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 $T':Y^* \to X^*$?


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.

  1. 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 $X$.

  2. 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 $Y$.

  • $\begingroup$ For 2. you have $x \mapsto \langle y,Tx \rangle_{X',X}$ but both arguments here are in $Y$ so this would read $x \mapsto \langle y,Tx \rangle_{Y,Y}$ which doesn't make sense. $\endgroup$ – eurocoder Jan 23 '17 at 15:07
  • $\begingroup$ @eurocoder: Thank you for spotting this typo. $\endgroup$ – gerw Jan 23 '17 at 21:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.