Adjoint of an adjoint of an operator not the operator itself

Question

In sadri hassani book, the author claims that $(A^\dagger)^\dagger = A$ is only valid in finite dimensional spaces. To me this seems also true in infinite dimensions as can be proven from the definition that $\langle x|A|y\rangle^* = \langle y|A^\dagger|x\rangle$ for all $x$ and $y.$ This can be seen by taking the complex conjugate of both sides. The only way this can be false is if the complex conjugate of a complex number is not the number itself in infinite dimensions. What am I missing here?

Answer 1

The following extends a little bit of your concern:

For normed spaces $X,Y$ and a bounded operator $A:X\rightarrow Y$, we defined $A^{\ast}:Y^{\ast}\rightarrow X^{\ast}$ by $\left<x,Ay^{\ast}\right>=\left<Ax,y^{\ast}\right>$ and we can further define $A^{\ast\ast}:X^{\ast\ast}\rightarrow Y^{\ast\ast}$ by $\left<y^{\ast},A^{\ast\ast}x^{\ast\ast}\right>=\left<A^{\ast}y^{\ast},x^{\ast\ast}\right>$.

Technically it will never be the case that $A=A^{\ast\ast}$ because literally $X\ne X^{\ast\ast},Y\ne Y^{\ast\ast}$. But when $Q:X\rightarrow X^{\ast\ast},Y\rightarrow Y^{\ast\ast}$ is surjective, where $\left<x^{\ast},Qx\right>=\left<x,x^{\ast}\right>$, then $X$ and $X^{\ast\ast}$, $Y$ and $Y^{\ast\ast}$ could be viewed as the same, and this is exactly when $X,Y$ are reflexive.

Hilbert space is reflexive because of the Riesz Representation Theorem.