# In a Hilbert space $H$, a Linear operator $A:H\to H$ satisfies $A ^{**}=A$

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.

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

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