1. Apr 8, 2013

### fluidistic

1. The problem statement, all variables and given/known data
I must show several properties about linear operators using the definition of the adjoint operator.
A and B are linear operator and $\alpha$ is a complex number.
The first relation I must show is $(\alpha A + B)^*=\overline \alpha A^*+B^*$.

2. Relevant equations
The definition I have an an adjoint is: $A^*$ is the adjoint of $A$ if $\langle g,Af \rangle = \langle A^* g ,f \rangle$ where f and g are any vectors in a Hilbert space.

3. The attempt at a solution
Let $C^*=(\alpha A+B)^*$. Using the definition of adjoint I get: $\langle C^*g,f \rangle=\langle g, Cf \rangle \Rightarrow \langle (\alpha A+B)^*g ,f \rangle =$$\langle g, (\alpha A+B)f \rangle =\langle g, \alpha A \cdot f \rangle + \langle g, Bf \rangle = \alpha \langle g, Af \rangle + \langle g ,Bf \rangle = \alpha \langle A^*g, f \rangle + \langle B^*g ,f \rangle$.
But I'm getting lost. I've no idea how I can obtain A, B, A^* and B^* using the definition of the adjoint.

Oh wait, on my draft I think I have finished the "proof". The last expression is worth $\langle \overline \alpha A^*g ,f \rangle + \langle B^* g, f \rangle = \langle (\overline \alpha A^* + B^*)g ,f \rangle$. Then by associativity $(\alpha A+B)^*=\overline \alpha A^* + B^*$.
Does this look right?

2. Apr 8, 2013

### micromass

Staff Emeritus
OK, so you proved

$$<(\alpha A + B)^* g,f> = <(\overline{\alpha}A^* + B^*)g,f>$$

But how do you obtain from there that $(\alpha A + B)^* = \overline{\alpha}A^* + B^*$?

3. Apr 8, 2013

### fluidistic

Good question, I don't know. To simplify the notation I must show that if $\langle Dg, f \rangle = \langle Eg,f \rangle$ then D=E, where D and E are linear operators; for any inner product in a Hilbert space.
When I look at the definition of the inner product, I only see the positive definiteness, linearity and conjugate symmetry. I don't see how that would help me.

4. Apr 8, 2013

### micromass

Staff Emeritus
Something that I think would help is the axiom $<x,x>=0$ then $x=0$. Because this is the only axiom that allows you to begin with the inner product and end with a statement about a vector.

Let's prove a more general thing. Let x,y be vectors such that $<x,z> = <y,z>$ holds for all z, then x=y.

To prove this, note that the above is equivalent to saying $<x-y,z>=0$ for all z. Does this give you a hint?

5. Apr 8, 2013

### fluidistic

I'm not 100% sure.
Here is my attempt: $\langle x ,z \rangle =\langle y, z \rangle$ holds for all z.
I add "-y" in the first argument of each side (can I really do that without demonstrating that the equality still hold?) to get $\langle x-y,z \rangle =\langle y-y ,z \rangle = \langle 0 ,z \rangle =0$
So that $\langle x-y, z \rangle =0$ must hold for all z. This implies that $x-y=0$. End of proof.

P.S.:Yes I can add -y in the first argument of the inner product. To prove it is simple, using the linearity of the inner product.

6. Apr 8, 2013

### micromass

Staff Emeritus
This implication is not very clear. why does $<x-y,z>=0$ for all z imply that $x-y=0$?

7. Apr 8, 2013

### dextercioby

It is of utmost importance to mention that the operators are bounded and defined everywhere on the Hilbert space, else one needs a new reasoning.

8. Apr 8, 2013

### fluidistic

Because in particular $z=x-y$. So we have $\langle x-y , x-y \rangle =0$, which is possible if and only if $x-y=0$. Thus $x=y$. End of proof?
Is is the same as showing that the operators are continuous over the whole Hilbert space?

9. Apr 8, 2013

### dextercioby

Yes, good point.

No, continuity is an alternative assumption to boundedness, because for ∞-dim separable Hilbert spaces the 2 notions are equivalent.

10. Apr 8, 2013

### micromass

Staff Emeritus
There is a connection between continuity (and equivalent: boundedness) and being everywhere defined.

A linear operator $T:D\rightarrow H$ is usually defined on a dense subset $D$ of the hilbert space $H$. Now, it turns out that if $T$ is continuous on $D$, then there exists a unique extension of $T$ on the entire Hilbert space H. This is called the BLT theorem (BLT = bounded linear transformation). So if $T$ is continuous on a dense subset, then we can make it everywhere defined.

If $D$ is not dense, then we can only extend $T$ on $\overline{D}$.