Normed Dual Space of Infinite-Dimensional Normed Vector Space is Infinite-Dimensional

Theorem

Let $\GF \in \set {\R, \C}$.

Let $\struct {X, \norm {\, \cdot \,} }$ be an infinite dimensional normed vector space.

Let $X^\ast$ be the normed dual space of $X$.

Then $X^\ast$ is infinite dimensional.

Proof

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Aiming for a contradiction, suppose $X^\ast$ is finite dimensional.

Then from Normed Dual Space of Finite-Dimensional Vector Space is Isometrically Isomorphic to Original Space:

$X^{\ast \ast} \cong X^{\ast}$

From Isometric Isomorphism between Normed Vector Spaces preserves Dimension:

$\dim X^{\ast \ast} = \dim X^\ast$

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

Then $\iota$ is a linear isometry from Evaluation Linear Transformation on Normed Vector Space is Linear Isometry.

Hence $\iota : X \to \iota X$ is an isometric isomorphism.

From Isometric Isomorphism between Normed Vector Spaces preserves Dimension:

$\dim X = \dim \iota X$

From Image of Vector Subspace under Linear Transformation is Vector Subspace:

$\iota X$ is a linear subspace of $X^{\ast \ast}$.

From Dimension of Proper Subspace is Less Than its Superspace:

$\dim \iota X \le \dim X^{\ast \ast}$

So:

$\dim X \le \dim X^{\ast \ast} = \dim X^\ast$

But $\dim X$ is not finite.

Hence $\dim X^\ast$ is not finite.

This contradicts our assumption that $X^\ast$ is finite dimensional.

Hence by Proof by Contradiction $X^\ast$ is infinite dimensional.

$\blacksquare$