Proof Riesz representation theorem I have a question regarding the proof of the Riesz representation theorem. Why do we declare the isomorphism $\Phi: H \rightarrow H'$ in an antilinear way? I mean if, this isomorphism would pick the other coordinate of the inner product, then this isomorphism should be perfectly linear? I cannot imagine why Mathematicians would deliberately be messy, so I guess there is a meaning behind this. 
 A: If you want a linear isomorphism between $H$ and $H'$, you can have it. Just pick an orthonormal basis $(e_n)$ in $H$, take its dual basis $(f_n)$ in $H'$, and define 
$$\Phi(\sum c_n e_n)=\sum c_n f_n\tag{1}$$
There's a catch, though: $\Phi$ is not canonical, it depends on the choice of basis $(e_n)$.  Such isomorphisms are of little use; thinking of them can even be counterproductive.  
Indeed, replace $e_n$ with $\hat e_n=\alpha e_n$, where $\alpha$ is a unimodular complex number. The corresponding coefficient $\hat c_n$ is $\alpha^{-1}c_n$, and the corresponding $\hat f_n$ is $\bar \alpha f_n$. So, $$\Phi(\sum \hat c_n \tilde e_n)=\sum \hat c_n \hat f_n = \alpha^{-1} \bar \alpha \sum c_n f_n$$
which is different from (1). 
The antilinear isomorphism $$\Phi(\sum c_n e_n)=\sum \bar c_n f_n\tag{2}$$
does not have this problem. Indeed,   it satisfies $$\langle x,\Phi(x)\rangle =\|x\|^2\tag{3}$$ and there is only one linear functional with norm $\|x\|$ that is equal to $\|x\|^2$ on $x$. The property (3) is what we really need from an isomorphism between $H$ and its dual: the isomorphism should respect the  duality relation.
