Difficulty with EASY equivalence statement

[Emieno]

Truly I was not joking.
All of what I was trying to speak about was what you mentioned in your #22nd post.

So we have to prove $\langle Sx\,|\,x \rangle = \langle Tx\,|\,x \rangle \Rightarrow \langle Sx\,|\,y \rangle = \langle Tx\,|\,y \rangle$.

$$(\Rightarrow)$$
Suppose $\langle Sx\,|\,x \rangle = \langle Tx\,|\,x \rangle$. Then rearrangining gives $\langle Sx\,|\,x \rangle - \langle Tx\,|\,x \rangle =0$ which implies $\langle (S-T)x\,|\,x \rangle = 0$. Now let $x=y$(?). Then $\langle (S-T)x\,|\,y \rangle = 0$. and reverse the process to solve.

Is this what you mean?

Yes, here, your problem is All x,y in H, so x,y are arbitrary, you can replace x with y or vice versa (which also usually happens with some integral problems)

 

[Oxymoron]

So what you quoted is correct?

 

[Emieno]

where do you think the mistake is ?

 

[Oxymoron]

There's a mistake? Well, I am not so sure you can just let x=y. Thats the only thing that worries me

 

[Emieno]

That idea is applied from integral calculations. Again, it is always possible since we are dealing with arbitrary *letters*. Proofs in Sets, Mappings etc also have a lot of similar cases.

 

[Emieno]

Again before I go to bed, I would like to only say that I was not saying that x=y, but just *because they are arbitrary, so we can replace the x with y or y with x*.

I am sure you can also say <Sy|x>=<Ty|x> , because S=T;

 

[Hurkyl]

I think you have the quantifiers wrong, Oxymoron, and that's the source of your problem:

$$\left( \forall x \in \mathcal{H}: Ax = 0 \right) \implies A = 0$$

is the statement, and is trivial to prove. (What's the definition of equality for two operators?)

 

[Oxymoron]

Two operators are equal if and only if they do the same thing to an arbitrary element. That is

$$S=T \quad \Leftrightarrow \quad Sx = Tx$$

 

[Oxymoron]

Ok, so using the same ideas I can prove $S=T \Rightarrow \langle Sx\,|\,x \rangle = \langle Tx\,|\,x \rangle$ like this:

Suppose $S=T$. Then by the equality of operators we have $Sx = Tx$. Then we have $\langle Sx\,|\,x \rangle = \langle Tx\,|\,x \rangle$ and we are done.

How does this look?

 

[Hurkyl]

Yes, that looks like a good proof of that statement.

(But, as emieno was saying, you could skip the step saying that Sx=Tx)