TITLE: Adjoint of a linear transformation in an infinite dimension inner product space
QUESTION [14 upvotes]: We learned that if $V$ is a finite inner product space then for every linear transformation $T:V\to V$, there exists a unique linear transformation $T^*:V\to V$ such that $\forall u, v \in V: (Tv, u)=(v, T^*u)$.
The construction of $T^*$ used the fact that $V$ is finite and therefore has an orthonormal basis, which is not the case had it been infinite.
Are there infinite dimension inner product spaces such that not all linear transformations have an adjoint? Or is it somehow possible to extend this definition to infinite spaces as well?

REPLY [4 votes]: Let's look at an example. Take the vector space $\mathbb{R}[x]$ of all real polynomials in one variable and define an inner product as 
$$\left( \sum_{j=0}^n a_jx^j, \sum_{k=0}^m b_kx^k \right)=\sum_{h=0}^{\min(n,m)}a_hb_h.$$
Now let $T$ be the linear operator such that 
$$T\sum_{j=0}^n a_jx^j=\sum_{j=0}^n a_j+\left(\sum_{j=1}^na_j\right)x+\ldots + (a_{n-1}+a_n)x^{n-1}+a_nx^n.$$ 
Think $T$ like the operator represented by the infinite matrix below: 
$$\begin{bmatrix} 
1 & 1 & 1 & \ldots \\
0 & 1 & 1 & \ldots \\
0 & 0 & 1 & \ldots \\
\vdots & \vdots & \vdots & \ddots \\
\end{bmatrix}$$
Should $T$ have an adjoint with respect to the inner product $(,)$, it should be somehow associated to this infinite matrix: 
$$\begin{bmatrix} 
1 & 0 & 0 & \ldots \\
1 & 1 & 0 & \ldots \\
1 & 1 & 1 & \ldots \\
\vdots & \vdots & \vdots & \ddots \\
\end{bmatrix}$$
but this makes no sense in $\mathbb{R}[x]$. Formally, let's suppose such an adjoint operator $T^\star$ exists. Fix $k \in \mathbb{N}$: who is $T^\star x^k$? For all $n=0, 1,\ldots$, we should have 
$$(x^n, T^\star x^k)=(T x^n, x^k)=(1+\ldots+ x^n, x^k)=\begin{cases}1 & n \ge k \\ 0 & n <k \end{cases};$$
this means that $T^\star x^k$ should be a polynomial with degree $\ge n$ for all $n\in \mathbb{N}$. So, $T$ hasn't got an adjoint. 

The reason for it is that the mapping 
$$P \in \mathbb{R}[x] \mapsto (P, \cdot) \in \mathbb{R}[x]^\star$$
(here $\mathbb{R}[x]^\star$ is the algebraic dual space of $\mathbb{R}[x]$) is not an isomorphism, because it is not surjective. In fact it can be shown that $\mathbb{R}[x]^\star$ can be represented as $\mathbb{R}[[x]]$, the space of formal power series with real coefficients.