Dual operators between Hilbert spaces : With or without riesz representation - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T07:51:50Z http://mathoverflow.net/feeds/question/71040 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/71040/dual-operators-between-hilbert-spaces-with-or-without-riesz-representation Dual operators between Hilbert spaces : With or without riesz representation Martin 2011-07-23T02:54:48Z 2011-08-26T20:18:41Z <p>Let $X$ and $Y$ be Hilbert spaces over the real numbers (so complex conjugation plays no role, and everything will be linear in the strict sense). Let $f : X \rightarrow Y$ be a linear continuous mapping.</p> <p>By the Riesz representation theorem, Hilbert spaces are isometric isomorphic to their own dual spaces. This leads to different notions of duality, which confuses me.</p> <p>(i) The dual operator of $f^\ast$ is the operator $f^\ast : Y^\ast \rightarrow X^\ast$ defined by $y^\ast \mapsto ( x \mapsto y^\ast( f x ) )$</p> <p>(ii) The dual operator is the adjoint, i.e. the unique operator $f^\ast : Y \rightarrow X$ such that $\forall x \in X, y \in Y : \langle fx,y \rangle_Y = \langle x,f^\ast y \rangle_X$</p> <p>The transition between these two different notions is full and faithfully functiorial, to say it like that. - Nevertheless, I would like to differ between these two notions of duality; just think in analysis of the Hilbert space $H^1_0$ and its dual $H^{-1}$. But I even don't think speaking about "different dualities" is not a crime.</p> <p>So, I would to know whether there are different words for this, whether these are really different concepts (except whether I use the isometry or not) and in which contexts, generally, to use these two appropiately. Although I think at least in the less algebraic parts of mathematics it would be helpful not to implicitly use the Riesz representation, this seems to be always swept under the rug.</p> http://mathoverflow.net/questions/71040/dual-operators-between-hilbert-spaces-with-or-without-riesz-representation/73800#73800 Answer by kostja for Dual operators between Hilbert spaces : With or without riesz representation kostja 2011-08-26T20:18:41Z 2011-08-26T20:18:41Z <p>Hello!</p> <p>Like Mr Yuan suggested, call the first one 'dual' and write $f^*$ and the second one adjoint and write $f^\dagger$. Then a fairy simple calculation shows, that $f^*$ and $f^\dagger$ are closely related to each other:</p> <p>Let $i: X \to X^*$ and $j: Y \to Y^*$ be the operators coming from the Riesz's represantation theorem. Then for any $y' \in Y^*$ and $x \in X$ there holds:</p> <p>$\langle j^{-1}\cdot y', f \cdot x\rangle = \langle f^\dagger \cdot j^{-1} \cdot y, x\rangle$.</p> <p>On the right hand side we have: $\langle f^\dagger \cdot j^{-1} \cdot y', x\rangle = i \cdot f^\dagger \cdot j^{-1} \cdot y' \cdot x$,</p> <p>while on the right hand side there is: $\langle j^{-1} \cdot y', f \cdot x \rangle = y' \cdot f\cdot x = f^* \cdot y' \cdot x$</p> <p>That for we get: $f^* \cdot y' \cdot x = i \cdot f^\dagger \cdot j^{-1} \cdot y' \cdot x$. Since this holds for all $x \in X$, there must be $f^* \cdot y' = i \cdot f^\dagger \cdot j^{-1} y'$ for all $y' \in Y^*$ and we can conclude, that</p> <p>$f^* = i\cdot f^\dagger \cdot j^{-1}$.</p> <p>If you don't destinguish between $X$ and $X^*$ and $Y$ and $Y^*$ respectively, then $f^* = f^\dagger$.</p> <p>Kind regards Konstantin</p>