# Discontinuous linear functional

I’m trying to find a discontinuous linear functional into $\mathbb{R}$ as a prep question for a test. I know that I need an infinite-dimensional Vector Space. Since $\ell_2$ is infinite-dimensional, there must exist a linear functional from $\ell_2$ into $\mathbb{R}$. However, I’m having trouble actually coming up with it.

I believe I’m supposed to find an unbounded function (although I’m not sure why an unbounded function is necessarily not continuous; some light in that regard would be appreciated too), so I thought of using the vectors $e^i$, which have all entries equal to zero, except for the $i$-th one. Then, you can define $f(e^i)=i$. That’d be unbounded, but I’m not sure if it’d be linear, and even if it is, I’m not sure how to define it for all the other vectors in $\ell_2$.

A friend mentioned that at some point the question of whether the set $E=\{e^i:i\in\mathbb{Z}^+\}$ is a basis would come up, but I’m not sure what a basis has to do with continuity of $f$.

I’m just learning this topic for the first time, so bear with me please.

The space of sequences that are eventually zero (suggested by a few people) turned out to be exactly what I needed. It also helped to cement the notions of Hamel basis, not continuous, etc.

$\newcommand{\Zobr}[3]{#1:#2\to#3}\newcommand{\R}{\mathbb R}$ 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

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 $\Zobr fX{\R}$.

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.