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

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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 $Y$.

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  • $\begingroup$ 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

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