# 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 the limit of a convergent sequence in $\map J X$.

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 $L$.

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:

\(\ds \norm {j_n - j_m}_{X^{\ast \ast} }\) | \(=\) | \(\ds \norm {J x_n - J x_m}_{X^{\ast \ast} }\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds \norm {\map J {x_n - x_m} }_{X^{\ast \ast} }\) | Evaluation Linear Transformation on Normed Vector Space is Linear Transformation from Space to Second Normed Dual | |||||||||||

\(\ds \) | \(=\) | \(\ds \norm {x_n - x_m}_X\) | Evaluation Linear Transformation on Normed Vector Space is Linear Isometry |

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:

\(\ds L\) | \(=\) | \(\ds \lim_{n \mathop \to \infty} J x_n\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds \map J {\lim_{n \mathop \to \infty} x_n}\) | Continuous Mappings preserve Convergent Sequences | |||||||||||

\(\ds \) | \(=\) | \(\ds J x\) |

Since:

- $J x \in \map J X$

we have:

- $L \in \map J X$

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

- $\map J X$ contains each limit of a convergent sequence in $\map J X$

So, from Subset of Metric Space contains Limits of Sequences iff Closed:

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

$\blacksquare$

## Sources

- 2020: James C. Robinson:
*Introduction to Functional Analysis*... (previous) ... (next) $26.1$: The Second Dual