Dual operator

Remarks

The map $T'$ is well defined since $f \circ T$ is linear and continuous, and $T'$ is linear and bounded. By the Hahn–Banach theorem, its operator norm is $\norm{T'} = \norm{T}$. It is the restriction of the algebraic dual map to continuous functionals; other notations are $T^*$ and $T^t$.

Example: for $1 < p < \infty$ and $1/p + 1/q = 1$, identify the dual of $\ell^p$ with $\ell^q$ via $f(x) = \sum_n f_n x_n$. Let $S$ be the right shift on $\ell^p$. Then $f(Sx) = \sum_{n \ge 2} f_n x_{n-1} = \sum_{m \ge 1} f_{m+1} x_m$, so $S' f = (f_2, f_3, \ldots)$ is the left shift on $\ell^q$.

Non-example: the rule $f \mapsto f \circ T$ does not give a map $Y' \to X'$ when $T$ is not continuous. Let $X = Y$ be the space of polynomials on $[0, 1]$ with the supremum norm, $T p = p'$, and $f(p) = p(0)$. The polynomials $p_n(x) = x (1 - x)^n$ have $\norm{p_n} \le 1/(n+1)$, but $(f \circ T)(p_n) = p_n'(0) = 1$, so $f \circ T$ is not continuous.

The rule $(ST)' = T' S'$ holds, and $\ker T'$ is the annihilator of the range of $T$.