little bit,

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a similar thing we will now

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do for operators

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here.

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The analogon of finite would

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be linear operators between

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finite dimensional vector

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spaces.

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So let's look at an operator

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from FN to

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FM on both sides,

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we can just choose the normal

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Euclidean norm, the standard

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norm.

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However, everything I say

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now does not depend on the

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norm.

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It holds no matter which

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norm you choose here.

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The first step is not so

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hard to show, it tells us

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that the linear operator

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between finite dimensional

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norm spaces is always

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continuous.

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And of course, you already

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know continuous means the

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same as bounded for linear

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operators.

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In other words, the image

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of the unit ball under T

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is a bounded set.

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Now, this is a bounded set

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in our finite dimensional

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norm space FM.

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And there you know this is

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one half of the things you

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need for being compact.

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However, the second ingredient

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we just get when we form

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the closure of this set,

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this means that we just can

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use a bar over the set.

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And then we get a set that

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is always compact in

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FM.

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With this essentially, you

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now know what a compact

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operators.

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When we look at the image

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of the unit ball and form

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the closure.

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And this set is compact,

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we speak of a compact operator.

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Of course, this always holds

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in this case, but not in

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general.

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In order to see that let's

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consider the identity

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operator.

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In LP.

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Of course, identity operator

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just means we take a sequence

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X and send it to itself.

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Therefore, calculating the

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image of the unit ball is

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not so hard, it stays the

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unit ball.

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Now we are interested in

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the closure of the set.

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So we use the bars again.

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So what you have on the right

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hand side is just a closed

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unit ball in LP.

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And there we know from the

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last video it's closed and

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bounded but not compact.

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Hence, we have something

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here.

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We would not call a compact

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operator because the image

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is just too large in this

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case.

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So you should always remember

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compact operators is what

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we get when we extend the

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finite dimensional operators

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here a little bit.

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OK, then let's write down

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the formal definition.

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So we need two norm spaces

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and often there will be Banach

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spaces then a

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bounded linear operator

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T from X to

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Y is called

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compact.

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If we have the thing we discussed

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before that this set

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here is compact.

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In Y therefore, in the

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case that Y is a finer

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dimensional norm space, this

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here is nothing special.

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However, in the infinite

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dimensional case, it really

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is.

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Hence, I would suggest that

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we now look at a common

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example, let's

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look at an integral operator

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defined for the continuous

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functions.

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So it should take a function

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defined from 0 to 1 and

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send that to another function

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in the same space.

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And as often the space of

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continuous functions should

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carry the supreme norm.

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So what we can do is apply

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the operator T to a function

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F and then we get out a new

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function.

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Hence, we can look what the

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function does at a given

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point S where S comes from

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the unit interval.

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Now the number that comes

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out here should be given

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by an integral from 0 to

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1 where we have the function

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F involved, but

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also a fixed function

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K.

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And this function should

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get the variable S and also

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the integration variable

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T.

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So we have a function with

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two variables and for us

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it should be also a continuous

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function.

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So we have it from C defined

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on the unit interval times

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the unit interval.

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Now, since the function K

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goes into the definition

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of T I put it into the

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index here.

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OK, then let's check if

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TK is indeed a

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compact operator.

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An important fact we will

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need here is that the function

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K is defined on a

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compact set.

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So it's not just continuous,

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it's uniformly continuous

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to refresh your memory.

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Let's write down what this

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exactly means.

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For all epsilon crater zero,

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there exists a delta crater

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zero such that for

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all points we put in and

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now we need two variables.

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And for them, it should hold

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if the distance is less than

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delta, the distance of the

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images should be less than

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epsilon.

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On the left hand side, we

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measure the distance with

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the Euclidean norm in R two.

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And on the right hand side,

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we measure with the Euclidean

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norm in R one, which is the

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absolute value.

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Why we need the uniform continuity

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here we see in a moment.

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The first step we have to

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do when we see such an integral

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operator is to check if

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this integral defines

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indeed a continuous function.

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Simply because otherwise

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the operator wouldn't be

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well defined.

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It really should map continuous

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functions to continuous

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functions.

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Checking the continuity.

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Then means we look at the

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difference of the images

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when we put in different

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points.

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So this should be small when

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the points as one and two

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are close together.

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Therefore, we first calculate

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and then we can look what

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happens?

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Now, the first thing you

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should see here is that we

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can put that into one

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integral.

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So use some parentheses here

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and delete this integral

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here.

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Then of course, we pull in

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the absolute value then we

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get an inequality.

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And with this, we have everything

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we need because this one

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here is less than a supremum

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norm of F.

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And this one by the uniform

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continuity of K can be as

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small as we want.

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And exactly this is what

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we should formally write

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in front of the whole calculation.

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So for a given epsilon greater

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than zero, we choose the

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delta in such a way that

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this whole thing holds.

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Therefore, we can choose

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as one as two from the unit

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interval such that the distance

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is less than delta

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for this as one as two here.

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And the same T on both sides,

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we can apply what we know

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that this is less than epsilon

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by using also that this one

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is less than the supreme

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norm.

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We are finished with the

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whole integral, it's

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simply less than epsilon

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times the supreme norm.

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And because the supreme norm

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of F is just a constant in

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the whole calculation, we

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now know that this function

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is indeed continuous.

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So we can note our

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operator as written as

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here is well defined.

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However, our calculation

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here shows us even more.

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If we define the set A

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as the image of the unit

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ball, then we see

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by this whole estimate here,

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that the set A is

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uniformly equi

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continuous.

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If you don't know what this

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means anymore, let's write

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it down again.

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It just means that for all

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epsilon greater zero, there

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exists a delta such that

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for all S one S two and all

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G in A, we have the

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uniform continuity

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implication here.

