The Adjoint of a Bounded Linear Operator Between Banach Spaces

# The Adjoint of a Bounded Linear Operator Between Banach Spaces

Definition: Let $X$ and $Y$ be Banach spaces and let $T : X \to Y$ be a bounded linear operator. The Adjoint of $T$ is the linear operator $T^* : Y^* \to X^*$ defined for all $f \in Y^*$ by $T^*(f) = f \circ T$. |

Note that for every $f \in Y^*$, the map $f \circ T : X \to \mathbb{R}$ is a bounded linear functional on $X$ since for every $x \in X$ we have that:

(1)\begin{align} \quad |(f \circ T)(x)| = |f(T(x)| \leq \| f \|\| T(x) \| \leq \| f \| \| T \| \| x \| \end{align}

So indeed, $T^*$ is a map with domain $Y^*$ and codomain $X^*$.

It is also easy to verify that $T^*$ is a linear operator. The following proposition tells us that if $T$ is a bounded linear operator from $X$ to $Y$ then $T^*$ is a bounded linear operator from $Y^*$ to $X^*$ and that moreover, $\| T^* \| = \| T \|$.

Proposition 1: Let $X$ and $Y$ be Banach spaces and let $T : X \to Y$ be a bounded linear operator. Then $T^* : Y^* \to X^*$ is a bounded linear operator and $\| T^* \| = \| T \|$. |

**Proof:**For each $f \in Y^*$ we have that:

\begin{align} \quad \| T^*(f) \| = \| f \circ T \| \overset{(*)} \leq \| f \| \| T \| = \| T \| \| f \| \end{align}

- Where the inequality at $(*)$ comes from the fact that $f$ is a bounded linear functional on $X$ and $T$ is a bounded linear operator on $X$. Thus $\| T^* \| \leq \| T \|$.

- For the reverse inequality, let $x_0 \in X$ with $\| x_0 \| = 1$. By one of the corollaries on the Corollaries to the Hahn-Banach Theorem page we have that for the point $T(x_0) \in Y$ there exists an $f_0 \in Y^*$ with $\| f_0 \| = 1$ such that $f_0(T(x_0)) = \| T(x_0) \|$. Therefore:

\begin{align} \quad \| T^* \| = \sup_{f \in Y^*,\| f \| = 1} \| T^*(f) \| \geq \| T*(f_0) \| = \| f_0 \circ T \| = \sup_{x \in X, \| x \| = 1} |(f_0 \circ T)(x)| \geq |(f_0 \circ T)(x_0)| = \| T(x_0) \| \end{align}

- So for every $x_0 \in X$ with $\| x_0 \| = 1$ we have that $\| T(x_0) \| \leq \| T^* \|$. So given any $x \in X$ with $x \neq 0$, consider the point $\frac{x}{\| x \|}$. It has norm $1$ and so:

\begin{align} \quad \left \| T \left ( \frac{x}{\| x \|} \right ) \right \| \leq \| T^* \| \end{align}

- Hence for all $x \in X$ with $x \neq 0$:

\begin{align} \quad \| T(x) \| \leq \| T^* \| \| x \| \end{align}

- And of course the above inequality also holds if $x = 0$. Thus $\| T \| \leq \| T^* \|$. Hence $\| T \| = \| T^* \|$. $\blacksquare$