00:01
All right, so in this question we're given a bounded linear operator between two norm spaces x and y.
00:06
We're told it is onto y, so it is surjective, and it satisfies this estimate of the form.
00:11
So there exists a b such that this holds for every x, and then we want to prove that then this implies that t is invertible and the inverse is bounded.
00:21
So first step, let's show that the inverse exists.
00:25
So you know an operator is invertible if it is injective and surjective, so if it is one -to -one and onto, we already have onto, so we only have to prove one -to -one.
00:36
So first step, the inverse exists, so we need to show that t is one -to -one or injective.
00:45
So you know how this goes.
00:47
Let's suppose x in x is such that t of x is equal to zero, and we want to show that x must be zero.
01:01
Then let's see what we can do.
01:03
I mean, we only have one assumption on x, so we must have to use it.
01:07
So i want to show that x is zero.
01:09
I will do so by showing that the norm of x is equal to zero.
01:13
So let's take the norm of x.
01:15
So we can use this estimate in here, in red.
01:20
We can use this estimate in here to write that norm of x is less than or equal to one over b times the norm of t of x.
01:31
This is valid for every x.
01:33
And now we're done, right? because by assumption, t of x is a zero vector.
01:37
The zero vector has norm equal to zero.
01:40
So this is equal to one over b times zero, which is zero.
01:45
So now the norm of x, this is always positive, right? non -negative.
01:48
It is above zero, but also below zero.
01:52
So we conclude that the norm of x is equal to zero, and then this implies that x is zero itself.
01:58
So we show that if a number is such that its image is zero, that number is zero...