We'll go through the first sheet's exercises and we start with exercise one, where we consider

a linear operator between two banal spaces X and Y.

The first part of this exercise is to prove that the following three conditions are equivalent.

And K being continuous at the origin, K being continuous everywhere, and K being bounded.

We will prove this by, so we need to prove that everything is equivalent to everything

else and we will prove this by cyclical implication.

Now the easiest thing to look at is obviously B implies A because if a function or this

mapping is everywhere continuous then it's in particular continuous at the origin.

So B implies A is trivial.

The next part is to show that the continuous, an operator being continuous at X equal to

zero, a linear operator is also bounded.

So let K be linear and continuous at X equal to zero and let's assume that K is unbounded.

This means that there exists a sequence in X such that the norm of Xn is bounded but

the norm of the result of applying K to the sequence is unbounded.

So let's say it converges to infinity or diverges to infinity by taking us to the sub-sequence.

So this follows from unboundedness.

Now we want to lead this into a contradiction with linearity and continuity at X equal to

zero.

And the trick here is to define Xn tilde as Xn divided by the norm of K of Xn.

So this is not an element in X, this is a real number, so this is actually an element

in X, nothing problematic.

We can take this vector here, scale it by one over the norm of K of Xn and that gives

us a new element and we have a new sequence Xn tilde in X.

So what properties does this Xn tilde have?

First interesting thing is the norm of Xn tilde is the norm of Xn divided by K of Xn

and because this is just a scalar we can put it outside of the norm so this becomes X norm

of Xn divided by Y norm of K of Xn.

This is bounded, this goes to infinity, so this converges to zero for n to infinity.

This means that Xn tilde converges to zero in X for n to infinity.

Now let's look at what we have.

We know the case linear and it's also continuous at X equal to zero.

This means by continuity at zero that the limit of n to infinity of K of Xn is K of

the limit n to infinity of Xn tilde and well that is zero in X and that is zero in Y.

So, this means that K of Xn tilde converges for n going to infinity to the zero vector

in Y.

But the norm of K Xn tilde in Y is well the norm of K of Xn divided by norm of K of Xn.

It looks slightly messy but now we can use linearity of K.

So we have 1 over norm of K of Xn times K of Xn.

Now we can pull out this positive scalar outside of the norm.

So then this becomes norm of K of Xn divided by the norm of K of Xn.

This is equal to 1.

So this means that in particular K of Xn tilde does not converge to zero because the difference

between K of Xn tilde and zero which is exactly this quantity here is always constant it does

not converge to zero.

So we have proven that there is a contradiction here between that part and this part here.

That means that a function cannot be at the same time, so a linear operator cannot at

the same time be continuous at X equal to zero and unbounded.

So one of those things has to be false.

So it means that if it's linear and continuous in the origin then it's also bounded.

It remains to show that C implies B so that bounded operators are continuous at the origin.