Discontinuous linear functional

$\newcommand{\Zobr}[3]{#1:#2\to#3}\newcommand{\R}{\mathbb R}$ A different approach to show existence of unbounded functionals is using the notion of Hamel basis.

Definition: Let $V$ be a vector space over a field $K$. We say that $B$ is a Hamel basis in $V$ if $B$ is linearly independent and every vector $v\in V$ can be obtained as a linear combination of vectors from $B$. (By linearly independent we mean that if a finite linear combinations of elements of $B$ is zero, then all coefficients must be zero.)

This is equivalent to the condition that every $x\in V$ can be written in precisely one way as $$\sum_{i\in F} c_i x_i$$ where $F$ si finite, $c_i\in K$ and $x_i\in B$ for each $i\in F$.

This is probably better known in the finite-dimensional case, but many properties of bases remain true in the infinite-dimensional case as well:

• Every vector space has a Hamel basis. In fact, every linearly independent set is contained in a Hamel basis.
• Any two Hamel bases of the same space have the same cardinality.
• Choosing images of basis vector uniquely determines a linear function, i.e., if $B$ is a basis of $V$ then for any vector space $W$ and any map $\Zobr gBW$ there exists exactly one linear map $\Zobr fVW$ such that $f|_B=g$.

Claim: If $X$ is an infinite-dimensional linear normed space, then there exist non-continuous linear function $\Zobr fX{\R}$.

See also Example 4.2 in Heil: A basis theory primer.

Proof. Choose an infinite linearly independent set $\{x_n; n\in\mathbb N\}$ such that $\|x_n\|=1$. (An infinite linearly independent set exists, since $X$ is infinite-dimensional. Normalizing the vectors does not influence the linear independence.) There is a Hamel basis $B$ containing this set.

Then there is a linear function $\Zobr fX{\R}$ such that $f(x_n)=n$ and $f(b)=0$ for $b\in B\setminus\{x_n; n\in\mathbb N\}$. This function is obviously unbounded. $\square$

In fact, Srivatsan's comment above is a special case of this result, since $\{e^i; i\in\mathbb N\}$ is a Hamel basis of the space $c_{00}$ of sequences that are eventually zero.

This answer is very similar to this one.

The above was taken from these notes of mine. Several more results and references can be found there. I have also mentioned some basic facts about Hamel basis in another answer at this site.

You can also find much more information about Hamel bases at other posts at this site: "Hamel basis", hamel basis site:math.stackexchange.com.

You won't find an explicit example of a discontinuous linear functional defined everywhere on a Banach space: these require the Axiom of Choice. However, you can find a discontinuous linear functional on a normed linear space. A typical scenario would be that you have Banach space $X$ (whose norm I'll denote $\|.\|_X$) which is a dense linear subspace of Banach space $Y$ (under a different norm $\|.\|_Y$, where $\|x\|_X \ge \|x\|_Y$ for all $x \in X$), and a linear functional $\phi$ on $X$ which is continuous for the norm $\|.\|_X$ but not for the norm $\|.\|_Y$. Thus if you take $X$ with the norm $\|.\|_Y$, you have a normed linear space with a discontinuous linear functional $\phi$. For example, take $X = \ell_2$, $Y = \ell_\infty$, and $\phi(x) = \sum_{i=1}^\infty x_i/i$.

As Robert Israel already mentioned, you cannot write down an explicit (free of the axiom of choice) unbounded linear functional on a Banach space. But it's generally not hard for incomplete normed spaces. Nobody has mentioned my favorite example: the functional $\ell: C^1[-1,1] \to \mathbb{R}$ given by $\ell(f) = f'(0)$.