# Mathematics Prelims

## July 20, 2008

### The Continuous Dual Space

Filed under: Analysis,Functional Analysis — cjohnson @ 8:35 pm

An important subspace of the algebraic dual space of a normed space $X$ is the space which consists only of bounded linear functionals on $X$.  This space is called the continuous space (or just dual space, and the conjugate space in some older texts, such as Kolmogorov), and is denoted $X'$.  Note that we can define the algebraic dual space for any vector space, but require a norm for the (continuous) dual.  This space is itself a normed space where functionals are given the operator norm.  In fact, regardless of the normed space $X$, the dual $X'$ is a Banach space.

Let $(f_n)_{n \in \mathbb{N}}$ be a Cauchy sequence in $X'$.  Then for each $\epsilon > 0$ there exists a $N \in \mathbb{N}$ such that $\| f_n - f_m \| < \epsilon$ for all $m, n > N$.  This means that

$\displaystyle \| f_n - f_m \| = \sup_{x \in X, \, \|x\| = 1} |f_n(x) - f_m(x)| < \epsilon$

So for a given $x \in X$ with $\|x \| = 1$, $(f_n(x))$ forms a Cauchy sequence in $\mathbb{R}$, and so converges.  We will then define a functional $f$ pointwise as follows.

$\displaystyle f(x) = \lim_{n \to \infty} f_n(x)$.

We must show that this functional is linear and bounded.  Linearity follows easily from the linearity of each $f_n$ and properties of limits.

$\displaystyle f(\alpha x + \beta y) = \lim_{n \to \infty} f_n(\alpha x + \beta y) = \alpha \lim_{n \to \infty} f_n(x) + \beta \lim_{n \to \infty} f_n(y)$.

For boundedness,

$\displaystyle |f(x)| = \lim_{n \to \infty} |f_n(x)| \leq \lim_{n \to \infty} \| f_n \| \|x\| \leq M \|x\|$

for some $M$ (this is because every Cauchy sequence is bounded).

So $f$ is a bounded linear operator on $X$, and we just need to show $f_n \to f$.

Let $\epsilon > 0$ be given.  As $(f_n)$ is Cauchy, there is an $N \in \mathbb{N}$ such that for all $m, n > N$ we have $\|f_m - f_n\| < \epsilon$.  Letting $m \to \infty$, we have $\|f - f_n\| \leq \epsilon$ for all $n > N$, so $f_n \to f$.

And so the dual of any normed space is Banach, regardless of whether the original space was or not.

Note that for every finite dimesional normed space, the continuous and algebraic duals are in fact the same as all linear operators are bounded in finite dimensions.

(This proof is a modified version of a proof in Eidelman.)

1. You write “Letting $m \to \infty$, we have $\|f - f_n\| \leq \epsilon$“. I don’t think so because we have to prove that $f_n \to f$.