# Adjoint of an operator definition

1. Sep 29, 2007

### jostpuur

When my lecture notes discuss the adjoint of an operator in Banach spaces, it is defined like this. The adjoint of an operator

$$T:X\to Y$$

is an operator

$$T^*:Y^*\to X^*$$

so that for all $f\in Y^*$ and $x\in X$

$$(T^* f)(x) = f(T x)$$.

But we get into Hilbert spaces, it is said to be given by the equation

$$(Tf|g) = (f|T^*g)$$

The Hilbert space is also a Banach space, so these definitions seem to be contradicting.

In fact my lecture notes are unfortunately messy. I cannot tell for sure what precisely are the definitions, but this is what it says, approximately. Any major misunderstandings could be pointed out. I can conclude that I'm understanding something wrong, because I'm not understanding what the adjoint really is.

2. Sep 29, 2007

### Hurkyl

Staff Emeritus
Don't forget for a Hilbert space H, we have the natural isomorphism

$$H \cong \left(H^*\right)^*.$$

Because of this, it is common to silently substitute back and forth between elements of H and the corresponding linear functional on $H^*$. Or equivalently, to drop the functional notion altogether and instead use the bra-ket product instead.

(You didn't say exactly what you found contradictory, so I have to guess)

3. Sep 29, 2007

### jostpuur

Ooh... I was so close...

4. Sep 29, 2007

### Hurkyl

Staff Emeritus
By the way... I was assuming by (Tf | g) you were using the bra-ket notation: you meant that $f \in H^*$ and $g \in H$.

Were they supposed to both be in H, and this notation mean the inner product?

5. Sep 29, 2007

### jostpuur

Yes it was an inner product between two vectors f and g of some Hilbert space H. T was an operator T->H.

Mentioning the Riesz's representation theorem in this context would have been enough. I couldn't make the connection. It could be I'm too tired, also... but that could be an excuse. Although I am tired, actually...

6. Sep 29, 2007

### cogito²

It's also worth noting that the form for general Banach spaces $$(\cdot|\cdot)$$ is bilinear while for Hilbert Spaces it's taken as sequilinear. This does lead to small differences. For example, in a general Banach space, $$\sigma(T) = \sigma(T^*)$$ while for a Hilbert space, $$\sigma(T) = \sigma(T^*)^*$$ (where the second "*" denotes complex conjugation).

7. Sep 29, 2007

### Hurkyl

Staff Emeritus
Could you clarify? The Banach space structure doesn't include an inner product. And if a complex Banach space happened to be a inner product space, then the definition of inner product space says the inner product must be sesquilinear.

8. Sep 29, 2007

### cogito²

Well it's not an inner product. It's the bilinear form

$$( \cdot | \cdot ) : X \times X^* \to \mathbb{C}$$​

given by

$$(x, \phi) \mapsto \phi(x)$$.​

In that notation, the adjoint condition on Banach spaces is given exactly by what the original poster said (except now the form is bilinear and not sesquilinear):

$$(T^* \phi)(x) = (x,T^*\phi) = (Tx,\phi) = \phi(T x).$$​

So this rephrases the adjoint condition in familiar language:

$$(x,T^*\phi) = (Tx,\phi).$$​

But since this form is bilinear in Banach spaces and the form of Hilbert spaces is sequilinear, it shows that the definition of adjoint is actually slightly different in Hilbert spaces as opposed to Banach spaces (i.e. the adjoint in a Hilbert space is not the same as the one you get when you consider the Hilbert spaces as Banach spaces). You get this pesky complex-conjugation going on with the identification of $$H^*$$ with $$H$$. Actually Hilbert spaces could be studied with the form of Banach spaces and you really wouldn't lose anything, except that you'd have to deal with the complex conjugation in other places in the theory. But the sesquilinear approach is entrenched and there's not getting rid of that.

9. Sep 29, 2007

### Hurkyl

Staff Emeritus
Fortunately, the bra-ket product is bilinear too! (And it should be, since it's semantically the same thing as you wrote! Except with the two factors reversed)

$$(a \langle \phi |) | \psi \rangle = \langle \phi | (a | \psi \rangle = a \langle \phi | \psi \rangle$$

The only oddity arises, as you mention, when you reinterpret as an inner product, because transposition is conjugate-linear:

$$(a |\psi \rangle)^* = \bar{a} \langle \psi |$$

I guess I never really think of it because I essentially never think of Hilbert space arithmetic in terms of the inner product. I always think of the product as multiplying forms with vectors.

Last edited: Sep 29, 2007