Image of Evaluation Linear Transformation on Banach Space is Closed

Theorem
Let $\struct {X, \norm \cdot_X}$ be a Banach space.

Let $\struct {X^{\ast \ast}, \norm \cdot_{X^{\ast \ast} } }$ be the second normed dual of $\struct {X, \norm \cdot_X}$.

Let $J : X \to X^{\ast \ast}$ be the evaluation linear transformation.

Then:


 * $\map J X$ is closed in $X^{\ast \ast}$.

Theorem
Let $L$ be a limit point of $\map J X$.

Let $\sequence {j_n}_{n \mathop \in \N}$ be a sequence in $\map J X$ such that:


 * $\sequence {j_n}_{n \mathop \in \N}$ converges to $\map J X$.

Note that for each $n \in \N$ there exists $x_n \in X$ such that:


 * $j_n = J x_n$

From Evaluation Linear Transformation on Normed Vector Space is Linear Isometry, we have:


 * $J$ is a linear isometry.

Specifically:


 * $J$ is a bounded linear transformation.

From Continuity of Linear Transformations, we have:


 * $J$ is continuous.

We show that $\sequence {x_n}_{n \mathop \in \N}$ converges to some $x \in X$.

We will then have:


 * $L = J x \in \map J X$

from continuity.

Note that:

Let $\epsilon > 0$.

Since:


 * $\sequence {j_n}_{n \mathop \in \N}$ is convergent

we have, from Convergent Sequence in Normed Vector Space is Cauchy Sequence:


 * there exists $N \in \N$ such that for all $n, m \ge N$, we have $\norm {j_n - j_m}_X < \epsilon$.

So, for $n, m \ge N$, we have:


 * $\norm {x_n - x_m}_X < \epsilon$

Since $\epsilon$ was arbitrary, $\sequence {x_n}_{n \mathop \in \N}$ is Cauchy.

Since $X$ is a Banach space, we have:


 * $\sequence {x_n}_{n \mathop \in \N}$ converges to a limit $x \in X$.

Since $J$ is continuous, we obtain:

Since:


 * $J x \in \map J X$

we have:


 * $L \in \map J X$

Since $L$ was an arbitrary limit point of $\map J X$, we have:


 * $\map J X$ contains all its limit point.

So:


 * $\map J X$ is closed in $X^{\ast \ast}$.