Unbounded Linear Functionals

# Unbounded Linear Functionals

Recall that if $(X, \| \cdot \|_X)$ is a normed linear space then a linear functional $f$ on $X$ is said to be bounded if there exists an $M \in \mathbb{R}$, $M \geq 0$ such that for every $x \in X$:

(1)
\begin{align} \quad |f(x)| \leq M \| x \|_X \end{align}

And we define the least such $M$ to get $\| f \| = \inf \{ M : |f(x)| \leq M \| x \|_X, \: \forall x \in X \}$.

 Definition: Let $(X, \| \cdot \|_X)$ be a normed linear space and let $f$ be a linear functional on $X$. Then $f$ is Unbounded if it is not bounded.

If $f$ is an unbounded linear functional on $X$ then for every positive real number $M \in \mathbb{R}$, $M > 0$ there exists an $x_M \in X$ such that:

(2)
\begin{align} \quad |f(x_M)| > M \| x_M \|_X \end{align}

In particular, for each $n \in \mathbb{N}$ there exists an $x_n \in X$ such that:

(3)
\begin{align} \quad |f(x_n)| > n \| x_n \|_X \end{align}

We will use this property in the following proposition.

 Proposition 1: Let $(X, \| \cdot \|_X)$ be a normed linear space and let $f$ be a linear functional on $X$. If $f$ is unbounded then $\ker f$ is a dense subspace of $X$.
• Proof: We already know that $\ker f$ is always a subspace of $X$ so all that we need to show is that $\ker f$ is dense in $X$. We do this by showing that for each $x \in X$ there is a sequence $(z_n)$ in $\ker f$ that converges to $x$.
• Since $f$ is unbounded, for each $n \in \mathbb{N}$ there exists an $x_n \in X$ such that $|f(x_n)| > n \| x_n \|_X$. In particular, observe that if $x_n = 0$ then this implies that $0 = |f(x_n)| > n \| x_n \|_X = n \cdot 0 = 0$, a contradiction. So $x_n \neq 0$ for every $n \in \mathbb{N}$. For each $n \in \mathbb{N}$ let $\displaystyle{y_n = \frac{x_n}{\| x_n \|_X}}$. Then $\| y_n \|_X = 1$ for each $n \in \mathbb{N}$ and also:
(4)
\begin{align} \quad |f(y_n)| = \left | f \left ( \frac{x_n}{\| x_n \|_X}\right ) \right | > n \frac{\| x_n \|_X}{\| x_n \|_X} = n \end{align}
• That is, for each $n \in \mathbb{N}$ we have that:
(5)
• Now for $x \in X$ let $(z_n)$ be the sequence defined for each $n \in \mathbb{N}$ by:
• Note that each $z_n$ is well defined since $f(y_n) \neq 0$ by $(*)$. Furthermore we see that each $z_n \in \ker f$ since:
• Also, $(z_n)$ converges to $x$ since:
• Where the inequality at $(**)$ comes from the fact that $|f(y_n)| > n$ for each $n \in \mathbb{N}$. Thus, for each $x \in X$ there is a sequence $(z_n)$ in $\ker f$ that converges to $x$. So $\ker f$ is dense in $X$. $\blacksquare$