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.

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.

(i) The dual operator of $f^\ast$ is the operator $f^\ast : Y^\ast \rightarrow X^\ast$ defined by $y^\ast \mapsto ( x \mapsto y^\ast( f x ) )$

(ii) The dual operator is the adjoint, i.e. the unique operator $f^\ast : Y \rightarrow X$ such that $\forall x \in X, y \in Y : \langle fx,y \rangle_Y = \langle x,f^\ast y \rangle_X$

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.

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.

  • 4
    $\begingroup$ The first is defined completely in terms of the linear structure while the second requires the inner product. Call the first one the dual and the second the adjoint. Use $\ast$ for the first one and $\dagger$ for the second one. $\endgroup$ Commented Jul 23, 2011 at 3:09
  • $\begingroup$ I use the notion "dual" for the first one and "Hilbert space dual" for the second one... $\endgroup$
    – Dirk
    Commented Jul 23, 2011 at 6:41

2 Answers 2


Like Mr Yuan suggested, call the first one 'dual' and write $f^\ast$ and the second one adjoint and write $f^\dagger$. Then a fairy simple calculation shows, that $f^\ast$ and $f^\dagger$ are closely related to each other:

Let $i: X \to X^\ast$ and $j: Y \to Y^\ast$ be the operators coming from the Riesz's represantation theorem. Then for any $y' \in Y^\ast$ and $x \in X$ there holds:

$\langle j^{-1}\cdot y', f \cdot x\rangle = \langle f^\dagger \cdot j^{-1} \cdot y, x\rangle$.

On the right hand side we have: $\langle f^\dagger \cdot j^{-1} \cdot y', x\rangle = i \cdot f^\dagger \cdot j^{-1} \cdot y' \cdot x$,

while on the right hand side there is: $\langle j^{-1} \cdot y', f \cdot x \rangle = y' \cdot f\cdot x = f^\ast \cdot y' \cdot x$

That for we get: $f^\ast \cdot y' \cdot x = i \cdot f^\dagger \cdot j^{-1} \cdot y' \cdot x$. Since this holds for all $x \in X$, there must be $f^\ast \cdot y' = i \cdot f^\dagger \cdot j^{-1} y'$ for all $y' \in Y^\ast$ and we can conclude, that

$f^\ast = i\cdot f^\dagger \cdot j^{-1}$.

If you don't destinguish between $X$ and $X^\ast$ and $Y$ and $Y^\ast$ respectively, then $f^\ast = f^\dagger$.

Kind regards Konstantin


The Riesz-Fischer theorem about all continuous functionals on a Hilbert space $V$ being of the form $v\to \langle v,w\rangle_V$ is not quite as "functorial" as one might have anticipated. The understandable misconception is not about complex conjugation or not, so suppose that all vector spaces are real. That is, for $T:V\to W$, in order for the square with edges $T:V\to W$, $V\to V^*$ by $v\to \langle -,v\rangle_V$, $W\to W^*$ by $w\to \langle-,w\rangle_W$, $T^*:W^*\to V^*$ to commute, it is necessary and sufficient that (tracing through the diagram) $$ \langle v,v'\rangle\;=\; \langle Tv,Tv'\rangle $$ for all $v,v'\in V$. Otherwise, the "functor" of Riesz-Fischer fails. The real point is that it is a "functor" only on a particular category of Hilbert spaces and maps, namely, isometries-to-image.


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