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I have encountered two different definitions for the adjoint of a linear operator, and I can't seem to see how they are equivalent.

The first definition is as follows: Let $(V,\|\cdot\|_V)$, and $(W, \|\cdot\|_W)$ be normed vector spaces.

Define the dual of $V$ as $V^*=\left\{\phi:V\rightarrow \mathbb{C}| \phi\text{ is a bounded linear transformation}\right\}$.

The adjoint of $T$ is the linear transformation: $$T^*:W^*\rightarrow V^* \text{, given by } (T^*\phi)\,(v)=\phi(T(v))$$

The second defintion is as follows:

Let $(H_1,\langle\cdot\rangle_1)$, and $(H_2,\langle\cdot\rangle_2)$, be Hilbert Spaces and let $T:H_1\rightarrow H_2$ be a bounded linear transformation.

The adjoint of T is the function $T^*:H_2\rightarrow H_1$, given by $\langle Tx,y\rangle_2=\langle x,T^*y\rangle_1$

I know the first definition is more general, it applies to any normed vector space, while the second only applies to Hilbert Spaces. This makes me think that somehow, the first definition can be specialized to the the scond for the case of a Hilbet Space. I've been trying this for a while, and I think it involves the Riesz Representation Theorem, but I'm not quite sure how to go about it.

The Riesz Representation Theorem says, that for a Hilbert Space $(H,\langle\cdot\rangle)$:

$\Psi:H\rightarrow H^*$, $x \mapsto \Psi_x$, where:

$\Psi_x:H\rightarrow \mathbb{C}$, $h\mapsto\langle h,x\rangle$,

Is a skew linear bijective isometry with operator norm of $1$.

If anyone could offer some help on how to get from the first definition to the second, I'd be much appreciative, and much less confused about this whole business!

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The Riesz theorem gives you an anti-linear isometry between $H$ and $H^*$. Lets call it $i: H\to H^*$ and its inverse $i^*:H^*\to H$. You have that $i(x)\ (y) = \langle x,y\rangle$. The inverse is actually the adjoint of the map in the topological vectorspace sense, which is why I used the notation $i^*$.

So let $T:H_1\to H_2$ be continuous linear. The topological vectorspace adjoint is a map $T^*: H_2^*\to H_1^*$. Using our anti-linear isometries $i_1^*, i_2$ we can get a map

$$i_1^*\circ T^*\circ i_2: H_2\to H_2^*\to H_1^*\to H_1$$ as a composition of two anti-linears and one linear, this is a linear map. By putting in the definitions you can explicitly check that it satisfies the defining equation of the Hilbert space adjoint:

$$\langle i_1^*T^*i_2 x,y\rangle_1 =(i_1i_1^*T^*i_2)(x)\,(y)= T^*i_2(x)\,(y)=i_2(x)\,(Ty)=\langle x,Ty\rangle_2$$

Alternatively you can look at $i_2^* \circ T\circ i_1: H_1^*\to H_2^*$ and take the adjoint of that, but its the same as the above: $$(i_2^* \circ T\circ i_1)^*=i_1^*\circ T^*\circ i_2$$

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  • $\begingroup$ Thanks for the help! This makes 100% sense and is explained really well. My only issue is that the result is slightly different to definition 2, as it still includes the isometries on the left hand side. I imagine theres probably a simple reason for this which I'm missing, but I'd really appreciate an explanation if possible. $\endgroup$ – CoffeeCrow Sep 28 '17 at 11:51
  • $\begingroup$ Strictly speaking, the two adjoints are different things, the Hilbert adjoint differs from the TVS adjoint by conjugation with the isometries. However in dealing with Hilbert spaces one usually pretends that $H$ and $H^*$ are not canonically isomorphic, but are actually the same space. From this point of view it is unfriendly to view the adjoint as a map $H_2^*\to H_1^*$, as writing $H^*$ reminds us that we have sinned in pretending that canonically isomorphic means "identical". So one uses the identifications to obtain a map $H_2\to H_1$ and calls that the adjoint. $\endgroup$ – s.harp Sep 28 '17 at 13:18
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    $\begingroup$ In physics it is usual to use the symbol $^\dagger$ to refer to the adjoint of an operator between Hilbert spaces. This might be convenient to differ between the two notions when you are in the situation that you actually want to look at the TVS adjoint of an operator between Hilbert spaces (I have never been in this situation). In other words for $T:H_1\to H_2$: $$T^\dagger: H_2\to H_1\qquad T^*: H_2^*\to H_1^*$$ and the answer above is saying: $$T^\dagger = i_1^*\circ T^*\circ i_2$$ $\endgroup$ – s.harp Sep 28 '17 at 13:20
  • $\begingroup$ Okay, that makes sense! There's two concepts of adjoint, which although they are closely related and for most intents and purposes the same, they're technically different, but people often pretend they're not, got it! Thanks for all your help, I really appreciate it. $\endgroup$ – CoffeeCrow Sep 29 '17 at 1:27

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