Let $H$ be a Hilbert space and let $A:H\to H$ be a linear operator. Show that $A^{**}=A$.

For a linear functional $f\in X^*$ ($X$ not necessarily being a Hilbert space), $f(x)$ is expressed through $\langle x,f\rangle$, an inner product of pair of elements of $X$ and $X^*$ respectively. A dual operator is defined through the relation $\langle Ax,f\rangle=\langle x,A^*f\rangle$ and like in $\langle x,f\rangle$, $\langle x,A*f\rangle $ stands for $[A^*f](x)$, which is $[A^*f](x)=f(A(x))$. So the dual operator here, $A^*:H^*\to H^*$ and $[A^*f]:H\to \Bbb{R}$ is a linear functional.

According to the definition of operator and dual operators through an inner product relation between a space and its dual, for $\phi \in H^{**}$, $\langle A^*f,\phi\rangle =\phi(A^*f)$ and $\langle A^*f,\phi\rangle =\langle f,A^{**}\phi\rangle =[A^{**}\phi](f)$, but I don't see how $A^{**}=A$.

Maybe it has to do with reflexivity of $H$? But even if $H$, and $H^{**}$ are isomorphic, what algebraic way can one use to show it?

Speaking of which, if $H=H^{**}$, then $J:H \to H^{**}$, $J(x)(f)=f(x)$, is a clear map, but $J^*:H^{**}\to H$, $J^*\phi(f)$ is unclear and doesn't seem to return an element of $H$. I could use some help here.

  • $\begingroup$ I may have mixed up dual of $A^*$ with adjoint of $A^*$, $\endgroup$
    – Meitar
    Dec 13, 2021 at 14:24

1 Answer 1


In general, the equality $$\tag1f(x)=\langle x,f\rangle$$ is just notation; there is no inner product. But it is the inner product when you are doing it in a Hilbert space. The reason is the Riesz Representation Theorem, which tells you that a Hilbert space is way more than reflexive, as you actually have $H=H^*$, via the identification of $h\in H$ with $\langle \cdot,h\rangle\in H^*$.

So when $A:H\to H$ is a linear operator, you can also think of $A^*$ as a linear operator $H\to H$ via the identification above. And the same with $A^{**}$.

So now you start with $\phi\in H^{**}$. By definition, $$\tag2 (A^{**}\phi)f=\phi(A^*f),\qquad f\in H^*. $$ As mentioned above, for $f\in H^*$ there exists a unique $k\in H$ such that $f=\langle\cdot,k\rangle$. And similarly there exists $z\in H$ such that $\phi(f)=\langle k,z\rangle$. Using $(2)$, $$\tag3 \langle k,A^{**}z\rangle=(A^{**}\phi)f=\phi(A^*f)=\langle A^*k,z\rangle. $$ Continuing, $$\tag4 \langle k,A^{**}z\rangle=\overline{\langle z,A^*k\rangle}=(A^*f)z=\overline{f(Az)} =\overline{\langle Az,k\rangle}=\langle k,Az\rangle. $$ As $k$ can be taken to be any element in $H$, we get $A^{**}z=Az$.

  • $\begingroup$ Thank you for the detailed answer! I am wondering how $<k,A^{**}z>$ is defined, is it because of the reflexivity? $\endgroup$
    – Meitar
    Dec 13, 2021 at 15:20
  • 1
    $\begingroup$ It's because we identity $H^*$ with $H$, and so $H^{**}$ with $H$ too. $\endgroup$ Dec 13, 2021 at 15:28
  • $\begingroup$ This is very accurate and definite, I appreciate it. Is it direct that $<\cdot ,z>(f)=<\cdot , z>(<,\cdot , k>)=<k,z>$? $\endgroup$
    – Meitar
    Dec 13, 2021 at 17:15
  • $\begingroup$ $\phi$ is defined by an element of $H^*$ and this element, $g$, is defined by $z\in H$, but I end up with $\phi=\langle \cdot ,g\rangle =\langle \cdot ,\langle \cdot ,z\rangle\rangle$ $\endgroup$
    – Meitar
    Dec 13, 2021 at 17:25

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