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From the definition of the

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last video, I only had to

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change some names.

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I used the name G for the

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function here because we

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already had a function F

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and of course, we needed

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the names as one as two for

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the variables.

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However, what you should

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see is that this one is the

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definition of a being

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uniformly equi continuous.

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And it's the same thing as

301
00:07:48,130 --> 00:07:49,119
we have written here.

302
00:07:50,109 --> 00:07:51,079
In other words, with the

303
00:07:51,089 --> 00:07:52,559
calculation above, we have

304
00:07:52,570 --> 00:07:54,279
proven that A is indeed

305
00:07:54,290 --> 00:07:56,220
uniformly equi continuous

306
00:07:56,970 --> 00:07:58,029
at this point, you might

307
00:07:58,040 --> 00:07:59,609
already guess that we want

308
00:07:59,619 --> 00:08:01,350
to apply the Arzelà–Ascoli

309
00:08:01,369 --> 00:08:02,079
theorem here.

310
00:08:03,029 --> 00:08:04,329
Therefore, another step we

311
00:08:04,339 --> 00:08:06,019
have to do is showing that

312
00:08:06,029 --> 00:08:07,519
the whole set is bounded.

313
00:08:07,529 --> 00:08:08,720
Or in other words that the

314
00:08:08,730 --> 00:08:09,869
operator is bounded.

315
00:08:10,570 --> 00:08:12,109
Hence, let's calculate the

316
00:08:12,119 --> 00:08:13,070
operator norm.

317
00:08:13,619 --> 00:08:14,989
So we have the supreme of

318
00:08:15,000 --> 00:08:16,839
all the norms of the images

319
00:08:16,850 --> 00:08:18,600
where F has norm one

320
00:08:19,040 --> 00:08:20,230
by the definition of the

321
00:08:20,239 --> 00:08:21,230
supreme norm.

322
00:08:21,239 --> 00:08:22,769
This is the supreme over

323
00:08:22,779 --> 00:08:24,670
all s where we calculate

324
00:08:24,679 --> 00:08:25,690
the absolute value of the

325
00:08:25,700 --> 00:08:26,299
integral.

326
00:08:26,760 --> 00:08:28,630
As often, we can just pull

327
00:08:28,640 --> 00:08:29,929
in the absolute value into

328
00:08:29,940 --> 00:08:31,910
the integral and get an inequality

329
00:08:31,920 --> 00:08:32,308
out.

330
00:08:32,820 --> 00:08:34,308
And with this, we are almost

331
00:08:34,320 --> 00:08:35,840
finished because the last

332
00:08:35,849 --> 00:08:37,590
part here is less or

333
00:08:37,599 --> 00:08:38,849
equal than the supreme norm

334
00:08:38,859 --> 00:08:39,229
of F.

335
00:08:39,969 --> 00:08:41,260
And this is by assumption

336
00:08:41,270 --> 00:08:43,028
just one, this

337
00:08:43,039 --> 00:08:44,239
means that we don't need

338
00:08:44,249 --> 00:08:45,708
the outer supreme anymore

339
00:08:45,778 --> 00:08:47,179
and can just write down

340
00:08:48,000 --> 00:08:49,789
everything is less or equal

341
00:08:49,799 --> 00:08:50,979
than this integral.

342
00:08:51,380 --> 00:08:52,940
However, this is also less

343
00:08:52,950 --> 00:08:54,580
or equal than the supremum

344
00:08:54,659 --> 00:08:56,419
norm of our function K.

345
00:08:56,859 --> 00:08:58,320
As often, it's not important

346
00:08:58,330 --> 00:09:00,080
what the number here is exactly.

347
00:09:00,140 --> 00:09:01,599
It's only important that

348
00:09:01,609 --> 00:09:03,500
it is constant because then

349
00:09:03,510 --> 00:09:05,280
we know that TK is a

350
00:09:05,289 --> 00:09:06,500
bounded operator.

351
00:09:06,950 --> 00:09:08,780
And now finally comes our

352
00:09:08,789 --> 00:09:10,609
conclusion, we can apply

353
00:09:11,090 --> 00:09:11,590
ATSA Ascoli.

354
00:09:12,280 --> 00:09:13,830
This said the image of the

355
00:09:13,840 --> 00:09:15,690
unit ball is uniformly

356
00:09:15,700 --> 00:09:17,630
equi continuous and bounded.

357
00:09:18,340 --> 00:09:19,940
Therefore, both things also

358
00:09:19,950 --> 00:09:21,109
hold for the closure of the

359
00:09:21,119 --> 00:09:21,630
set.

360
00:09:21,820 --> 00:09:23,669
And by Arzelà–Ascoli, we now

361
00:09:23,679 --> 00:09:25,020
have a compact set.

362
00:09:25,530 --> 00:09:27,070
And by our definition, we

363
00:09:27,080 --> 00:09:28,909
also know TK

364
00:09:29,010 --> 00:09:30,750
the integral operator is

365
00:09:30,760 --> 00:09:32,020
a compact operator.

366
00:09:32,799 --> 00:09:34,270
So you see this was a long

367
00:09:34,280 --> 00:09:36,030
example, but it is a standard

368
00:09:36,039 --> 00:09:37,489
example for a compact

369
00:09:37,500 --> 00:09:38,150
operator.

370
00:09:38,599 --> 00:09:40,070
Of course, we will talk about

371
00:09:40,080 --> 00:09:41,669
compact operators in later

372
00:09:41,679 --> 00:09:42,869
videos even more.

373
00:09:43,580 --> 00:09:45,330
Therefore, as always, I hope

374
00:09:45,340 --> 00:09:46,150
I see you there.

375
00:09:46,380 --> 00:09:46,950
Bye.