The next topic we're going to look at is differentiation and integration.

The first question which I'll pose is, is differentiation easier or integration easier?

Hopefully we'll have an answer at the end of this lecture.

We start by considering the following function spaces, x which is the space of continuous

functions on 0,1.

So continuous functions on 0,1 going to R.

And y will be all functions from 0,1 to R which are C1, so they are one times continuously

differentiable and we'll say f of 0 is equal to 0, which is a normalization thing.

Then we construct the following mapping from x to y.

We attach the supremum norm on x and also the supremum norm to y.

And we map f to a different function, this function maps x to the definite integral 0

to x of f of y dy.

So we could call this the integration functional.

This yields the primitive f such that the derivative of f, capital F is f.

So if we differentiate this we get f.

And this f of 0 is equal to 0 thing, this is just supposed to make this primitive unique.

So this f is not this f because it's the other space.

Maybe I'll write this as g just to make this clear.

This is the one defined mapping and t is linear, it's invertible and it's continuous.

So let's check that.

Well it's clear that it's linear, so if we take f plus g, we have f plus g in here, this

by linearity of the integral this decouples into two integrals and things like that.

So it's clear that it's linear.

Invertible, well, is it invertible?

Or given g is equal to t of f, i.e. g of x is 0x f of y dy, pick g prime as the

inverse of g.

So we call this the inverse of g, well maybe the other way around so the definition goes

along this line.

So we can call this t of minus g, g prime and by plugging in we say that it's invertible,

t minus 1 of t of f of g, so this is t minus 1 of the mapping 0x f of y dy.

Well, we have defined the inverse as being the derivative by, you know, we know how we

differentiate integrals like that, so this is a mapping x goes to f of x and this is

indeed the same as f, right, okay, so this is nice.

It has an inverse and it also continues as a mapping from this space, from this bar space

to this bar space, which we have to check really quickly.

So the image tf in the supremum's norm is the supremum x in 0, 1, 0x f of y dy.

Now what do we do with that?

We can bound this from above by using the triangle inequality.

The triangle inequality says that we can put the absolute value inside the integral f of

y dy.

Now because this is larger than, well, larger equal than 0, we can, without making any mistakes,

augment this to the integral from 0 to 1, of course, and then we can just drop this.

That's nothing to take the supremum off anymore.

So this is the same as the integral from 0 to 1 of f of y dy and this is less or equal

than the supremum, but we can still integrate over that, supremum of f of y, where y is

in 0, 1, dy and this is equal to the supremum of f of y dy.

And this is the definition of the supremum norm of f.

And this shows that t is continuous.

This shows boundedness and we know that linear operators are bounded if and only if they're

continuous.