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
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.