I'm a bit confused about adjoint operators. Let $T:X \to Y$ be a linear isomorphism between Hilbert spaces. Then is it true that $(Tx,y)_Y = (x,T^*y)$ exists (does $T^*:Y \to X$ always exist)? What if these Hilbert spaces are part of Hilbert triples? Then what is the deal with the operator $T':Y^* \to X^*$?
[Math] Relation between adjoint operators/dual operators
functional-analysishilbert-spaces
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These are not the usual definitions as I know them.$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$
First, I am only familiar with the situation that $H$ is a Hilbert space and $D(T)$ is dense in $H$ (which entails no loss of generality, as we can replace $H$ with the completion of $D(T)$.)
I would say:
$T$ is symmetric if $\inner{Tx}{y} = \inner{x}{Ty}$ for all $x,y \in D(T)$. (Note your definition doesn't make sense, because you are applying $T$ to vectors that may not be in $D(T)$.)
$T$ is Hermitian if it is symmetric and bounded. (If $T$ is bounded then it has a unique bounded extension to all of $H$, so we may as well assume $D(T) = H$ in this case.) Since a symmetric operator is always closable, the closed graph theorem implies that a symmetric operator with $D(T) = H$ is automatically bounded.
$T$ is self-adjoint if the following, more complicated condition holds. Let $D(T^*)$ be the set of all $y \in H$ such that $|\inner{Tx}{y}| \le C_y ||x||$ for all $x \in D(T)$, where $C_y$ is some constant depending on $y$. If $T$ is symmetric, one can show that $D(T) \subset D(T^*)$; $T$ is said to be self-adjoint if it is symmetric and $D(T) = D(T^*)$.
With these definitions, we have Hermitian implies self-adjoint implies symmetric, but all converse implications are false.
The definition of self-adjoint is rather subtle and this may not be the place for an extended discussion. However, I'd recommend a textbook such as Reed and Simon Vol. I. Perhaps I'll just say that symmetric operators, although the definition is simple, turn out not to be good for much, per se. One needs at least self-adjointness to prove useful theorems.
Given a bounded linear operator $T$ on a tensor product $H_1 \otimes H_2$, the operator $T$ is not necessarily of the form $T_1 \otimes T_2$, where $T_i:H_i \to H_i$. Even in finite dimensions, $T$ is not necessarily of this form.
Here is a simple counterexample. Let $H_1 = H_2 = \mathbb C^n$. Let $T$ be the "swap" map, the linear extension of $$ T(x \otimes y)= y \otimes x. $$ After a little bit of thought, it should become intuitively clear that $T \neq T_1 \otimes T_2$ for any maps $T_1, T_2$; the maps $T_i$ depend only on the space $H_i$, but the map $T$ needs to "work with" both inputs $x$ and $y$ simultaneously.
You can easily obtain a proof by fixing orthonormal bases $(e_i)$ for the Hilbert spaces $H_i$. If $T_1$ corresponds to the matrix $(t_{ij}^1)$ and $T_2$ to the matrix $(t_{kl}^2)$, then $T_1 \otimes T_2$ corresponds to the matrix $(t_{ij}^1t_{kl}^2)$. We will then have, $$(T_1 \otimes T_2)(e_s \otimes e_t) =\sum_{is}t_{ij}^1t_{kt}^2 \ e_i \otimes e_k. $$ If this is to equal $e_t \otimes e_s$, we must have $t_{is}^1 = \delta_{it}$ and $t_{kt}^2=\delta_{ik}$. But these constraints need to hold for every choice of $s$ and $t$ simultaneously and are thus contradictory. (For example, the conditions imposed on $t_{ij}^1$ by insisting that $e_1 \otimes e_2 \mapsto e_2 \otimes e_1$ and also that $e_1 \otimes e_3 \mapsto e_3 \otimes e_1$ are contradictory.)
Another way of thinking about this is as follows. Let $L(H)$ denote the bounded linear operators on a Hilbert space $H$. We have an inclusion $L(H_1) \otimes L(H_2)$ into $L(H_1 \otimes H_2)$ (as described in the first paragraph of your question). If the spaces $H_i$ are finite-dimensional, we even have $$ L(H_1) \otimes L(H_2) = L(H_1 \otimes H_2). $$ But not every element in $L(H_1) \otimes L(H_2)$ is an elementary tensor, i.e., of the form $T_1 \otimes T_2$. Even when $H_i$ is finite-dimensional, the most we can ask for in general is that $T$ is of the form $$ T=\sum_i T_{1,i} \otimes T_{2,i}. $$
Best Answer
If you have a linear operator $T : X \to Y$ between the Hilbert spaces $X$ and $Y$, you have to distinguish between to notions of the dual operator.
The Hilbert space adjoint: Define $T^* : Y \to X$ by $( T^*(y), x)_X = (y, T x)_Y$. Note that $x \mapsto (y, Tx)_Y$ is a linear functional on $X$ and, hence, can be identified with an element in $X$.
The "usual" adjoint: Define $T^* : Y' \to X'$ by $\langle T^*(y^*), x \rangle_{X',X} = \langle y^*, T x\rangle_{Y',Y}$. Note that $x \mapsto \langle y^*,Tx \rangle_{Y',Y}$ is a bounded linear functional on $X$, hence an element of $X'$.
Here, $(\cdot,\cdot)_X$ refers to the scalar product in $X$, whereas $\langle \cdot, \cdot \rangle_{X',X}$ refers to the duality product between $X'$ and $X$.
Of course, both adjoints are linked via the Riesz isomorphisms of $X$ and $Y$.