A Norm. Lin. Sp. is Fin.-Dim. IFF Alg. Dual equals the Topo. Dual

# A Normed Linear Space is Finite-Dimensional IFF the Algebraic Dual and Topological Dual are the Same

Recall from The Topological Dual of a Normed Linear Space page that if $X$ is a normed linear space then the topological dual denoted $X^*$ is the normed linear space of all continuous linear functionals on $X$.

We will now prove a very important result which states that $X$ is a finite-dimensional normed linear space if and only if $X^{\#} = X^*$.

Theorem 1: Let $X$ be a normed linear space. Then $X$ is finite-dimensional if and only if $X^{\#} = X^*$. |

**Proof:**$\Rightarrow$ Suppose that $X$ is finite-dimensional. From the theorem on the Every Linear Operator on a Finite-Dimensional Normed Linear Space is Bounded page we must have that every linear operator $\varphi : X \to \mathbb{C}$ is continuous. That is, every linear operator on $X$ is continuous. So:

\begin{align} \quad X^{\#} = X^* \end{align}

- $\Leftarrow$ We prove the converse of the theorem by proving the contrapositive of the converse by assuming that $X$ is infinite-dimensional and then exhibiting a linear functional on $X$ that is not continuous.

- Suppose that $X$ is infinite-dimensional. Let $\mathcal B \subset X$ be a basis of $X$. We may assume that $\mathcal B$ consists entirely of unit vectors (for if $b \in \mathcal B$ is such that $\| b \| \neq 1$ then replace $b$ with $\displaystyle{\frac{b}{\| b \|}}$).

- Since $X$ is infinite-dimensional, there exists a subset $\{ b_n \}_{n=1}^{\infty}$ of $\mathcal B$ that is infinite.

- Now since $\mathcal B$ is a basis of $X$, for every $x \in X$ we can uniquely write:

\begin{align} \quad x = \sum_{b \in \mathcal B} a_b b \end{align}

- where each $a_b \in \mathbb{C}$. Observe that $a_b = 0$ for all but finitely many $b \in \mathcal B$. Now define a function $\varphi : X \to \mathbb{C}$ for each $x \in X$ by:

\begin{align} \quad \varphi \left ( \sum_{b \in \mathcal B} a_b b \right ) = \sum_{n=1}^{\infty} n a_{b_n} \end{align}

- From the remark made above, $\varphi$ is well-defined. Furthermore, $\varphi$ is a linear functional, so $\varphi \in X^{\#}$. For each $n \in \mathbb{N}$ we have that:

\begin{align} \quad \varphi (n x_{b_n}) = n \end{align}

- Therefore $\| \varphi \| \geq n$ for every $n \in \mathbb{N}$ and so $\varphi$ is not continuous. Hence:

\begin{align} \quad X^{\#} \neq X^* \end{align}