Hilbert space adjoint

Remarks

For fixed $y$, the map $x \mapsto \ip{Tx}{y}$ is a bounded linear functional on $H$, so the Riesz representation theorem gives a unique vector $T^* y$ representing it. The map $T^*$ is linear and bounded, with operator norm $\norm{T^*} = \norm{T}$ and $\norm{T^* T} = \norm{T}^2$. It satisfies $(S + T)^* = S^* + T^*$, $(aT)^* = \bar a T^*$, $(ST)^* = T^* S^*$ and $T^{**} = T$. For finite-dimensional spaces it is the adjoint of a linear map.

Example: let $S(x_1, x_2, \ldots) = (0, x_1, x_2, \ldots)$ be the right shift on $\ell^2$, and $L(y_1, y_2, \ldots) = (y_2, y_3, \ldots)$ the left shift. Then $\ip{Sx}{y} = \sum_{n \ge 2} x_{n-1} \overline{y_n} = \sum_{m \ge 1} x_m \overline{y_{m+1}} = \ip{x}{Ly}$, so $S^* = L$.

Non-example: $S$ itself does not satisfy the defining identity in place of $S^*$. With unit vectors $e_n$, $\ip{S e_1}{e_2} = \ip{e_2}{e_2} = 1$, while $\ip{e_1}{S e_2} = \ip{e_1}{e_3} = 0$. So $S^* \ne S$, and $S$ is not self-adjoint.

Over $\R$, the same definition applies, and then $(aT)^* = a T^*$.