# 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:

$j_n = J x_n$
$J$ 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
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 $\map J X$
$\map J X$ is closed in $X^{\ast \ast}$.

$\blacksquare$