Criterion for the Adjoint of a Bounded Linear Operator to be Injective

Recall from The Adjoint of a Bounded Linear Operator Between Banach Spaces page that if $X$ and $Y$ are Banach spaces and if $T : X \to Y$ is a bounded linear operator then the adjoint of $T$ is the linear operator $T^* : Y^* \to X^*$ defined for all $f \in Y^*$ by:

(1)

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

We noted that $T^*$ is also a bounded linear operator from $Y^*$ to $X^*$ and that $\| T^* \| = \| T \|$.

The following proposition tells us exactly then $T^*$ is injective.

Proposition 1: Let $X$ and $Y$ be Banach spaces and let $T : X \to Y$ be a bounded linear operator. Then the adjoint $T^* : Y^* \to X^*$ is injective if and only if $T(X)$ is a dense subspace of $Y$.

• Proof: $\Rightarrow$ Let $T^* : Y^* \to X^*$ be injective. Note that $T(X)$ is always a subspace of $Y$. So $\overline{T(X)}$ is a closed subspace of $Y$.

• Suppose that $T(X)$ is not dense in $Y$. Then $\overline{T(X)} \neq Y$. Let $y_0 \in Y \setminus \overline{T(X)}$. Then $\delta = d(y_0, \overline{T(X)}) > 0$. By one of the corollaries on the Corollaries to the Hahn-Banach Theorem page there exists an $F \in Y^*$ such that $F(y_0) = \delta$ and $F(y) = 0$ for all $y \in \overline{T(X)}$.

• So for every $x \in X$ we have that $T(x) \in T(X)$ and so $F(T(x)) = 0$. But observe that $0 \in Y^*$ is such that for all $x \in X$, $0(T(x)) = 0$. Therefore $T^*(F) = T^*(0)$. Since $T^*$ is injective, $F = 0$. But this is not true since $F(y_0) = \delta \neq 0$. So the assumption that $T(X)$ is not dense is false. Hence $T(X)$ is a dense subspace of $Y$.

• $\Leftarrow$ Let $T(X)$ be a dense subspace of $Y$. Let $f \in \ker (T^*)$. Then $f \in Y^*$ and $T^*(f) = 0$, i.e., $f \circ T = 0$. So for every $x \in X$ we have that $f(T(x)) = 0$. So $T(x) \in \ker f$ for every $x \in X$.

• Now since $T(X)$ is dense in $Y$ we have that for each $y \in Y$ there exists a sequence $(x_n)$ in $X$ for which $(T(x_n))$ converges to $y$. Since $f$ is continuous, $(f(T(x_n))$ converges to $f(y)$. But for $(f(T(x_n)) = (0)$. So $f(y) = 0$. Since this is true for every $y \in Y$ we have that $f = 0$. Therefore:

(2)

\begin{align} \quad \ker (T^*) = \{ 0 \} \end{align}

• Hence $T^*$ is injective. $\blacksquare$